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How to use your
Tatva Practice Book
1
8. Quantization of charge
When a physical quantity can
have only discrete values rather
than any value, the quantity is
said to be quantised. The
smallest charge that can exist in Scan the QR Code in each
nature is the charge of an chapter’s theory section to
electron.
view micro concept videos
related to the chapter, on
SCAN CODE the Vedantu app
Electrostatics
Exercise - 2:
2
Solve all types of
1. The rate constant for a first
order reaction is 4.606 ×
10–3s–1. The time required to
reduce 2.0g of the reactant to
0.2g is:
exercise questions (JEE 2020)
based on the latest JEE
pattern. (a) 500s (b) 1000s
(c) 100s (d) 200s
Answer Key
3
CHAPTER-1: ELECTROSTATICS
For extra exam
Exercise-1: Basic Objective preparation content,
visit the Vedantu app.
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TABLE OF CONTENTS
ELECTROSTATICS
Theory ................................................................................................................................................ 8
CAPACITANCE
Theory ................................................................................................................................................ 86
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ELECTROSTATICS
Chapter 01 9
ELECTROSTATICS
ELECTROSTATICS
1. ELECTRIC CHARGE
1.1 Definition
(v) Charge is conserved: Charge can neither be
Charge is the property associated with matter due to which it
produces and experiences electrical and magnetic effects. created nor be destroyed.
1mC 10 3
C, 1C 106 C, 1nC 109 C electric and magnetic fields but also radiates energy in
C.G.S. unit of charge is e.s.u. 1C = 3 × 109 e.s.u. the space surrounding the charge in the form of
(iv) i.e., charge can not exist without mass though mass ne with n = 0, 1, 2, 3 ...... Charge on a body can never
can exist without charge
be 0.5 e, ±17.2 e or ± 10–5 e etc.
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1.6 Comparison of Charge and Mass charged while the silk becomes negatively charged.
We are familiar with role of mass in gravitation, and we However, ebonite on rubbing with wool becomes
have just studied some features of electric charge. We negatively charged making the wool positively charged.
can compare the two as shown below: Clouds also become charged by friction. In charging by
friction in accordance with conservation of charge, both
Charge Mass
positive and negative charges in equal amounts appear
1. Electric charge can be Mass of a body is a simultaneously due to transfer of electrons from one
A body can be charged by following methods: involves the contact of a charged object to a neutral
(i) By friction: In friction when two bodies are rubbed object. Hence when an uncharged conductor is brought
together, electrons are transferred from one body to the in contact with a charged conductor, charge is shared
other. As a result of this one body becomes positively between the two conductors and hence the uncharged
charged while the other negatively charged, e.g., when a conductor gets charged.
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Fig 1.4
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(a) The direction of force is always along the line joining the Fair 1 Q1Q 2
Fm .
two charges. K 40K r 2
(b) The force is repulsive if the charges have the same sign
0 K 0r
and attractive if their signs are opposite.
(c) This force is conservative in nature. Medium K
(d) This is also called inverse square law.
Vacuum/air 1
2.1 Variation of k Water 80
Constant k depends upon system of units and medium
Mica 6
between the two charges.
Glass 5-10
NOTE: 1 q1q 2
F12 r1 r2
40 r1 r2 3
Absolute permittivity of air or free space
Dimension is M 1L3 T 4 A 2
1 q1q 2
F21 r2 r1
40 r2 r1 3
Fig 1.7
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rˆ12 rˆ21 by F F12 F22 2F1F2 cos and the force direction is
F2 sin
1 q1 q 2 given by tan
F12 rˆ12 F1 F2 cos
4 0 r 2
1 q 1q 2 1 q 2 q1
F12 rˆ12 rˆ21
4 0 r 2 4 0 r 2
1 q 2 q1
r̂21 F21
4 0 r 2
Fig 1.11
3. ELECTRIC FIELD
on a charge Q Net force on Q will be affect the source charge Q and its electric field is not
changed, therefore expression for electric field intensity can
Fnet F1 F2 ......... Fn 1 Fn
F
be better written as E lim
q 0 q0
0
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ELECTROSTATICS
Unit and Dimensional formula: Its S.I. unit – 3.4 Point Charge
Newton volt Joule Point charge produces its electric field at a point P which is
and
coulomb meter coulomb meter
distance r from it given by
C.G.S. unit = Dyne/stat coulomb.
Q
EP (Magnitude)
Dimension: [E] = [MLT–3 A–1] 40 r 2
Direction of electric field: Electric field (intensity)
E is a vector quantity. Electric field due to appositive
charge is always away from the charge and that due to Fig. 1.16
a negative charge is always towards the charge For + ve point charge, E is directed away from it.
For – ve point charge, E is directed towards it.
Fig 1.15
E 2 sin
tan
E1 E 2 cos
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ELECTROSTATICS
r kQ kQ
= sin
1
= sin 1 sin 2
Lr 2 Lr
Thus electric field at a general point in the
dx surrounding of a uniformly charged rod which
x
Fig. 1.18 subtend angles 1 and 2 at the two corners of rod
As shown in figure, if P is any general point in the can be given as
surrounding of rod, to find electric field strength at P,
kQ
we consider an element on rod of length dx at a
in x–direction : Ex cos 2 cos 1 and in y–
Lr
distance x from point O as shown in figure. Now if kQ
dE be the electric field at P due to the element, then
direction E y sin 1 sin 2
Lr
kdq Q
dE = Here dq = = dx
x r
2 2
L 3.5.2 Electric Field Strength Due to a Charged
Circular arc at its Centre
Electric field strength in x–direction due to dq at P is
Figure shows a circular arc of radius R which subtend
kdq kQ sin
dEx = dE sin = 2 2
= dx
x r L x r
2 2 an angle at its centre. To find electric field strength
at C, we consider a polar segment on arc of angular
Here we have x = r tan and dx = r secd
width dθ at an angle θ from the angular bisector XY as
Thus,
shown.
kQ r sec 2 d kQ
dEx sin = sin d
L r sec
2 2
Lr
Net electric field strength due to dq at point P in x–
direction is
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x
Rd 3.6 Properties of Electric Field Lines
++++++++++ Electric field lines originate from a positive charge &
++++
+
+++ terminate on a negative charge.
+
+
++
d
++
++
+
++
R Fig 1.20
C The number of field lines originating/terminating on a
dEsin charge is proportional to the magnitude of the charge.
dE dEcos
Y
Fig. 1.19
The length of elemental segment is Rd, the charge
Q
on this element d is dq = .d
Fig 1.21
Due to this d, electric field at centre of arc C is
The number of Field Lines passing through
kdq
given as dE perpendicular unit area will be proportional to the
R2
Now electric field component due to this segment magnitude of Electric Field there.
dEsin which is perpendicular to the angular bisector Tangent to a Field line at any point gives the direction
gets cancelled out in integration and net electric field of electric field at that point. This will be
at centre will be along angular bisector which can be instantaneous path charge will take if kept there.
EC dE cos
/2 /2 Fig 1.22
kQ kQ
2
cos d
R 2 /2
cos d Two or more field lines can never intersect each other.
/2 R
[Net electric field cannot have multiple directions].
kQ /2 kQ
sin /2 2
sin sin Uniform field lines are straight, parallel & uniformly
R 2 R 2 2
placed
2kQ sin
2 Field lines cannot form a loop.
2
R
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Electric Flux, A EA
Fig 1.25
5. GAUSS’S LAW
Fig 1.23
5.1 Definition
As increases, flux through area A decreases. If we draw a
According to Gauss’s law, total electric flux through a
vector of magnitude A along the positive normal, it is called
1
closed surface enclosing a charge is times the magnitude
the area vector, A corresponding to the area A. 0
1
of the charge enclosed. i.e. net (Qenc.)
0
Q en
i.e. . E .dA 0
NOTE:
Gauss’s law is only applicable for a closed surface.
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Fig 1.28
net E.dA E dA E dA E 4r 2
Fig 1.26
Lets take a spherical gaussian surface with charge ‘+Q’ kept Qenclosed = q
at the centre.
q q
We know field lines for a +ve charge are always radially Thus, E 4 r 2 or E
0 4 0 r 2
outward. Angle between dA & E is zero.
From the definition of the electric field, the force on a point
charge q0 located at a distance r from the charge q is F = q0E.
Therefore, which is Coulomb’s law.
Hence Net flux = Q/0. The cylindrical symmetry tells us that the field strength will
be the same at all points at a fixed distance r from the line.
NOTE: Thus, if the charges are positive. The field lines are directed
Although we derived gauss law for a spherical radially outwards, perpendicular to the line charge.
surface it is valid for any shape of Gaussian surface
and for any charge kept anywhere inside the surface
Fig 1.29
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ELECTROSTATICS
Fig. 1.31
According to Gauss law,
Q
Fig 1.30
E .ds or E 4r Q
2
0 0
Consider a large plane sheet of charge with surface charge
Electric field at P (Outside sphere)
density (charge per unit area) . We have to find the electric
field E at a point P in front of the sheet. 1 Q R 2 Q A
E out .
40 r 2 0 r 2 4R
2
NOTE:
If the charge is positive, the field is away from the
plane 6.3.2 At the Surface of Sphere
At surface r = R
To calculate the field E at P. Choose a cylinder of area of 1 Q
So, E s . 2
cross-section A through the point P as the Gaussian surface. 4 0 R 0
The flux due to the electric field of the plane sheet of charge
passes only through the two circular caps of the cylinder. 6.3.3 Inside the Sphere
Inside the conducting charged sphere electric field is zero
According to Gauss law E . dS q in / 0
and potential remains constant everywhere and equals to the
A potential at the surface. Ein = 0
I circular
E.dS
II circular
E.dS
cylindrical
E.dS
0
surface surface surface
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ELECTROSTATICS
Graphical variation of electric field with distance 6.4.2 At the Surface of Sphere
At surface r = R
1 Q R
Es .
4 0 R 2 3 0
q in Qr 3 Qr 3
E in .ds
0 0 R 3
E in4 r 2
0 R 3
or
1 Qr r
E in . E in r
Fig 1.32 4 0 R 3 3 0
total ch arg e Q 3Q
total volume 4 R 3 4R 3
3
Fig 1.34
Graphical variation of electric field with distance
Fig 1.33
6.4.1 Outside the Sphere at P (r > R)
According to Gauss’s law
Q Q
E .ds
or E 4r 2 0
0
Fig 1.35
1 Q
E out .
4 0 r 2
Q R 3
using Eout
4 3 30 r 2
R
3
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ELECTROSTATICS
kQ sin / 2
E
R2 / 2
kQx
E=
R
1/ 2
2
+ x2
2kQ x
E = 2 1–
3/2
R R x2
2
E=0
For x R
kQ
E
x2
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ELECTROSTATICS
1J
1eV 1.6 1019 C 1.6 1019 J 1.6 1012 erg
C
7.3 Potential Energy of a System of n Charges
Fig 1.36 In a system of n charges electric potential energy is
The electric force on the charge q2 is calculated for each pair and then all energies so obtained are
r
2
q1q 2 q1q 2 1 1 8. ELECTRIC POTENTIAL
W 40 r 2
dr
40
r1 r2
r
1 Suppose, a test charge q is moved in an electric field from a
The change in potential energy U(r2) – U(r1) is, therefore, point A to a point B while all the other charges in question
remain fixed. If the electric potential energy changes by UB
q q 1 1
U r2 U r1 W 1 2 – UA due to this displacement, we define the potential
40 r2 r1
difference between the point A and the point B as
As the potential energy of the two-charge system is assumed
to be zero when they have infinite separation.
q1 q 2 1 1 q 1q 2
U(r) = U(r) – U()
4 0 r 4 0 r
Fig 1.37
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ELECTROSTATICS
U U UA Wext NOTE:
V i.e. VB VA B KE 0
q q q At the centre of two equal and opposite charge
Joule
S.I. unit = volt
Coulomb
[V] = [ML2T–3A–1]
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ELECTROSTATICS
where E x V , E y V and E z V
x y z
Fig 1.42
B B
are a family of plane perpendicular to the field lines.
So, V B VA E . dr cos 0 E . dr Ed A metallic surface of any shape is an equipotential
A A
surface e.g. When a charge is given to a metallic
9.1 Equipotential Surface or Lines
surface, it distributes itself in a manner such that its
If every point of a surface is at same potential, then it
every point comes at same potential even if the object
is said to be an equipotential surface. Regarding
is of irregular shape and has sharp points on it.
equipotential surface following points should be kept
in mind :
The direction of electric field is perpendicular to the
equipotential surfaces or lines.
The equipotential surfaces produced by a point charge
or a spherical charge distribution are a family of
concentric spheres.
Fig 1.43
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ELECTROSTATICS
(iv) At the axis of Disc 2kQ
VP R2 x2 x
R2
kQ
VP
R
(ii) For x = R
kQ
VP
R
(ii) For x R
kQ
VP
x
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(ii) For x = R
kQ
VP
R
kQ
VP
x
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ELECTROSTATICS
k( q) k( q)
VP
10. ELECTRIC DIPOLE AP BP
r >> d (distance ‘r’ is large as compared to d);
10.1 General information
AP OP BP OP
System of two equal and opposite charges separated by a
small fixed distance is called a dipole. rd rd
OP O P
2 cos 2 cos
Dipole axis: Line joining negative charge to positive
charge of a dipole is called its axis. It may also be k( q) k( q)
termed as its longitudinal axis. Vp
(r d / 2 cos ) (r d / 2 cos )
Equatorial axis: Perpendicular bisector of the dipole
is called its equatorial or transverse axis as it is 1 1
k( q) .
perpendicular to length. r d / 2 cos r d / 2 cos
Dipole length: The distance between two charges is
known as dipole length (d)
Dipole moment: It is a quantity which gives r d / 2 cos r d / 2 cos
kq
information about the strength of dipole. It is a vector 2 d2 2
r cos
quantity and is directed from negative charge to 4
positive charge along the axis. It is denoted as p and
is defined as the product of the magnitude of either of kq d cos
d2
the charge and the dipole length. i.e. p q d r2
4
cos 2
(a) Electric Potential due to a dipole (b) Electric Field due to dipole
(i) For points on the axis
Let the point P be at distance r from the centre of the dipole
on the side of the charge q, as shown in fig (a). Then
q
E q pˆ
40 (r a) 2
q
q). Also E q pˆ
4 0 (r a) 2
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From Eqs. (i) and (ii), it is clear that the dipole field at large
For r > > a
distances does not involve q and a separately ; it depends on
2kp
E (r >> a) ...(i) the product qa. In terms of p, the electric field of a dipole at
r3
large distances takes simple forms :
(ii) For points on the equatorial plane At a point on the dipole axis
The magnitudes of the electric fields due to the two 2p 2kp
E
3
r a
charges + q and –q are given by 40 r r3
q 1 At a point on the equatorial plane
Eq
4 0 r 2 a 2
p kp
E (r a)
q 1 40 r 3
r3
E q and are equal.
4 0 r 2 a 2
Electric field due to an electric dipole
y
Enet
Er
E
O
x
-q +q
Fig 1.46
(b) a point on the equatorial plane of the dipole. p is the of dipole electric field at O is in radially outward direction
dipole moment vector of magnitude p = q × 2a and and due to the negative charge it is radially inward as shown
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V 2kp cos (ii) Work: From the above discussion it is clear that in
Er
r r3 an uniform electric field dipole tries to align itself in
If the direction of Enet is at an angle α from radial it is again turn so that it makes an angle 2 with the
Fig 1.49
Fig 1.47
W = U = U – U 90° = – pE cos
(a) Net force on electric dipole Fnet 0
U = – pE cos
(b) Torque pE sin p E
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I
T = 2π
pE
(I = moment of inertia)
field must be normal to the surface at every point This follows from results 1 and 2 above. Since E = 0
inside the conductor and has no tangential component
If E were not normal to the surface, it would have
on the surface, no work is done in moving a small test
some non-zero component along the surface. Free
charge within the conductor and on its surface. That is,
charges on the surface of the conductor would then
there is no potential difference between any two points
experience force and move. In the static situation,
inside or on the surface of the conductor. Hence, the
therefore, E should have no tangential component.
result. If the conductor is charged, electric field normal
Thus electrostatic field at the surface of a charged
to the surface exists; this means potential will be
conductor must be normal to the surface at every point.
different for the surface and a point just outside the
(For a conductor without any surface charge density,
surface.
field is zero even at the surface).
In a system of conductors of arbitrary size, shape and
charge configuration, each conductor is characterized
by a constant value of potential, but this constant may
differ from one conductor to the other.
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E nˆ
0
kQ1 k Q1 k Q1 Q 2
Va
r1 r2 r2
kQ1 k Q1 k Q1 Q 2
Vb
r2 r2 r2
Fig 1.52
Fig 1.53
Fig 1.56
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Fig 1.57
Charge is flown from outer surface because as long as Q1
remains on inner shell, ‘– Q1’ will be induced on inner shell.
Fig 1.59
–q
q
Fig 1.60
Fig 1.58
(iii) Resultant field, due to q (which is inside the cavity)
k Q1 x k Q1 x k Q1 Q2 x
VP 0 and induced charge on S1, at any point outside S1 (like
r1 r2 r2
B,C) is zero. Resultant field due to q + Q on S2 and
Q x Q1 x Q1 Q2 x any other charge outside S2 , at any point inside of
1 0
r1 r2 r2 surface S2 (like A, B) is zero
Q1 Q2 x r S2
x Q1 Q2 1 .B .C
r1 r2 r1 r2
S1
.q q+Q
NOTE: .A
–q
As it can be seen not all charge on the surface flows to
ground. When the outermost conductor is earthed then
the charge residing on the outermost surface of outer
conductor will flow to ground.
Fig 1.61
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Fig 1.65
(E)
Fig 1.63
S2
(B)
S2
S1
•q ·q
S1 •C C
Fig 1.67
Fig 1.64
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(F)
S2
C
S1 • •q
Case A B C
S1 Uniform Non uniform Non uniform
S2 Uniform Uniform Uniform
Fig 1.71
Case D E F
S1 Non uniform Uniform Non uniform Q Q2
Final common potential = k . 1
r1 r2
S2 Uniform Non uniform Non uniform
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Fig 1.75
E
0
Net electric field at point P, near a conducting surface,
having surface charge distribution ‘’ is given by [/0].
Fig 1.73
13.2 Parallel Plate Theory
3Q 2
U To find charge distribution on each surface of plates
200 R
Fig 1.76
Fig 1.74
Two conducting plates having area ‘A’ (area is large as
Q2 compared to distance, so that field is uniform) and the
U
8 0 R thickness of plates is small so that charge only appears on
parallel faces.
The energy stored per unit volume around a point in an
U 1
electric field is given by u e 0 E 2 . If in place
Volume 2
of vacuum some medium of dielectric constant K is present
1
then u e K0 E 2 .
2
SCAN CODE
ELECTROSTATICS
37
ELECTROSTATICS
Hence it can be said that net charge enclosed will be zero 13. FORCE ON A CHARGED CONDUCTOR
which implies the charges appearing on the facing surfaces
are equal & opposite to each other. To find force on a charged conductor (due to repulsion of
like charges) imagine a small part XY to be cut and just
separated from the rest of the conductor MLN. The field in
the cavity due to the rest of the conductor is E2, while field
due to small part is E1. Then
Fig 1.78
Net electric field at any point ‘P’ or ‘R’ has to be zero.
(Enet)P = 0
Fig 1.80
There are 4 distributions, the net field at P should be zero.
Inside the conductor
Q q
EP 1 1 E P 2 q E = E1 – E2 = 0 or E1 = E2
2 0 2A0 2A0
Outside the conductor E E1 E 2 .
q Q q 0
E P 3 E P 4 2
2A0 2A 0
Thus, E1 E2
2 0
E P 1 E P 3 E P 2 E P 4
To find force, imagine charged part XY (having charge
Q1 q q q Q q dA placed in the cavity MN having field E2). Thus force
This shows 2
2A 0 2A 0 2A 0 2A 0 2
dF = (dA)E2 or dF dA . The force per unit area or
2 0
Q1 Q2
q so final distributions would be dF 2
2 electric pressure is P (Electrostatic pressure)
dA 2 0
2
The force is always outwards as (+) is positive i.e.,
whether charged positively or negatively, this force will try
to expand the charged body.
Fig 1.79
NOTE:
When charged conducting plates are placed parallel to
each other, the two outermost surfaces get equal
charges and the facing surfaces get equal and opposite
charges
SCAN CODE
ELECTROSTATICS
ELECTROSTATICS 38
SOLVED EXAMPLES
Example - 1
Example -2
A particle of mass m and charge (–q) enters the
Three equal charges each +Q, placed at the corners of
region between the two charged plates initially,
an equilateral triangle of side a. What will be the
moving along X-axis with speed vx as shown in fig.
force on any charge ? k 1
The length of each plate is L and a uniform electric 40
2 2
field E is maintained between the plates. Show that (a) kQ (b) 2 kQ
a2 a2
the vertical deflection of the particle at the far edge
(c) 2 kQ 2 (d) 3 kQ 2
2
a2 a2
of the plate is. qEL .
2mv 2x Ans. (d)
Sol. Suppose net force is to be calculated on the charge
which is kept at A. Two charges kept at B and C are
applying force on that particular charge, with
directions as shown in the figure.
qE
ay
m
Example - 5
- 5 A cube of side b has a charge q at each of its vertices.
Determine the potential and electric field due to this
charge-array at the centre of the cube.
Sol. O is the centre of cube ABCDEFGH. Charge q is
placed at each of eight corners of the cube.
Surface charge density on inner surface = q
4r12 Electric Potential :
= 1 q 1 2q
40 ( 3 b / 2) 4 0 3 b
Net potential at O due to all 8 charges at corners of the
cube V 8 1 2q 1 . 16q
40 3 b 4 0 3 b
Electric Field : The electric field at O due to charges
at all corners of the cube is zero, since, electric fields
due to charges at opposite corners such as A and H, G
and D, B and E, F and C are equal and opposite.
ELECTROSTATICS 40
Example - 6 2 106
| E C | 9 109 2
1.44 107 N / C
Four charges are placed on corners of a square as 5
102
2
shown in figure having side of 5 cm. If Q is one
micro coulomb, then electric field intensity at centre 10 6
| E D | 9 109 0.72 107 N / C ;
will be 5
2
102
2
E CA E BD ...(i)
2 2
E
2 106
| E B | 9 109 2
1.44 107 N / C ;
5
102
2
ELECTROSTATICS 41
1 q 3 q 2 q 3 q 1 q 2 q1 Example - 9
0
40 (r32 )i (r31 )i (r21 )i A point particle of mass M is attached to one end of
a massless rigid non-conducting rod of length L.
1 q 3 q 2 q 3 q 1 q 2 q1 1
mv
2
(4)(2) (4)(2) that the rod makes a small angle θ (say of about 5
(9.0 109 ) 10 (9.0 10 )
12 9
(5.0) (5.0) degrees) with the field direction (see figure). What
will be the minimum time, needed for the rod to
(4) (2) (4) (2) 1 become parallel to the field after it is set free
(3.0) (3.0) 10 2 10 v
12 3 2
16 16 1
(9 103 ) (9 103 ) 103 v2
∴ 5 3 2
2 1
(9 103 ) (16) 103 v2
15 2
∴ v = 6.2 m/s ML
(a) t 2
Example - 8 2 pE
Ans. (a)
ELECTROSTATICS 42
Sol. In the given situation, the system oscillates in electric A according to the figure show below. The total field
field with maximum angular displacement θ. Its time at C must be zero. The field at C due to the point
period of oscillation (similar to dipole)
charge is E = k Q
r2
I
T 2 towards left. The field at C due to the induced charges
pE
where I = moment of inertia of the system and must be kQ towards right i.e. directed towards Q.
Hence the minimum time needed for the rod becomes r2
Example - 11
parallel to the field is t T I A long thin rod lies along the x-axis with one end at
4 2 pE
the origin. It has a uniform charge density λ C/m.
2 2 2 Assuming it to infinite in length, the electric field at
Here, I M L M L ML
2 2 2 point x = – a on the x-axis will
(a) (b)
ML2
t 0 a 2 0 a
⇒ 2 2 qL E
ML (c) (d) 2
2 2qE 4 0 a 0 a
Ans. (c)
Example - 10 Sol.
A point charge Q is placed outside a hollow
spherical conductor of radius R, at a distance (r > R)
from its centre C. The field at C due to the induced
E
(c) k Q directed towards Q 4 0 a
r2
(d) k Q directed away from Q Example - 12
r2 Two infinite plane parallel sheets separated by a
Ans. (c)
distance d have equal and opposite uniform charge
Sol.
densities σ. Electric field at a point between the
sheets is
(c) (d) 2
2 0 0
Ans. (b)
ELECTROSTATICS 43
Sol. Sol.
E1
2 0
At equilibrium of Pendulum
EII
2 0 2 0 2 0 0
T sin θ = Fe
T cosθ = mg
Example - 13
Fe QE Q
An infinite plane with uniformly distributed positive tan tan
mg mg 2 0 mg
charge has surface charge density σ. A small metallic
Q
sphere S of mass m and charge +Q is attached to a tan 1
thread and tied to a point P on the sheet AB. The 2 0 mg
(a) tan 1 Q
0 mg
(b) cot 1 Q
2 0 mg
Inside the solid sphere E k . Q .x E x
R3
(c) tan 1 Q
2 0 mg Q'
Q 4
x3
4
R 3
3
(d) cot 1 Q 3
0 mg
Q Q' Q
Ans. (c) Q' 3
.x 3 E k. 2
k. x
R x R3
ELECTROSTATICS 44
Example - 15 Example - 17
Two very long line charges of uniform charge An electrostatic field line cannot be discontinuous,
density λ and –λ are placed along the same line with why?
the separation between the nearest ends being 2a as
Sol. Electrostatic field line originates from positive charge
shown in figure . The electric field intensity at the
point O is and terminates on negative charge i.e., it cannot
abruptly start or end in space. So the electric field line
is always continuous (directed from higher to lower
potential).
Example - 18
(a) a (b) 4 a
Find the electrostatic potential at equatorial point of
an electric dipole.
(c) 3 a (d) 8 a
Sol. Let P be a point on the equator of an electric dipole
Ans. (a)
formed of charges –q and +q at separation 2a.
Sol.
The distance of point P from centre of dipole = r
AP = BP = r 2 a2
Electrostatics potential at
P, V 1 q q
P
40 BP AP
E1 = E2 = ; Net - electric field at 0
4 0 a
= E1 E 2 2 =
4 0 a 2 0 a
Example - 16
Depict the equipotential surfaces for a system of two
identical positive point charges placed a distance ‘d’
apart.
Sol. Equipotential surfaces due to two identical charges is 1 q q
VP 0
40 r 2 a 2 r a2
2
shown in fig.
Thus, electrostatic potential at each equatorial point of
an electric dipole is zero.
Example - 19
A regular hexagon of side 0.10 m has a charge 5 μC
at each of its vertices. Calculate the potential at the
centre of the hexagon.
ELECTROSTATICS 45
Sol. The potential due to similar charges is additive. If we take (+) sign x = 2 (1 – x).
Let O be the centre of the hexagon.
This gives x 2 m = 66.7 cm
3
6 Example - 21
V = 6 × 9 × 109 × 5 × 10 = 2.7 × 106 volt
0.10 Two point charges 10 × 10–8 C and – 4 × 10– 8 C are
Example - 20 separated by a distance of 70 cm in air as shown in
(i) Two point charges 4Q and Q are separated by a figure.
distance 1 m in air. At what point on the line joining
the charges is the electric field intensity zero?
(ii) Also calculate the electrostatic potential energy
(i) Find at what distance from point A would the
of the charges if Q = 2 × 10–7 C.
electric potential be zero.
Sol. (i) Let x be the distance of point P from charge 4Q
(ii) Also calculate the electrostatic potential energy
where electric field intensity is zero, then of the system.
Sol. (i) The electric potential due to opposite charges is
E E1 E 2 0
subtractive. To find the electric potential, the
sign of the charge is retained in the formula.
Let x be the distance of point P from A at which
q electric potential is zero
1 4Q 1 Q 4 1 1 10 10 8 1 (4 108 )
2 0
40 x 2 40 (1 x) 2 x (1 x)2 40 x 40 (0.70 x)
x2 = 4(1 – x)2
x = 2 (1 – x)
ELECTROSTATICS 46
10 4
x 0.70 x
7.0 – 10x = 4x
10 10 6 2
or 14 x = 7.0 or x 17.0 0.5m U 9 109 3 27 J
0.1
14
Example - 23
Also V 1 q1 q2
0
40 x x 0.70
Infinite charges are lying at x = 1, 2, 4, 8…meter on
X-axis and the value of each charge is Q, find the
value of intensity of electric field at point x = 0 due
to these charges. Also find the value of potential at
this point if the value of Q is 1 micro coulomb.
8 8
10 10 4 10 0 10 4 (a) 12 × 109 Q N/C, 1.8 × 104 V
x x 0.70 x x 0.70
(b) Zero, 1.2 × 104 V
(c) 6 × 109 Q N/C, 9 × 103 V
x 7 m = 1.167 m
6 (d) 4 × 109 Q N/C , 6 × 103 V
Thus electric potential is zero at distances x = 0.5 m Ans. (a)
–8
and 1.167 m from charge q1 = 10 × 10 C. Sol. By the superposition, Net electric field at origin
(ii) Electrostatic potential energy of system 1 1 1 1
E kQ 2 2 2 2 ...
1
9 8 8
q1q2 = 9 × 10 × (10 10 )(4 10 ) 1 2 4 8
=U
4 0 r 0.70 1 1 1
E kQ 1 ...
4 16 64
U = –5.14 × 10–5 J
1 1 1
1 ...
Example - 22 4 16 64
is an infinite geometric progression. Its sum can be
Three particles, each having a charge of 10 micro
coulomb are placed at the corners of an equilateral
obtained by using the formula S a ;
triangle of side 10cm. The electrostatic potential 1 r
energy of the system is Where a = First term, r = Common ratio.
(Given 1
9 109 N m 2 / C )
4 0 Here a = 1 and r 1 so,
4
(a) Zero (b) Infinite
(c) 27 J (d) 100 J 1 1 1 1 4
1 ... .
4 16 64 1 1 / 4 3
Ans. (c)
ELECTROSTATICS 47
4 Example - 25
E 9 109 Q 12 109 Q N / C
3 An arc of radius r carries charge. The linear density
of charge is λ and the arc subtends an angle π/3 at the
Hence, Electric potential at origin
centre. What is electric potential at the centre
1 1106 1 106 1106 110 6
V ...
40 1 2 4 8 (a) 4 0 (b) 8 0
1 1 1
9 109 106 1 .............. (c) 12 0 (d) 16 0
2 4 8
Ans. (c)
Sol.
1
3 4
9 10
1 = 1.8 × 10 volt
1
2
Example - 24
A conducting sphere of radius R is charged to a
potential of V volt. Then the electric field at a
distance r (> R) from the centre of the sphere would Length of the arc = r
kq Kr
(a) RV (b) V V k =
r2 r r r 3
2 1
(c) rV (d) R V V = K V= . V=
R2 r3 3 4 0 3 12 0
Ans. (a) Example - 26
Sol. Conducting sphere (a) Plot a graph comparing the variation of
potential ‘V’ and electric field ‘E’ due to a point
charge ‘Q’ as a function of distance ‘R’ from
the point charge.
(b) Find the ratio of the potential differences that
must be applied across the parallel and the
series combination of two identical capacitors
If r > R so that the energy stored, in the two cases,
becomes the same.
1 Q Q Sol. (a) The graph of variation of potential and electric
V= . = k.
4 0 R R field due to a point charge Q with distance R
from the point charge is shown in fig.
V .R Q
Q= E = k.
k r2
VR VR
E = k. 2
E= 2
kr r
ELECTROSTATICS 48
In series arrangement, net capacitance C C . The distance of the field point on the axis of a small
s
2 dipole is doubled. By what factor will the electric
In parallel arrangement, net capacitance, Cp = 2C field, due to the dipole change ?
Energy stored U 1 CV 2
2 Sol. For a small dipole, Eaxis = 1 2p 1
3
If VS and VP are potential differences applied 40 r 3 r
across series and parallel arrangements, then When the distance r is doubled, the electric field
given Us = Up 1 C V 2 1 C V 2 strength becomes 1/8 times the original field.
S S p p
2 2
Example - 30
Vp Cs
C/2 1
VS Cp 2C 2
A positive charge is fixed at the origin of
Example - 27 coordinates. An electric dipole, which is free to move
A 500 μC charge is at the centre of a square of side and rotate, is placed on the positive x–axis. Its
10 cm. Find the work done in moving the charge of moment is directed away from the origin. The dipole
10 μC between two diagonally opposite points on will :
the square. (a) Move towards the origin
(b) Move away from the origin
(c) Rotate by /2
(d) Rotate by
Sol. The points A and B are equidistant from the centre of Sol.
square where charge q = 500 μC is located ; therefore, -q charge is near the +Q as compare to +q, so
points A and B are at the same potential i.e., VA = VB. attractive force will be greater than repulsive force.
Work done in moving charge q0 = 10μC from A to Therefore, dipole moves towards the origin.
B is W = q0 (VB – VA) = 0
Example - 31
Example - 28
An electric dipole with a dipole moment 4 × 10–9 A small metallic charged sphere is placed at the
Cm is aligned at 30º with the direction of a uniform centre of large uncharged spherical shell and the two
electric field of magnitude 5 × 104 NC–1. Calculate are connected by a wire. Will any charge flow on
the magnitude of the torque acting on the dipole. the outer shell?
Sol. A dipole placed in a uniform electric field experiences Sol. Yes, the whole charge will flow to the outer shell ;
a torque τ =pE sin θ which tends to align the dipole because charge always resides on outer surface of a
Here, p = 4 × 10–9 C-m, E = 5 × 104 NC–1, θ = 30º A hollow charged conductor has a tiny hole cut into
Torque τ = 4 × 10–9 × 5 × 104 sin 30º its surface. Show that the electric field in the hole is,
where is unit vector in the outward normal direction
4 × 10–9 × 5 × 104 × 1 = 10–4 Nm
2 and is the surface charge density near the hole.
ELECTROSTATICS 49
Sol. Example - 33
The electric field component in the fig. shown are
Ex = x1/2, Ey = Ez= 0, in which = 800 N/Cm1/2.
Calculate (a) the electric flux through the cube
(b) the charge within the cube. The side of cube a =
0.1 m.
Sol. According to the question only the electric field has
only X-component, Y and Z-components are zero; so
electric flux diverging along Y and Z-axis is zero
Consider a hollow conductor (of any shape) having
since E S = ES cos 90º = 0. The electric flux
hole AB at its surface. Consider a point P near the
diverging along X-axis from faces (1) and (2) exists.
hole just outside the conductor. First we consider the
The electric field at face (1) (where x = a) is E1 =
conductor with hole filled up.
The electric field at point P due to whole conductor x1/2= a1/2
Electric charge and its properties and Force between charges 9. Two point charges q1 = + 2C and q2 = – 1C are separated by
a distance d. The position on the line joining the two charges
1. 1012 particles (nuclei of helium) fall per second on a where a third charge = + 1C will be in equilibrium is at a
neutral sphere. Calculate the time (in second) in which distance
the sphere gets charged by 2 C . (a) d / 2 from q1 between q1 & q2
2. The ratio of the electric force between two electrons to (b) d / 2 from q1 away from q2
the gravitational force between them is of the order of
(c) d / 2 – 1 from q2 between q1 & q2
(a) 1042 (b) 1040
(c) 1036 (d) 1032 (d) d / 2 – 1 from q2 away q1
3. Two point charges +2 coulomb and +6 coulomb repel each 10. A charge Q is placed at each of the opposite corners of a
other with a force of 12 N. If a charge of –4 coulomb is square. A charge -q is placed at each of the other two corners.
given to each of these charges then the force will be If the net electrical force on is zero, then the Q/q
(a) 4 N (repulsive) (b) 4 N (attractive)
(c) 8 N (repulsive) (d) 8 N (attractive) (a) 2 2 (b) –1
14. Two charges, each equal to q, are kept at x = –a and x = a on 19. An infinite number of charges, each equal to q, are placed
q along the x axis at x 1, x 2, x 4, x 8, etc. The
the x–axis. A particle of mass m and charge q 0 is
2 electric field at the point x 0 due to the set of charges is
placed at the origin. If charge q 0 is given a small
q
displacement y (y << a) along the y –axis, the net force .
acting on the particle is proportional to n 0 Find n .
(a) y (b) –y 20. A charged particle enters at point A and comes out from
point B. Its velocity vector makes angle and with
1 1
(c) y (d) y electric field at these two points, respectively. The ratio of
kinetic energy of the charged particle at these two points
Electric field KB
will be (Given 60 and 30
4 KA
15. A charged oil drop is suspended in uniform field of 3×10
V/m, so that it neither falls nor rises. The charge on the
drop will be
–15 2
(take the mass of the charge = 9.9 × 10 kg and g = 10 m/s )
–18 –18
(a) 3.3 × 10 C (b) 3.2 × 10 C
–18 –18
(c) 1.6× 10 C (d) 4.8 × 10 C
16. Two point charges q1 = 4 C & q2 = 9 C are placed 20 cm Electric field due to continuous charge distribution
apart. The electric field due to them will be zero on the line
joining them at a distance of 21. The magnitude of the average electric field normally
present in the atmosphere just above the surface of the
(a) 8 cm from q1 (b) 8 cm from q2
Earth is about 150 N/C, directed inward towards the center
(c) 80/13 cm from q1 (d) 80/13 cm from q2 of the Earth. This gives the total net surface charge carried
by the Earth to be:
17. Charge of +q, +2q, + and –are placed at the corners of a
square ABCD, calculate the electric field at the intersection [Given 0 = 8.85 × 10–12 C2/Nm2, RE = 6.37 × 106 m]
of the diagonals of the square when q = 5/3 × 10–9 C and
(a) +670 kC (b) – 670 kC
each side square of is 10 cm.
(c) – 680 kC (d) + 680 kC
22. In a uniformly charged sphere of total charge Q and radius
R, the electric field E is plotted as function of distance from
the centre. The graph which would correspond to the above
will be
(a) (b)
23. The electric field strength due a uniformly charged ring of 27. A spherically symmetric charge distribution is
radius R at a distance x from its centre on its axis carrying characterised by a charge density having the following
charge has somewhere maximum value on the axis. This is variation:
at distance
r
(r) O 1 for r R
(a) x = R (b) x 2 R R
Qr12 Qr12
(c) (d)
4 0 R 4 3 0 R 4
29. A uniformly charged sphere is placed inside a charged
hollow sphere as shown in the figure. O is the centre of
hollow sphere and C is the centre of solid sphere. The
(a) cos (b) cot
3
(c) sin (d) tan magnitude of charge on both the spheres is 4 7 C .
26. A thin semi-circular ring of radius r has a positive charge
distributed uniformly over it. The net field E at the centre The electric field at point P which lies just outside the
O is hollow sphere is given by 3 K N / C . Find the value of
1
. (Here, K 4
0
q q ˆj
(a) ĵ (b)
42 0 r 2 4 2 0 r 2
q ˆj q
(c) (d) ĵ
2 0 r 2
2
2 0 r 2
2
ELECTROSTATICS 53
Electric potential energy 36. Three charges q, q and Q are located at the vertices of
30. Two equal charges are placed at a distance of ‘2 a’ and a a right-angled isosceles triangle as shown in the figure. If
third charge – 2is placed at the midpoint. The potential q
energy of the system is the total interaction energy is zero, then Q . Find
p r
q2 6 q2
(a) (b) r2 ?
80 a 80 a
7 q2 9 q2
(c) (d)
80 a 80 a
1 2qQ 1
(a) 1 (b) zero
40 a 5
1 2qQ 1 1 2qQ 2
(c) 4 a 1 (d) 4 a 1
0 5 0 5
34. Two charge particles having charges 1C and –1C and
of mass 50 gm each are held at rest while their separation 4 2q
(a) zero, 4q/a2 (b) zero ,
is 2 meter. Find the speed of the particles when their a2
separation is 1 meter.
(a) 0.20 m/s (b) 0.6 m/s
(c) 4 2q / a ; 4q / a 2 (d) 4 2 q ; 4 2q / a 2
(c) 0.3 m/s (d) 0.4 m/s
35. A charged particle with q is shot towards another charged 39. Two charges of 4 C each are placed at the corners A and B
particle with charged q which is fixed, with a speed v. it of an equilateral triangle of side length 0.2 m in air. The
approaches upto a closest distance r and then returns. If q 1 N m2
was given a speed 2v, the closest distance of approach electric potential at C is 9 109
would be 40 C2
–3
40. An electric charge 10 C is placed at the origin (0, 0) of X- 47. The variation of potential is maximum if one goes
Y coordinate system. Two points A and B are situated at
(a) along the field line
( 2, 2) and (2, 0) respectively. The potential difference (b) Perpendicular to field line
between the points A and B will be (c) in any direction
(a) 9 V (b) zero (d) at 45° with the direction of field line
(c) 2 V (d) 4.5 V 48. Which of the following may be discontinuous across a
41. Two concentric, thin metallic spheres of radii R1 and R2 (R1 > R2) charged conducting surface ?
bear charges Q1 and Q2 respectively. Then the potential at
(a) Electric potential
1 (b) Electric intensity
radius r between R1 and R2 will be K 4
0 (c) Both electric intensity and potential
(a) K (Q1 + Q2) / r (b) K (Q1/r + Q2/R2) (d) None of these
(c) K (Q2/r + Q1/R1) (d) K (Q1/R1 + Q2/R2) 49. Charges are placed on the vertices of a square as shown.
42. A hollow hemisphere of radius R is charged uniformly with Let E be the electric field and V the potential at the centre.
surface density of charge . What will be the potential at If the charges on A and B are interchanged with those on D
centre ? and C respectively, then
R
(a) 2 (b) 4
0 0
4R
(c) 2 (d) 3
0 0
10
(a) q 3iˆ 2jˆ kˆ N.m
(b) q 3iˆ 2ˆj kˆ N.m
(d) V / m and in the +ve x direction
9
(c) q 3iˆ 2ˆj kˆ N.m (d) q ˆi 2ˆj kˆ N.m
56. An electric field (30iˆ 20 ˆj )Vm 1 exists in space. If the
63. An electric dipole of dipole moment p is placed in uniform
potential at the origin is zero, then find the potential (in
electric field E , with p parallel to E . It is then rotated by
volt) at (5 m,3 m).
an angle of . The work done is
Electric dipole & Dipole in electric field (a) pE sin (b) pE cos
(c) pE (1 – cos ) (d) pE (1 – sin)
57. Two short dipoles each of dipole moment p are placed at
64. An electric dipole is placed at an angle of 30º to a non-
origin. The dipole moment of one dipole is along x axis,
uniform electric field. The dipole will experience
while that of other is along y axis. The electric field at point
(a, 0) is given by (a) a translational force only in the direction of the field
(b) a translational force only in a direction normal to the
2p p direction of the field
(a) 4 a (b)
0 40 a 3 (c) a torque as well as a translational force
(d) a torque only
5p
(c) (d) zero
40 a 3
56
ELECTROSTATICS
2 1
(c) (d) (1 + 2) 0
0
70. A positive point charge is placed just outside the centre of
the flat face of a hemispherical surface in air. The electric
flux through the flat face is
(b) in the region between the plates
68. A cube of side 10 cm encloses a charge of 0.1 C at its 0
centre. What is number of field lines through each face of
the cube. (c) in the region outside the plates
0
(a) 1.13 × 1011 (b) 1.13 × 106
(d) zero in the region between the plates
(c) 1.13 × 1013 (d) 1883
ELECTROSTATICS 57
Q Q
(a) (b)
2 3
2Q Q
(c) (d)
3 4
79. Two charged metallic spheres of radii r1 and r2 are touched
and separated. What is the ratio of their surface charge
density.
75. The field at a distance r from a long string of charge per unit 1 r2 1 r1
length is (a) r (b) r
2 1 2 2
k k
(a) (b) 1 1 r2 r1
r2 r (c) 1 (d) r r
2 2 1 2
k 2k 80. What is the largest charge a metal ball of 1mm radius can
(c) (d)
2r r hold? Dielectric strength of air is 3×106 Vm–1.
(a) 3nC (b) 2nC
(c) 1/2 nC (d) 1/3nC
ELECTROSTATICS 58
(a) 2 b2 a 2
(b) a 2 b2
1 , 2 , 3 and 4 . Then : (2017)
(a) 1 2 3 4 (b) 1 2 3 4
2Q Q
(c) (d) (c) 1 2 3 4 (d) 1 3 ; 2 4
a 2 2 a 2
ELECTROSTATICS 59
2qE R R
(d) The new equilibrium position is at a distance (a) (b)
k 5 2
from x =0.
(c) R (d) R 2
ELECTROSTATICS 60
(a) 63iˆ 27 ˆj 10
2
(b) 63iˆ 27 ˆj 10
2
(c) 81iˆ 81 ˆj 10
2
(d) 81iˆ 81 ˆj 10
2
R 2R
(a) (b)
2 1 2 1
R 2R
(c) (d)
2 1 2 1
18. Charges –q and +q located at A and B, respectively, (a) Electric field of a uniformly charged sphere
constitute an electric dipole. Distance AB = 2a, O is the (b) Potential of a uniformly charged spherical shell
midpoint of the dipole and OP is perpendicular to AB. A (c) Potential of a uniformly charged sphere
charge is placed at P where OP = y and y . The charge (d) Electric field of a uniformly charged spherical shell
experiences an electrostatic force F. If is now moved along
21. An electric field of 1000 V /m is applied to an electric
y dipoleat angle of 45°. The value of electric dipole
the equatorial line to P’ such that OP’ = , the force moment is 10 –29Cm. What is the potential energy of the
3
electricdipole? (2019)
y
on will be close to 2a (2019) (a) –20 × 10–18 J (b) –7 × 10–27 J
3
(c) –10 × 10–29 J (d) –9 × 10–20 J
ELECTROSTATICS 61
22. Determine the electric dipole moment of the system of 26. An electric dipole is formed by two equal and opposite
three charges, placed on the vertices of an equilateral charges with separation d. The charges have same mass
triangle, as shown in the figure: (2019) m. It is kept in a uniform electric field E. If it is slightly
rotated from its equilibrium orientation, then its
angular frequency is : (2019)
qE 2qE
(a) (b)
md md
qE qE
(c) 2 (d)
md 2md
ˆj iˆ iˆ ˆj
(a) 3 q (b) q 27. The electric field in a region is given by
2 2
E Ax B iˆ, where E is in NC–1 and x is in metres.
(c) 2qljˆ (d) 3qljˆ The values of constants are A = 20 SI unit and B = 10 SI
23. The bob of a simple pendulum has mass 2 g and a charge unit. If the potential at x = 1 is V1 and that at x = –5 is V2,
of 5.0 μC. It is at rest in a uniform horizontal electric field then V1 V2 (2019)
of intensity 2000 V/m. At equilibrium, the angle that the (a) 320 V (b) –48 V
pendulum makes with the vertical is: (take g = 10 m/s2) (c) 180 V (d) –520 V
(2019) 28. A system of three charges are placed as shown in the
(a) tan (2.0)
–1
(b) tan (0.2)
–1
figure:
(c) tan (5.0)
–1
(d) tan–1(0.5)
24. A solid conducting sphere, having a charge Q, is
surrounded by an uncharged conducting hollow spherical
shell. Let the potential difference between the surface of
the solid sphere and that of the outer surface of the hollow
shell be V. If the shell is now given a charge of –4 Q, the If D >> d, the potential energy of the system is best given
new potential difference between the same two surfaces by (2019)
is nV, where n is: (2019)
25. A positive point charge is released from rest at a distance 1 q 2 qQd
(a) 4 d 2 D 2
r0 from a positive line charge with uniform density. Thes 0
peed (v) of the point charge, as a function of instantaneous
distance r from line charge, is proportional to: (2019) 1 q 2 2qQd
(b) 4 d D 2
0
1 q2 qQd
(c) 4 d D 2
0
1 q 2 qQd
(d) 2
4 0 d D
30. A uniformly charged ring of radius 3a and total charge +q 33. Shown in the figure is a shell made of a conductor. It has
is placed in xy-plane centred at origin. A point charge is inner radius a and outer radius b, and carries charge Q. At
moving towards the ring along the z-axis and has speed its centre is a dipole P as shown. In this case: (2019)
vat z = 4a. The minimum value of v such that it crosses the
origin is : (2019)
1/ 2
2 4 q2 2 1 q2
(a) (b)
m 15 4 0 a m 5 4 0 a
1/ 2 1/ 2
2 2 q2 2 1 q2
(c) (d)
m 15 4 0 a m 15 4 0 a
(a) surface change density on the inner surface is uniform
31. In free space, a particle A of charge 1 C is held fixed at a
Q
point P. Another particle B of the same charge and mas s
and equal to 2 2
4 g is kept at a distance of 1 mm from P. If B is released, 4 a
then its velocity at a distance of 9 mm from P is: (b) electric field outside the shell is the same as that of a
point charge at the centre of the shell.
1
Take 9 109 Nm2 C 2 (2019) (c) surface charge density on the outer surface depends
4 0 on P
(a) 1.0 m/s (b) 3.0×10 m / s
4
(d) surface charge density on the inner surface of the shell
(c) 2.0×10 m / s
3 (d) 1.5×10 m / s
2 is zero everywhere.
32. A simple pendulum of length L is placed between the plates 34. A point dipole p p0 xˆ is kept at the origin. The
of a parallel plate capacitor having electric field E, as shown potential and electric field due to this dipole on the y-axis
in figure. Its bob has mass m and charge q. The time period at a distance d are, respectively : (Take V = 0 at infinity)
of the pendulum is given by: (2019)
(2019)
P P P
(a) , (b) 0,
4 0 d 2 4 0 d 3 4 0 d 3
P P P
(c) 0, (d) ,
4 0 d 3 4 0 d 2 4 0 d 3
35. Let a total charge 2be distributed in a sphere of radius R,
(c) 2
L (d) 2 L
qE 1/2
1
R
2 qE 2 (c) a 2 R (d) a
g
4
g 3
m m
ELECTROSTATICS 63
36. Two infinite planes each with uniform surface charge 38. If finding the electric field around a surface is given by
density C / m are kept in such a way that the angle
2
q
E enclosed is applicable. In the formula 0 is permittivity
between them is 30 . The electric field in the region shown
0 0 A
between them is given by; (2020)
of free space, A is area of Gaussian surface and qenc is
charge enclosed by the Gaussian surface. This equation
can be used in which of the following situation? (2020)
(a) Only when the Gaussian surface is an equipotential
surface.
(b) Only when E constant on the surface..
3 1
(a) 2 1 y x
0 2 2 (c) Equipotential surface and E is constant on the
surface.
(d) for any choice of Gaussian surfaces.
3 1
1 y x 39. Consider two charged metallic spheres S1and S2 of radii
(b) 2
0 2 2 r1 and r2 respectively. The electric fields E 1(on S1) and
E1 r1
3 1 E2(on S2) on their surfaces are such that
(c) 2 1 2 y 2 x E2 r2 . Then
0 the ratio V1(on S1) /V2(on S2) of the electrostatic potentials
on each sphere is (2020)
3 1
1 y x r1
2
(d)
2 0 2 2 r1
(a) (b)
r2 r2
37. Three charged particles A, B and, C with charge –4q, +2q
and –2q present on the circumference of a circle of radius 3
‘d’. The charges particles A, C and centre O of the circle r2 r1
formed an equilateral triangle as shown in figure. Electric (c) (d)
r1 r2
field at O along x– direction is: (2020)
40. A particle of mass m and charge is released from rest in
uniform electric field. If there is no other force on the
particle, the dependence of its speed v on the distance x
travelled by it is correctly given by (graphs are schematic
and not drawn to scale) (2020)
(a) (b)
(c) (d)
3q 2 3q
(a) (b)
4 0 d 2 0 d 2
3q 3 3q
(c) (d)
0 d 2 4 0 d 2
ELECTROSTATICS 64
41. A solid sphere having a radius R and uniform charge 44. A charged particle (mass m and charge q) moves along X
density . If a sphere of radius R/2 is carved out of it as axis with velocity V0. When it passes through the origin it
shown in the figure. Find the ratio of the magnitude of enters a region having uniform electric field E Ejˆ
electric field at point A and B (2020)
which extend upto x = d. Equation of path of electron in
the region x > d is : (2020)
2 qEd d
17 18 (a) y qEd x (b) y x
(a) (b) mV02 mV02 2
54 54
qEd
18 21 (c) y x d (d) y qEd2 x
(c) (d) mV02 mV0
34 34
45. A small point mass carrying some positive charge on it, is
42. An electric dipole of moment C m at the origin (0,0,0). The
released from the edge of a table. There is a uniform electric
electric field due to this dipole at r iˆ 3 ˆj 5kˆ is field in this region in the horizontal direction. Which of
parallel to the following options then correctly describe the
trajectory of the mass ? (Curves are drawn schematically
[Note that r . p 0 ] p iˆ 3 ˆj 2 kˆ 10 29 (2020) and are not to scale). (2020)
Nm2
2 , then difference between 1 2 is (2020)
C
(c) (d)
ELECTROSTATICS 65
46. A charge is distributed over two concentric conducting 49. Two charged thin infinite plane sheets of uniform surface
thin spherical shells radii r and R (R > r). If the surface
charge densities on the two shells are equal, the electric charge density + and - , where + > - , intersect
potential at the common centre is : (2020) at right angle. Which of the following best represents the
electric field lines for this system: (2020)
(a)
1 (2R r)
(a) Q
4 0 (R 2 r 2 )
1 (R r)
(b) Q
4 0 (R 2 r 2 )
(b)
1 (R r)
(c) Q
4 0 2(R 2 r 2 )
1 (R 2r) Q
(d)
4 0 2(R 2 r 2 )
2
47. Two isolated conducting spheres S1 and S2 of radius R
3
1
and R have 12 C and 3C charges, respectively, and (c)
3
are at a large distance from each other. They are now
connected by a conducting wire. A long time after this is
done the charges on S1 and S2 are respectively : (2020)
(a) 6 C and 3 C (b) 4.5 C on both
3Q 2
(a) 4 R
0
3Q1
(b) 4 R
0
3Q1
(c)
16 0 R
Q2
(d) 4 R
0
ELECTROSTATICS 66
qQ
(c) 2 2y g
4 0 R (R y) m
qQ
(d) 2 2y g
2
4 0 R ym
q2 4q 2 53. Ten charges are placed on the circumference of a circle of
(a) decrease by (b) decrease by
40 d 30 d radius R with constant angular separation between
successive charges. Alternate charges 1, 3, 5, 7, 9 have
3q 2 2q 2 charge (+q) each, while 2, 4, 6, 8, 10 have charge (–q) each.
(c) increase by (d) increase by The potential V and the electric field E at the centreof the
40 d 30 d
circle are respectively. (Take V = 0 at infinity) (2020)
51. A particle of charge and mass m is subjected to an electric
(a) V = 0; E = 0
field E = E0 (1 – ax2) in the x-direction, where a and E0 are
constants. Initially the particle was at rest at x = 0. Other 10q 10q
than the initial position the kinetic energy of the particle (b) V ;E
4 0 R 4 0 R 2
becomes zero when the distance of the particle from the
origin is: (2020)
10q
(c) V 0; E
4 0 R 2
2
(a) (b) a
a 10q
(d) V 4 R ; E 0
0
3 1
(c) (d) 54. Charges Q1 and Q2 are at points A and B of a right angle
a a triangle OAB (see figure). The resultant electric field at
52. A solid sphere of radius R carries a charge +Q distributed point O is perpendicular to the hypotenuse, then Q1/Q2is
uniformly over its volume. A very small point like piece of proportional to: (2020)
it of mass m gets detached from the bottom of the sphere
and falls down vertically under gravity. This piece carries
charge q. If it acquires a speed when it has fallen through
a vertical height y (see figure), then : (assume the remaining
portion to be spherical). (2020)
x2 x22
(a) (b) 2
x1 x1
x13 x1
(c) 3 (d)
x2 x2
ELECTROSTATICS 67
55. Consider the force of on a charge ‘q’ due to a uniformly 58. A point charge of +12 C is at distance 6 cm vertically
charged spherical shell of radius R carrying charge
above the centre of a square of idea 12 cm as shown in
distributed uniformly over it. Which one of the following
figure. The magnitude of th elctric flux throught the square
statements is true for F, if ‘q’ is placed at distance r from
will be _____ ×103 Nm2/C. (2021)
the centre of the shell? (2020)
1 qQ
(a) F 0 for r R
4 0 R 2
(b) F
1 qQ
for r R
4 0 r 2
(c) F 1 qQ
for all r
4 0 r 2
59. 512 identical drops of mercury are charged to a potential of
1 qQ 2V each. The drops are joined to from a single drop. The
(d) F for r R
4 0 R 2 potential of this drop is ____ V. (2021)
60. Two identical conducting spheres with negligible volume
56. Two identical electric point dipoles have dipole moments
have 2.1nC and –0.1 nC charges, respectively. They are
p1 pi and p2 pi and are held on the x axis at brought into contact and then separated by a distance of
distance ‘a’ from each other. When released, they move 0.5 m. The electrostatie force acting between the spheres
along the x-axis with the direction of their dipole moments is ______ ×10–9N.
remaining unchanged. If the mass of each dipole is ‘m’,
their speed when they are infinitely far apart is: (2020) 1
Given : 4 0 9 109 SI unit (2021)
p 3 p 1
(a) (b) 61. Two small spheres each of mass 10 mg are suspended
a 2 0 ma a 0 ma from a point by threads 0.5m long. They are equally
charged and repel each other to a distance of 0.20 m. The
p 1 p 2 a
(c) a 2 ma (d) charge on each of the sphere is 10( 8) C . The value of
0 a 0 ma 21
‘ a‘ will be [Given g = 10 ms–2] (2021)
57.
An electric field E 4 xiˆ y 2 1 ˆj N / C , passes 62. 27 similar drops of mercury are maintained at 10V each. All
through the box shown in figure. The flux of the electric these spherical drops combine into a single big drop. The
potential energy of the bigger drop is how many times
field through surface ABCD and BCEF are marked as 1
that of a smaller drop. (2021)
Nm 2 63. Two electrons each are fixed at a distance ‘2d’. Proton
and 2 , then difference between 1 2 is placed at the midpoint is displaced slightly by a distance x
C (x < < d) perpendicular to the line joining the two fixed
(2020) charges. Proton will execute simple harmonic motion hav-
ing angular frequency : (m = mass of charged particle)
(2021)
1 1
q2 2 20 md 3 2
(a) 3 (c)
20 md
2
q
1 1
2q 2 2 0 md 3 2
(c) 3 (d)
0 md
2
2q
ELECTROSTATICS 68
64. Find the electric field at point P (as shown in figure) on the 69. A charge ‘q’ is placed at one corner of a cube as shown in
perpendicualr bisector of a uniformly charged thin wire of
length L carrying charge Q. The distance of the point P figure. The flux of electrstatic field E through the shaded
area is : (2021)
3
from the centre of the rod is a L (2021)
2
q q
(a) 24 (b) 4
Q 3Q 0 0
A. 4 L2 B.
0 40 L2
q q
(c) 48 (d) 8
Q Q 0 0
C. 3 L2 D.
0 2 3 0 L2 70. Given below are two statements:
65. Find out the surface charge density at the intersection of Statement I : An electric dipole is placed at the centre of a
point x = 3 plane and x-axis, in the region of uniform line hollow sphere. The flux of electric field through the sphere
charge of 8nC/m lying along the z-axis in free space. is zero but the electric field is not zero anywhere in the
sphere.
(2021)
Statement II: If R is the radius of a solid metallic sphere
(a) 0.07 nCm–2 (b) 47.88 C/m
and Q be the total charge on it.
(c) 0.424 nC m–2 (d) 4.0 nC m–2
The electric field at any point on the spherical surface of
66. An oil drop of radius 2 mm with a density 3 g cm–3 is held radius r in the light of the above statement, choose the
stationary under a constant electric field 3.55 × 105 V m–1 in correct answer from the options give below
the Milikan’s oil drop experiment. What is the number of
(2021)
excess electrons that the oil drop will possess?
(a) Bot statement I and Statement II are false
Consider g = 9.81 m/s2 (2021)
(b) Statement I is false but Statement II is true
(a) 48.8 × 1011 (b) 1.73 × 1012
(c) Statement I is ture but Statement II is false
(c) 1.73 5 1010 (d) 17.3 × 1010
(d) Both Statement I and Statement II are true
67. The electric field in a region is given by
71. Find out the surface charge density at the intersection of
3 4 point x = 3 plane and x-axis, in the region of uniform line
E E 0 i E 0 j N / C . The ratio of flux of reported
5 5 charge of 8nC/m lying along the z-axis in free space.
field reactangular surface of area 0.2 m2 ( parallel to y - z (2021)
plane) to that of the surface of area 0.3 m2 (parallel to x -z (a) 0.07 nCm –2
(b) 47.88 C/m
plane) is a : b, where a = ______. (c) 0.424 nC m–2 (d) 4.0 nC m–2
68. The electric field ina region is given by E 2 E 0 i 3 E 0 j
3 5
N
with E 0 4.0 103 . The flux of this field through a
C
reactangular surface area 0.4 m2 parallel to the Y - Z plane
is _____Nm2C–1 (2021)
ELECTROSTATICS 69
q In2 1 4Q2 1 2 Q2
(c) (d) 4 x (a) (b)
4 0 m 2 4 0 m 2
0 0
11. An electric dipole is situated in an electric field of uniform 16. A particle of specific charge (q/m) enters into uniform
intensity E whose dipole moment is p and moment of inertia electric field E along the centre line, with velocity
is I. If the dipole is displaced slightly from the equilibrium
position, then the angular frequency of its oscillations is qE
v 2q . After how much time it will collide with one
md
1/ 2 3/ 2
pE pE of the plates (figure)
(a) (b)
I I
1/ 2 1/ 2
I p
(c) (d)
pE IE
12. Two identical point charges are placed at a separation of
d. P is a point on the line joining the charges, at a distance
x from any one charge. The field at P is E, E is plotted
against x for values of x from close to zero to slightly less
than d. Which of the following represents the resulting
curve d
(a) (b) (a) Not possible (b) 2 V
md 2 md
(c) (d)
qE qE
17. A wire is bent in the form of a regular hexagon of side a
(c) (d) and a total charge is distributed uniformly over it. One
side of the hexagon is removed. The electric field due to
the remaining sides at the centre of the hexagon is
Q Q
(a) 2 (b)
12 3 0 a 16 3 0 a 2
13. A metallic solid sphere is placed in a uniform electric fied.
The lines of force follow the path (s) shown in figure as Q Q
(c) (d)
8 2 0 a 2 8 2 0 a 2
18. In the figure shown, if the linear charge density is , then the
net electric field at O will be
(a) 1 (b) 2
(c) 3 (d) 4
14. Two equal charges are separated by a distance d. A third
charge placed on a perpendicular bisector at x distance
will experience maximum coulomb force when k
(a) Zero (b)
d d R
(a) x (b) x
2 2
2 k 2 k
d d (c) (d)
R R
(c) x (d) x
2 2 2 3 19. An electron moves round a circular path of radius 0.1 m
15. If an electron has an initial velocity in a direction different about an infinite linear charge of density +1 C/m. The
from that of an uniform electric field, the path of the electron speed of the electron will be
is (a) 5.6 × 103 m/s (b) 2.8 × 105 m/s
(a) A straight line (b) A circle
(c) 5.6 × 107 m/s (d) 2.8 × 107 m/s
(c) An ellipse (d) A parabola
ELECTROSTATICS 71
20. An electron having charge e and mass m starts from lower 27. Figure shows two concentric, conducting shells of radii r
plate of two metallic plates separated by a distance d, if and 2r. The outer shell is given a charge Q. The amount of
the potential difference between the plates is V, the time charge that will appear on outer surface of inner shell if
taken by the electron to reach the upper plate is given by inner shell is grounded
(ignore gravity)
2md 2 md 2
(a) (b)
eV eV
md 2 2md 2
(c) (d)
2eV eV
21. A particle of mass ‘m’ and charge ‘q’ is accelerated through
a potential difference of V volt, its energy will be Q Q
(a) (b)
(a) qV (b) mqV 2 2
(c) – 2Q (d) + 2Q
q q
(c) V (d) Comprehensive Type Questions
m mV
Passage -1
22. A ring of charge with radius of 50 cm has gap of 0.002 m.
Using the following passage, solve Q. 28
If the ring carries a charge of 1 C. What is the electric field Consider an evacuated cylindrical chamber of height h
at the centre. having rigid conducting plates at the ends and an
(a) 8.5 × 107 N/C (b) 7.2 × 107 N/C insulating curved surface as shown in the figure. A number
(c) 3.2 × 10 N/C
7
(d) 4.5 × 107 N/C of spherical balls made of a light weight and soft material
23. A thin conducting ring of radius r has an electric charge + Q, and coated with a conducting material are placed on the
if a point charge is placed at the centre of the ring, then bottom plate. The balls have a radius r << h . Now a high
tension of the wire of ring will be voltage source (HV) is connected across the conducting
plates such that the bottom plate is at + V0 and the top
Qq Qq plate at –V0. Due to their conducting surface, the balls
(a) 2 (b)
80 r 4 0 r 2 will get charged, will become equipotential with the plate
and are repelled by it. The balls will eventually collide
Qq Qq with the top plate, where the coefficient of restitution can
(c) 2 2 (d) be taken to be zero due to the soft nature of the material
8 0 r 4 2 0 r 2
of the balls. The electric field in the chamber can be
24. n small drops of same size are charged to V volt each. If considered to be that of a parallel plate capacitor. Assume
they coalesce to form a single large drop, then its potential that there are no collisions between the balls and the
will be interaction between them is negligible. (Ignore gravity)
–1
(a) Vn (b) V n
1/3 2/3
(c) V n (d) V n
25. A solid conducting sphere having a charge is surrounded
by an uncharged concentric conducting hollow spherical
shell. Let the potential difference between the surface of
the solid sphere and that of the outer surface of the hollow
shell be V. If the shell is now given a change of –3Q, the
new potential difference between the same two surfaces
is (a) V (b) 2V 28. Two identical thin rings, each of radius R metres, are
(c) 4V (d) –2V coaxially placed a distance R metres apart. If Q1 coul, and
26. A hollow sphere of radius 2R is charged to V volt and Q2 coul, are respectively the charges uniformly spread on
the two rings, the work done in moving a charge from the
another small sphere of radius R is charged to V/2 volt.
centre of one ring to that of the other is
Then the smaller sphere is placed inside the bigger sphere
without changing the net charge on each sphere. The (a) zero (b) q(Q1 Q 2 ) ( 2 1)
potential difference between the two spheres would be (4 20 R)
(a) 3V/2 (b) V/4
(c) V/2 (d) V (c) q 2(Q1 Q 2 ) (d) q(Q1 Q 2 ) ( 2 1)
(40 R) (4 2 0 R)
ELECTROSTATICS 72
–3C 4C 5C
Subjective Questions
38. Three particles, each of mass 1 gm and carrying a charge
q, are suspended from a common point by insulated
massless strings, each 100 cm long. If the particles are in
equilibrium and are located at the corners of an equilateral
triangle of side length 3 cm, calculate the charge on each
2
particle. (Take g = 10 m/s ).
39. Two fixed, equal, positive charges, each of magnitude
–5
A B C 5 × 10 coulomb are located at points A and B separated
by a distance of 6 m. An equal and opposite charge moves
Then answer the following questions towards them along the line COD, the perpendicular
29. The charge that appears on the left surface of plate B is bisector of the line AB.
(a) –3C (b) 3C
(c) 6C (d) 5C
30. The charge on the inner surface of plate C if the plate B is
earthed
(a) –3C (b) 3C
(c) 6C (d) 5C
31. The charge on left surface of B if B and C are both earthed
(a) –3C (b) 3C
(c) 6C (d) 5C The moving charge, when it reaches the point C at a
True/False distance of 4m from O, has a kinetic energy of 4 joules.
Calculate the distance of the farthest point D which the
32. A small metal ball is suspended in a uniform electric field negative charge will reach before returning towards C.
with the help of an insulated thread. If high energy X- ray
40. A thin fixed ring of radius 1 metre has a positive charge
beam falls on the ball, the ball will be deflected in the –5
direction of the field. 1 × 10 coulomb uniformly distributed over it. A particle of
–6
33. A ring of radius R carries uniformly distributed charge + Q. mass 0. 9 gm and having a negative charge of 1 × 10
A point charge – is placed on the axis of the ring at a coulomb is placed on the axis at a distance of 1 cm from
distance 2R from the centre of the ring and released from the centre of the ring. Show that the motion of the
rest. The particle executes a simple harmonic motion along negatively charged particle is approximately simple
the axis of the ring. harmonic. Calculate the time period of oscillations.
Integer Type Questions 41. A non-conducting disc of radius a and uniform positive
surface charge density is placed on the ground with its
34. A solid sphere of radius R has a charge distributed in its
axis vertical. A particle of mass m and positive charge is
volume with charge density = kra, where k and a are
constant and r is the distance from its centre. If the electric dropped, along the axis of the disc from a height H with zero
field are at r = R/2 is 1/8 times that at r = R. What is the initial velocity. The particle has q/m = 40g/.
value of a. (a) Find the value of H if the particle just reaches the disc.
35. Two point charges 4C and –10C are placed 10 cm apart (b) Sketch the potential energy of the particle as a function
in air. An dielectric slab of large length and breadth but of of its height and find its equilibrium position.
thickness 5 cm is placed between them. Calculate the force
(in newton) of attraction between the charges if the relative 42. Three concentric spherical metallic shells, A, B and C of
permittivity of the dielectric is 9. radii a, b and c (a < b < c) have surface charge densities
36. The linear charge density on a dielectric ring of radius R is , – and respectively.
43. (a) A charge of is uniformly distributed over a spherical 49. A charge is placed at the centre of the line joining two
volume of radius R. Obtain an expression for the energy of equal charges Q. The system of the three charges will be
the system. in equilibrium if is equal to :
(b) What will be the corresponding expression for the (a) – Q/2 (b) – Q/4
energy needed to completely disassemble the planet earth (c) + Q/4 (d) + Q/2
against the gravitational pull amongst its constituent 50. A conducting sphere S1 of radius r is attached to an insulating
particles ? handle. Another conducting sphere S2 of radius R is
Assume the earth to be a sphere of uniform mass density. mounted on an insulating stand, S2 is initially uncharged.
Calculate this energy, given the product of the mass and S1 is given a charge brought into contact with S2 and
31
the radius of the earth to be 2.5 × 10 kg-m. removed. S1 is recharged such that the charge on it is
(c) If the same charge as in part (a) above is given to a again and it is again brought into contact with S2 and
spherical conductor of the same radius R, what will be the removed. This procedure is repeated n times.
energy of the system ? (a) Find the electrostatic energy of S2 after n such contacts
44. A circular ring of radius R with uniform positive charge with S1.
density per unit length is located in the y-z plane with its (b) What is the limiting value of this energy as n ?
centre at the origin O. A particle of mass m and positive
Multiple Choice Questions
charge is projected from the point P (R 3, 0, 0) on the 51. A small sphere of mass m and having charge is suspended
positive x-axis directly towards O, with an initial speed v. by a light thread
Find the smallest (non-zero) value of the speed v such (a) Tension in thread must reduce if another charged
that the particle does not return to P. sphere is placed vertically below it
45. A charge is distributed over two concentric hollow spheres (b) Tension in thread is greater that weight mg if another
of radii r and R (> r) such that the surface densities are equal. charged sphere is held in same horizontal line in which
Find the potential at the common centre. first sphere stays in equilibrium
46. Two fixed charges – 2and are located at the points with (c) Tension in thread is always equal to weight mg
coordinates (–3a, 0) and (+3a, 0) respectively in the x-y (d) Tension may increase to double its original value if
plane. another charge is below it
(a) Show that all points in the x-y plane where the electric 52. The correct options is(are)
potential due to the two charges is zero, lie on a circle Find
(a) charge cannot exist without mass but mass can exist
its radius and the location of its centre.
without charge
(b) Give the expression V (x) at a general point on the x-axis
(b) charge is conserved but mass is not conserved
and sketch the function V (x) on the whole x-axis.
(c) charge is independent of state of rest or motion
(c) If a particle of charge +starts form rest at the centre of
(d) mass is independent of state of rest or motion
the circle, show by a short quantative argument that the
particle eventually crosses the circle. Find its speed when 53. A deuteron and an alpha particle are placed in same
it does so. uniform electric field .The forces acting on them are F1
and F2 and their acceleration are a1 and a2 respectively.
47. Three point charges q, 2and 8are to be placed on a 9 cm
long straight line. Find the positions where the charges (a) F1 = F2 (b) a1 = a2
should be placed such that the potential energy of this (c) F1 F2 (d) a1 a2
system is minimum. In this situation, what is the electric 54. The electric field in a region is directed outward and is
field at the position of the charge due to the other two proportional to the distance r from the origin. Taking
charges ? electric potential to be zero at origion
48. Two isolated metallic solid spheres of radii R and 2R are (a) it is uniform in the region
charged such that both of these have same charge density (b) it is propotional to r
. The spheres are located far away from each other and (c) it is proportional to r2
connected by a thin conducting wire. Find the new charge
(d) it decreases as one goes away from origin
density on the bigger sphere.
ELECTROSTATICS 74
55. Ring with uniform charge and radius R is placed in y–z 61. In a uniformly charge dielectric sphere a very thin tunnel
plane with its centre at origin. Then has been made along the diameter shown in figure. A
(a) the electric field at origin is maximum particle with charge –having mass m is released from rest
at one end of the tunnel for the situation described, mark
kQ out the correct statements
(b) The potential at origin is
R
(a) Charged particle will perform SHM about the centre of
kQ sphere as mean position
(c) the field at the point (x, 0, 0) is
R 2 X2
20 mR 3
(d) maximum value of electric field on the axis will be (b) The time period of the particle is 2
Qq
q (c) speed of the particle crossing the mean position is
6 3 0 R 2
Qq
56. Two metallic spheres have same radii. One of them is solid
40Rm
and other is hollow. They are charged to same potential.
The charge on former is q1 and on latter is q2. We have (d) Particle will perform oscillations but not SHM
(a) q1 = q2
62. An electric charge (= 20 × 10–9C) is placed at a point (1, 2, 4).
(b) Electric field inside both of them is zero. At the point (3, 2, 1) the electric
(c) Electrostatic potential in both the spheres at an inside
(a) Field will increase by a factor K if the space between
point is same as on surface
the points is filled with a dielectric of dielectric constant
(d) Charge in both is effectively concentrated at the centre K
for field strength at an external point.
(b) Field will be along y axis
57. A spherical charged conductor has surface charge
density. The electric field on its surface is E and electric (c) Potential will be approximately 50 V volt
potential of the conductor is V. Now the radius of the (d) Field will have no y component.
sphere is halved keeping the charge to be constant.The
63. In the circuit shown in steady state
new value of electric field and potential would be
(a) 4E (b) 2V
(c) 2E (d) 4V
58. In uniform electric field equipotential surfaces must
(a) be plane surfaces
(b) be normal to the direction of the field
(c) be spaced that surfaces having equal difference in
potential are equally spaced
(d) have decreasing potential along field
59. When the separation between two charges is increased (a) charge across 4F capacitor is 20 C
(a) electric potential energy increases (b) charge across 4F capacitor is 10 C
(b) electric potential energy decreases
(c) potential difference across 4F capacitor is 5 Volt
(c) force between them decreases
(d) potential difference across 4F capacitor is 10 Volt
(d) electric potential energy may increase or decreases
64. Two conducting spheres of unequal radii are given
60. Two point charges 2 and 8 are placed at distance d apart.
charges such that they have the same charge density. If
A third charge –is placed at distance d/3 from 2on line
they are brought in contact
joining the charges 2and 8q. Then
(a) Electric potential energy of the system is maximum (a) some heat will be produced
(b) Electric potential energy of system is minimum (b) charge will flow from larger to smaller sphere
(c) charge –is in unstable equilibrium (c) charge will flow from smaller to larger sphere
(d) charge –is in stable equilibrium (d) no charge will be exchanged between the spheres
ELECTROSTATICS 75
Match the Column Type Questions 68. Five point charges, each of value + coul, are placed on
65. Different shaped charged bodies and their corresponding five vertices of a regular hexagon of side L metres. The
electric fields are mentioned Match the Column 1 with magnitude of the force on the point charge of value –
Column 2. coul. placed at the centre of the hexagen is ............ newton.
Column-1 Column-2
(a) Spherical charged (P) At centre electric field
conductor is zero
(b) Infinite plane sheet of (Q) Electric field is uniform
charge
(c) Uniformly charged (R) Electric field is
ring discontiunous at thesurface
(d) Sphere with uniform (S) At the surface electric
volume distribution of field is continous and
charge maximum.
Fill in the blanks 69. An infinite number of electric charges each equal to 5
nano-coulomb (magnitude) are placed along X-axis at
66. A point charge moves from point P to point S along the x = 1cm, x = 2 cm, x = 4 cm, x = 8 cm ………. and so on. In
path PQRS (fig.) in a uniform electric field E pointing parallel the setup if the consecutive charges have opposite sign,
to the positive direction of the x-axis.The coordinates of then the electric field in Newton/Coulomb at x = 0 is
the points P, Q, R and S are (a, b, O), (2a, O, O) (a, –b, O)
and (O, O, O) respectively. The work done by the field in 1
the above process is given by the expression.......... 9 109 N m 2 / c2
4 0
(a) 12 × 104 (b) 24 × 104
(c) 36 × 104 (d) 48 × 104
70. Two small balls having equal positive charges (coulomb)
on each are suspended by two insulating strings of equal
length L (metre) from a hook fixed to a stand. The whole
set up is taken in a satellite into space where there is no
gravity (state of weightlessness). The angle between the
two strings is ............ and the tension in each string is
.......... newtons.
67. The electric potential V at any point x, y, z (all in metres) in
2
space is given by V = 4x volts. The electric field at the
point (1m, 0,2m) is ................ V/m.
ELECTROSTATICS 76
(c) (d)
(c) (d)
7. A positive point charge is fixed at origin. A dipole with a
dipole moment p is placed along the x-axis far away from
the origin with p pointing along positive x-axis. Find : (a)
the kinetic energy of the dipole when it reaches a distance d
3. A uniform electric field pointing in positve x-direction exists from the origin, and (b) the force experienced by the charge
in a region. Let A be the origin, B be the point on the x-axis at this moment. (2003)
at x = +1 cm and C be the point on the y-axis at y = +1 cm. 8. Consider the charge configuration and a spherical Gaussian
Then the potentials at the points A, B and C satisfy : surface as shown in the figure. When calculating the flux of
(2001) the electric field over the spherical surface, the electric field
(a) VA < VB (b) VA > VB will be due to (2004)
(c) VA < VC (d) VA > VC
–3
4. A small ball of mass 2 × 10 kg having a charge of 1 C is
suspended by a string of length 0.8 m. Another identical
ball having the same charge is kept at the point of
suspension. Determine the minimum horizontal velocity
which should be imparted to the lower ball, so that it can
make complete revolution. (2001) (a) q2 (b) only the positive charges
(c) all the charges (d) +q1 and –q1
ELECTROSTATICS 77
9. There are two large parallel metallic plates S1 and S2 carrying Then which option(s) are correct
surface charge densities 1 and 2 respectively (1 > 2) (a) Total charge within 2R0 is q
placed at a distance d apart in vacuum. Find the work done (b) Total electrostatic energy for r R0 is zero
(c) At r = R0 electric field is discontinuous
by the electric field in moving a point charge a distance a (a
(d) There will be no charge anywhere except at r = R0
< d) from S1 towards S2 along a line making an angle /4 with
14. Positive and negative point charges of equal magnitude are
the normal to the plates. (2004)
a
kept at 0,0, and 0,0, respectively. The work done
a
10. Six charges of equal magnitude, 3 positive and 3 negative
2 2
are to be placed on PQRSTU corners of a regular hexagon,
such that field at the centre is double that of what it would by the electric field when another positive point charge is
moved from (–a, 0, 0) to (0, a, 0) is (2007)
have been if only one +ve charge is placed at R.
(a) positive
(2004) (b) negative
(c) zero
(d) depends on the path connecting the initial and final
positions
15. Consider a neutral conducting sphere. A positive point
charge is placed outside the sphere. The net charge on the
sphere is then (2007)
(a) +, +, +, –, –, – (b) –, +, +, +, –, – (a) negative and distributed uniformly over the surface of
the sphere
(c) –, +, +, –, +, – (d) +, –, +, –, +, –
(b) negative and appears only at the point on the sphere
11. A conducting bubble of radius a, thickness t (t << a) has closest to the point charge
potential V. Now the bubble collapses into a droplet. Find (c) negative and distributed non-uniformly over the entire
the potential of the droplet. surface of the sphere
(2005) (d) zero
16. A spherical portion has been removed from a solid sphere
12. Three infinitely long charge sheets are placed as shown in
having a charge distributed uniformly in its volume as shown
figure. The electric field at point P is (2005) in the figure. The electric field inside the emptied space is
( 2007)
2 2
(a) k̂ (b) k̂
0 0
q q 2q
19. Consider a system of three charges , and placed
3 3 3
at points A, B and C respectively, as shown in the figure. 21. The electric field at r = R is (2008)
Take O to be the centre of the circle of radius R and angle (a) independent of a
CAB = 60° (2008) (b) directly proportional to a
(c) directly proportional to a2
(d) inversely proportional to a
22. For a = 0, the value of d (maximum value of as shown in the
figure) is (2008)
3Ze 3Ze
(a) (b)
4R 3 R 3
4Ze Ze
(c) (d)
q 3R 3 3R 3
(a) The electric field at point O is directed along
80 R 2 23. The electric field within the nucleus is generally observed
the negative x–axis to be linearly dependent on r. This implies
(b) The potential energy of the system is zero (2008)
(c) The magnitude of the force between the charges at C R
(a) a = 0 (b) a
q 2 2
and B is
54 0 R 2 2R
(c) a = R (d) a
3
q
(d) The potential at point O is 24. Three concentric metallic spherical shells of radii R, 2R and
12 0 R
3R are given charges Q1, Q2 and Q3, respectively. It is found
20. Statement I : For practical purposes, the earth is used as a that the surface charge densities on the outer surfaces of
reference at zero potential in electrical circuits. the shells are equal. Then, the ratio of the charges given to
Statement II : The electrical potential of a sphere of radius R the shells, Q1 : Q2 : Q3, is (2009)
with charge uniformly distributed on the surface is given by (a) 1 : 2 : 3 (b) 1 : 3 : 5
Q (c) 1 : 4 : 9 (d) 1 : 8 : 18
40 R . (2008) 25. Under the influence of the coulomb field of charge +Q, a
charge – is moving around it in an elliptical orbit. Find out
(a) Statement–I is true, Statement–II is true; Statement–II is the correct statement(s). (2009)
the correct explanation for Statement–I. (a) The angular momentum of the charge – is constant
(b) If Statement–I is true, Statement–II is true; Statement–II is (b) The linear momentum of the charge –is constant
not a correct explanation for Statement–I.
(c) The angular velocity of the charge –is constant
(c) If Statement–I is true; Statement–II is false.
(d) The linear speed of the charge – is constant
(d) If Statement–I is false; Statement–II is true.
ELECTROSTATICS 79
26. Three concentric metallic spherical shells of radii R, 2R and (a) |Q1| > |Q2|
3R are given charges Q1, Q2 and Q3, respectively. It is found (b) |Q1| < |Q2|
that the surface charge densities on the outer surfaces of
the shells are equal. Then, the ratio of the charges given to (c) at a finite distance to the left of Q1 the electric field is zero
the shells, Q1 : Q2 : Q3, is (2009)
(a) 1 : 2 : 3 (b) 1 : 3 : 5 (d) at a finite distance to the right of Q2 the electric field is
(c) 1 : 4 : 9 (d) 1 : 8 : 18 zero
29. A tiny spherical oil drop carrying a net charge is balanced in
a
27. A disk of radius having a uniformly distributed charge 6C still air with a vertical uniform electric field of strength
4
81
a 105 Vm 1 . When the field is switched off, the drop is
is placed in the x–y plane with its centre at , 0, 0 . A 7
2 observed to fall with terminal velocity 2×10–3 ms–1. Given
rod of length a carrying a uniformly distributed charge 8C is g=9.8 ms–2, viscosity of the air = 1.8×10–5 Ns m–2 and the
a 5a density of oil = 900 kg m–3, the magnitude of charge is
placed on the x–axis from x to x . Two point
4 4 (2010)
(a) 1.6×10 C–19
(b) 3.2×10 C –19
1 2
(a) 1 2 R 2 (b) R
0 0
1 2
(c) 1
2
(d)
0 R 0 R 2
2C 2C 31. A spherical metal shell A of radius RA and a solid metal sphere
(a) (b)
0 0 B of radius RB (<RA) are kept far apart and each is given
charge +Q. Now they are connected by a thin metal wire.
10C 12C Then (2011)
(c) (d)
0 0
(a) Einside
A 0 (b) QA > QB
28. A few electric field lines for a system of two charges Q1 and
Q2 fixed at two different points on the x–axis are shown in A R B
the figure. These lines suggest that (c) R (d) E on
A
surface
E on
B
surface
B A
(2010)
32. Two large vertical and parallel metal plates having a
separation of 1 cm are connected to a DC voltage source of
potential difference X. A proton is released at rest midway
between the two plates. It is found to move at 45° to the
vertical just after release. Then X is nearly
(a) 1 × 10–5 V (b) 1 × 10–7 V
(c) 1 × 10–9 V (d) 1 × 10–10 V
ELECTROSTATICS 80
33. A spherical metal shell A of radius RA and a solid metal 36. A cubical region of side a has its centre at the origin. It
sphere B of radius RB (<RA) are kept far apart and each is encloses three fixed point charges, –at (0, –a/4, 0), +3at (0,
given charge +Q. Now they are connected by a thin metal 0, 0) and –at (0, +a/4, 0). Choose the correct option(s).
wire. Then (2011) (2012)
(a) Einside
A 0
(b) QA > QB
A R B
(c) R
B A
(2013)
2
E0 a 32
(c) E0a2 (d) (a) –4 (b)
2 25
32
(c) (d) 4
25
ELECTROSTATICS 81
39. Two non-conducting sphere of radii R1 and R2 and carrying 42. Four charges Q1, Q2, Q3 and Q4 of same magnitude are fixed
uniform volume charge densities + and –, respectively, along the x axis at x = – 2a, –a, +a and +2a, respectively. A
are placed such that they partially overlap, as shown in the positive charge is placed on the positive y axis at a distance
figure. At all points in the overlapping region. b > 0. Four options of the signs of these charges are given
in List I. The direction of the forces on the charge is given in
(2013)
List II. Match List I with List II and select the correct answer
using the code given below the lists. (2014)
(d) the electrostatic field has same direction Q. Q1, Q2 positive; Q3, Q4 negative 2. –x
R. Q1, Q4 positive; Q2, Q3 negative 3. +y
40. Let E1 (r), E2 (r) and E3 (r) be the respective electric fields at
a distance r from a point charge Q, an infinitely long wire S. Q1, Q3 positive; Q2, Q4 negative 4. –y
with constant linear charge density and an infinite plane Codes :
with uniform surface charge density . If E1 (r0) = E2 (r0) = E3
(a) P -3, Q-1, R-4, S-2 (b) P-4, Q-2, R-3, S-1
(r0) at a given distance r0, then (2014)
(c) P-3, Q-1, R-2, S-4 (d) P-4, Q-2, R-1, S-3
(a) Q 4r02 43. The figure below depict two situations in which two
infinitely long static line charges of constant positive line
charge density are kept parallel to each other. In their
(b) r0 resulting electric field, point charges and –are kept in
2
equilibrium between them. The point charges are confined
(c) E1(r0/2) = 2 E2 (r0/2) to move in the x-direction only. If they are given a small
displacement about their equilibrium positions, then the
(d) E2 (r0/2) = 4E3 (r0/2) correct statement(s) is (are)
41. Charges Q, 2Qand 4Q are uniformly distributed in three (2015)
dielectric solid spheres 1, 2 and 3 of radii R/2, R and 2R
respectively, as shown in figure. If magnitudes of the electric
fields at point P at a distance R from the centre of spheres 1,
2 and 3 are E1, E2 and E3 respectively, then
(2014)
(a) E1 > E2 > E3 (c) Charge +executes simple harmonic motion while charge
continues moving in the direction of its displacement.
(b) E3 > E1 > E2
(d) Charge –executes simple harmonic motion while charge
(c) E2 > E1 > E3 +continues moving in the direction of its displacement.
(d) E3 > E2 > E1
ELECTROSTATICS 82
44. Consider a uniform spherical charge distribution of radius 47. A length-scale (l) depends on the permittivity () of a
R1 centred at the origin O. In this distribution, a spherical dielectric material, Boltzmann constant (kB), the absolute
cavity of radius R2, Centred at P with distance OP = a = R1 – R2 temperature (T), the number per unit volume (n) of certain
(see figure) is made. If the electric field inside the cavity at charged particles, and the charge (q) carried by each of the
position r is E r , then the correct statement(s) is (are) particles. Which of the following expression(s) for l is(are)
dimensionally correct? (2016)
(2015)
nq 2 k B T
(a) l (b) l 2
k B T nq
q2 q2
n 2/3 k T
n1/3 k T
(c) l (d) l
B B
(a) E is uniform, its magnitude is independent of R2 but its
direction depend on r 48. The electric field E is measured at a point P(0, 0, d) generated
(b) E is uniform, its magnitude depends on R2 and its due to various charge distributions and the dependence of
direction depend on E on d is found to be different for different charge
distributions. List-I contains different relations between E
(c) E is uniform, its r magnitude is independent of a but
and d. List-II describes different electric charge distributions,
its direction depend on a along with their locations. Match the functions in List-I
with the related charge distributions in List-II.
(d) E is uniform and both its magnitude and direction
(2018)
depend on a
List-I List-II
45. An infinitely long uniform line charge distribution of charge
per unit length lies parallel to the y-axis in the y-z plane at (P) E is independent of d 1. A point charge at the origin
3 1
z= a (see figure.) If the magnitude of the flux of the electric (Q) E 2. A small dipole with point
2 d
field through the rectangular surface ABCD lying in the xy charges at (0, 0, l) and –at
L (0, 0, –l). Take 2l <<d
plane with its centre at the origin is
n0 (0 = permittivity of 1
free space), then the value of n is (2015) (R) E 3. An infinite line charge
d2
coincident with the x-axis
with uniform linear charge
density .
1
(S) E 4. Two infinite wires carrying
d3
uniform linear charge density
parallel to the x-axis. The one
46. Which one of the following statements is correct?
along (y = 0, z = l) has a
(2016)
charge density + and and
(a) The balls will execute simple harmonic motion between the one along (y = 0, z = -l)
the two plates has a charge density –.
(b) The balls will bounce back to the bottom plate carrying Take 2l<<d.
the opposite charge they went up with 5.Infinite plane charge
(c) The balls will bounce back to the bottom plate carrying coincident with the xy-plane
the same charge they went up with with uniform surface charge
(d) The balls will stick to the top plate and remain there
density.
ELECTROSTATICS 83
(a) The electric flux through the shell is 3R / 0 (a) The magnitude of total electric field on any two points of
the circle will be same.
(b) The z-component of the electric field is zero at all the
points on the surface of the shell
(b) Total electric field at point A is E A 2E0 i j
(c) The electric flux through the shell is 2R / 0 1/3
p0
(d) The electric field is normal to the surface of the shell at (c) R=
all points 4πε 0 E 0
50. A thin spherical insulating shell of radius R carries a
(d) Total electric field at point B is EB 0
uniformly distributed charge such that the potential at its
surface is V0. A hole with small area 4 R 2 ( 1) is 52. A charged shell of radius R carries a total charge Q. Given
made on the shell without affecting the rest of the shell. as the flux of electric field through a closed cylindrical
Which one of the following statement is correct? surface of height h, radius r and with its centre same as that
(2019) of the shell. Here, centre of the cylinder is a point on the
axis of the cylinder which is equidistant from its top and
(a) The ratio of the potential at the center of the shell to that bottom surfaces. Which of the following option(s) is/are
1
of the point at R from center towards the hole will be correct? [ 0 is the permittivity of free space]
2
1 (2019)
1 2
(a) If h 2R and r
4R Q
(b) The potential at the centre of shell is reduced by 2V0 . then
5 50
(c) The magnitude of electric field at the center of the shell
V0 3R Q
is reduced by (b) If h 2R and r then
2R 5 5 0
(d) The magnitude of electric field at a point, located on a
line passing through the hole and shell’s center, on a 8R 3R
(c) If h and r then 0
distance 2R from the center of the spherical shell will be 5 5
V0
reduced by Q
2R (d) If h 2R and r R then
0
ELECTROSTATICS 84
Find Answer key and Detailed Solutions at the end of this book
ELECTROSTATICS
CAPACITANCE
SCAN CODE
CAPACITANCE
87
CAPACITANCE
Chapter 02
CAPACITANCE
1. CAPACITANCE
2.2 Symbol
1.1 Definition The symbols of capacitor are shown below
We know that charge given to a conductor increases it’s
potential i.e., Q V Q = CV or
Smaller S.I. units are mF, F, nF and pF Capacitance of a capacitor is constant for the given
dimensions & medium.
1mF 10 3
F, 1F 10 F, 1nF 10 F, 1pF 10 F
6 9 12
NOTE:
In charging an uncharged capacitor by a battery half the
energy supplied is stored in the capacitor and remaining
Fig. 2.1 half energy (1/2 QV) is lost in the form of heat.
SCAN CODE
CAPACITANCE
88
CAPACITANCE
0 A ab 20
Capacitance: C Capacitance: C 4 0 Capacitance: C
d ba b
n
In the presence of dielectric medium In the presence of dielectric medium a
(dielectric constant K) (dielectric constant K) In the presence of dielectric
K0 A ab medium (dielectric constant K)
C C 4 0 K
d ba 20 K
C
b
ln
a
SCAN CODE
CAPACITANCE
89
CAPACITANCE
.
Fig. 2.7
Fig. 2.8
2 0 2 0 A
SCAN CODE
CAPACITANCE
90
CAPACITANCE
3. GROUPING OF CAPACITOR
1. Charge on each capacitor remains same and equals to Potential difference across each capacitor remains
the main charge supplied by the battery
Q = Q1 + Q2 + Q3
V = V1 + V2 + V3
+Q1 – Q1
+ –
C1 C2 C3 + –
+Q +Q + –
+Q –Q –Q –Q + –
+ – + – + –
+ – + – + – Q1
+ – + – + – +Q2 – Q2
Q + – + – + – + –
+ –
V1 V2 V3 + –
Q2 + –
+ – Q Q3 +Q3 – Q3
+ –
V + –
+ –
+ –
Fig. 2.9 V
Fig. 2.10
3. If two capacitors having capacitances C1 and C2 are If two capacitors having capacitance C1 and C2
connected in series then respectively are connected in parallel then
C2 C1 C1 C2
V1 .V and V2 .V Q1 .Q and Q2 .Q
C1 C2 C1 C2 C1 C2 C1 C2
4. If n identical capacitors each having capacitances C If n identical capacitors are connected in parallel
are connected in series with supply voltage V then
C Equivalent capacitance Ceq = nC and Charge on each
Equivalent capacitance Ceq and Potential
n Q
capacitor Q .
V n
difference across each capacitor V
n
NOTE:
NOTE:
Two capacitors are in parallel when their
Two capacitors are in series when charge leaving
positive plates are connected and negative plates
one capacitor directly enters into another capacitor,
are also connected with each other.
undivided and undisturbed.
In parallel combination, equivalent capacitance
In series combination equivalent capacitance is
is always greater than the individual capacitance
always lesser than that of either of the individual
capacitors.
SCAN CODE
CAPACITANCE
91
CAPACITANCE
(i)
Fig. 2.11
(ii)
Fig. 2.12
(iii)
Fig. 2.13
SCAN CODE
CAPACITANCE
92
CAPACITANCE
equal to the charge leaving the node” 4.2 Kirchoff's Voltage Law
Example: Kirchoff’s voltage law (2nd Law) states that in any complete
In the given circuit find out the charge on each loop within a circuit, the sum of all voltages across
components which supply electrical energy (such as cells or
capacitor. (Initially they are uncharged) generators) must equal the sum of all voltages across the
other components in the same loop.
Example:
Two capacitors of capacitance 1 μF and 2μF are charged to
potential difference 20 V and 15 V as shown in figure. If
now terminal B and C are connected together terminal A
with positive of battery and D with negative terminal of
battery then find out final charges on both the capacitor.
Fig. 2.14
Fig. 2.15
Sol. Let potential at A is 0, so at D it is 30 V, at F it is
Fig. 2.17
10 V and at point G potential is –25V. Now apply
Kirchhoff's Ist law at point E. (total charge of all Now applying Kirchoff voltage law
– 20 q 30 q
the plates connected to 'E' must be same as before – 30 0
1 2
i.e. 0) (x – 10) + (x – 30) 2 + (x + 25) 2 = 0 –40 – 2q – 30 – q = – 60
5x = 20 x=4 3q = – 10
10
Charge flow = – C
3
50
Charge on capacitor of capacitance 1F 20 q
3
80
Charge on capacitor of capacitance 2F 30 q
3
(Circuit Solving method)
Fig. 2.16
SCAN CODE
CAPACITANCE
93
CAPACITANCE
SCAN CODE
CAPACITANCE
94
CAPACITANCE
5.1 Ladder Problems Sol. Let equivalent capacitance x then capacitance after C,D
Example: point will be 2x because every capacitance becomes 2 times
as compared to A, B. So,
Fig. 2.21
Find equivalent capacitance between point A and B.
Fig. 2.24
Sol.
x
2x C C x
C
2x C C 2
Fig. 2.22
Ceq =
C eq
+C ×C
C eq =
5–1
C
C + C + Ceq 2
Example:
Find equivalent capacitance between points A and B.
Fig. 2.26
Ceq 2C
Example:
SCAN CODE
CAPACITANCE
95
CAPACITANCE
Fig. 2.31
Example:
Now Find out equivalent capacitance between A and B.
Fig. 2.28
3C
So, Ceq
2
Example:
Fig. 2.32
Sol.
Fig. 2.29
Example:. 6. DIELECTRIC
Find out equivalent capacitance between A and B.
Dielectrics are insulating (non-conducting) materials which
transmits electric effect without conducting. We know that
in every atom, there is a positively charged nucleus and a
negatively charged electron cloud surrounding it. The two
Fig. 2.30
oppositely charged regions have their own centres of charge.
Sol. Put numbers on the plates The charges will be as shown
The centre of positive charge is the centre of mass of
in the figure.
positively charged protons in the nucleus. The centre of
V12 = V32 = V34 negative charge is the centre of mass of negatively charged
So all the capacitors are in parallel combination.
electrons in the atoms/molecules.
Ceq = C1 + C2 + C3
SCAN CODE
CAPACITANCE
96
CAPACITANCE
6.1 Polarization of a Dielectric Slab 6.3 Dielectric Breakdown and Dielectric Strength
It is the process of inducing equal and opposite charges If a very high electric field is created in a dielectric, the
on the two faces of the dielectric on the application of outer electrons may get detached from their parent
electric field. atoms. The dielectric then behaves like a conductor.
This phenomenon is known as dielectric breakdown.
E
Then K where K is called dielectric constant. K
E net
SCAN CODE
CAPACITANCE
97
CAPACITANCE
Capacity C = KC C’ = KC
Charge Q = Q (Charge is conserved) Q’ = KQ
Potential V’ = V/K V’ = V (Since Battery maintains the potential
difference)
Intensity E’ = E/K E’ = E
Energy U’ = U/K U’ = KU
E
V E d t E t E d t .t
NOTE: K
If nothing is said it is to be assumed that battery is
t
disconnected. V E d t
K
t Q t
d t d t
7. WHEN DIELECTRIC IS PARTIALLY FILLED 0 K A 0 K
BETWEEN THE PLATES
Now capacitance of the capacitor
If a dielectric slab of thickness t (t < d) is inserted Q 0 A
between the plates as shown below, then E = Main C C
V dt
t
electric field between the plates, Ei= Induced electric
K
field in dielectric. E’ = (E – Ei) = The reduced value of
Dielectric slabs in series:
electric field in the dielectric. Potential difference
between the two plates of capacitor is given by
Fig. 2.39
Fig. 2.38
SCAN CODE
CAPACITANCE
98
CAPACITANCE
0 A ε0 A / 2
C
t t t d/2 d/2
d t1 t 2 t 3 ...... 1 2 3 ...... d–d/2–d/2+ +
K
1 K 2 K 3 K1 K3
ε0 A / 2 Kε 0 A
d/2 d/2 d
d–d/2–d/2+ +
K1 K3
K1 K 3 K 2 K 3
+
K1 K 3 K 2 K 3
Fig. 2.40
0 A
Dielectric slabs in parallel C
t1 t2 t t
3 4
K1 K 2 K 3 K 4
A parallel plate capacitor of area A, plate separation d and
capacitance C is filled with three different dielectric
materials having dielectric constants K1, K2 and K3 as Fig. 2.43
shown. If a single dielectric material is to be used to have
0A
the same capacitance C in this capacitor then its dielectric Capacitance C
d t
constant K is given by
Fig. 2.44
Fig. 2.41
C’ = (In this case capacitor is said to be short circuited)
ε0 A
Applying C =
t1 t
d – t1 – t 2 + + 2
k1 K 2 8. WHEN SEPARATION BETWEEN THE
PLATES IS CHANGING
We have,
If separation between the plates changes then it’s
1
capacitance also changes according to C . The effect on
d
other variables depends on the fact that whether the charged
capacitor is disconnected from the battery or battery is still
connected.
Fig. 2.42
SCAN CODE
CAPACITANCE
99
CAPACITANCE
Fig. 2.45
Fig. 2.46
1
Capacity Decreases because C i.e. C’ < C Decreases i.e., C’ < C
d
Remains constant because a battery is not present i.e., Decreases because battery is present i.e., Q’
Charge
Q’ = Q <Q
Q Q
Remains constant because E i.e., Decrease because E E Q i.e.,
Electric Field 0 A 0 A 0
E’ = E E’ < E
Decreases because
Q2 1
Energy Increases because U U i.e., U’ > U 1
2C C U CV2 U C i.e., U’ < U
2
SCAN CODE
CAPACITANCE
100
CAPACITANCE
Fig. 2.47
SCAN CODE
CAPACITANCE
CAPACITANCE 101
= n volume of a single
drop i.e.,
4 3 4
R n r3 ,
3 3
R n r
1/3
Q nq
(iv) Potential of big drop: V 1/3 V = n2/3 v
C n c
1
U CV2
2
(v) Energy of big drop:
1
(n1/3c)(n2/3v)2
2
U = n5/3 u
NOTE:
It is a very common misconception that a
capacitor stores charge but actually a capacitor
stores electric energy in the electrostatic field
between the plates.
Two plates of unequal area can also form a
capacitor but effective overlapping area is
considered.
SCAN CODE
CAPACITANCE
CAPACITANCE 102
SOLVED EXAMPLES
Example - 2
Example – 1
A charged air capacitor has stored energy U0. What
In the figure below, the capacitance of each
will be the energy stored when air is replaced by a
capacitor is 3F. The effective capacitance between
A and B is dielectric of dielectric constant K, charge Q
remaining the same.
2
Sol. Energy stored in dielectric filled capacitor U Q .
2C
1
U CV 2 .
2
C1, C2 and C3 is
Example- 6
Sol.
1 1 1 1 2 3 6 11
Ceq 3C 2C C 6C 6C
6C
Ceq
11
Assume a battery of potential V across A & B. Do the
Example- 5
charge distribution in the circuit.
Find the total energy stored in the capacitors in the
given network:
Variables : Q, q
Loop (A → B → C → F → A)
CAPACITANCE 104
Qq q
0 0 0 Q 4q
3C C ⇒ ...(i) 1 1 2 8 3C
C3
Loop (F → C → D → E → F) C C 3C 3C 8
q Q
V00
C 4C Þ 4 q Q 4 CV ...(ii)
Sol.
(a) Capacitance
(b) Capacitive reactance
(c) Magnetic field between the plates
(d) Energy stored in the capacitor
⇒ 11 2 C C Ans. (d)
1
C C C 2
Sol.
⇒
CAPACITANCE 105
Example - 9
Ans. (c)
1 1
U1 = C1 V12 + C 2 V22
2 2
Total energy after sharing charges:
1
U2 = C1 +C 2 V 2
2
Common potential is given by:
C1V1 +C 2 V2
V=
C1 +C 2
C V +C V C V +C V
2 2
1
U2 = C1 +C2 1 1 2 2 = 1 1 2 2
2 C1 +C 2 2 C1 +C 2
Loss in energy
C1C2 V1 -V2
2
U = U 2 -U1 =
2 C1 +C2
106
CAPACITANCE
Parallel Plate Capacitor & Spherical and cylindrical 7. In a parallel–plate capacitor, the region between the plates
capacitor is filled by a dielectric slab. The capacitor is charged from a
1. The capacity of a parallel plate capacitor is C. Its cell and then disconnected from it. The slab is now taken out.
capacitance when the separation between the plate is (a) The potential difference across the capacitor is reduced
halved will be (b) The potential difference across the capacitor is increased
(a) 4 C (b) 2 C (c) The energy stored in te capacitor is reduced
(c) C/2 (d) C/4 (d) No work is done by an external agent in taking the slab
2. The distance between two plates of a parallel plate out
capacitor is slowly increased. A graph is plotted between
capacitance C and the distance d. Which graph is correct? Dielectrics in capacitors
(a) A (b) B
(c) C (d) D
3. A parallel plate capacitor is charged and the charging
(a) 1 (b) 2
battery is then disconected. If the plates of the capacitor
are moved farther apart by means of insulating handles (c) 3 (d) 4
(a) the charge on the capacitor increases 9. A parallel plate condenser has a capacity C. The space
between the plates is half filled by oil of dielectric constant k
(b) the voltage across the plates decreases
as shown in the figure. The ratio of new capacity to the
(c) the capacitance increases original capacity is :
(d) the electrostatic energy stored in the capacitor
increases
4. A parallel plate capacitor is made of two circular plates
separated by a distance of 5 mm and with a dielectric of
dielectric constant 2.2 between them. When the electric
field in the dielectric is 3 × 104 V/m, the charge density of
the positive plate will be close to : (a) k : 1 (b) (1 + k) : 1
(a) 3 × 10–7 C/m2 (b) 3 × 104 C/m2
1+ k 1
(c) 6 × 104 C/m2 (d) 6 × 10–7 C/m2 (c) :1 (d) (1 + k) :
2 2
5. A spherical conductor has a capacity of 2F. Its radius
will be (0 = 8.85 × 10–12 S.I. units) 10. The capacities of two conductors are C1 and C2 and their
respective potentials are V1 and V2. If they are connected by
(a) 1.8 m (b) 1800 m
4
a thin wire then the loss of energy will be
(c) 1.8 × 10 m (d) 0.018 m
6. Capacitance (in F) of a spherical conductor having radius C1C 2 (V1 V2 ) C1C 2 (V1 V2 )
(a) (b)
1 m, is 2(C1 C 2 ) 2(C1 C 2 )
–10 –6
(a) 1.1 × 10 (b) 10
–9 –3 C1C 2 (V1 V2 ) 2 (C1 C 2 ) (V1 V2 )
(c) 9 × 10 (d) 10 (c) (d)
2(C1 C 2 ) C1C2
CAPACITANCE 107
11. A capacitor connected to a 10 V battery collects a charge 16. A parallel plate capacitor is made of two plates of length
of 40 microcoulomb with air as dielectric and 100 l, width and separated by distance d. A dielectric slab
microcoulomb with oil as dielectric. The dielectric constant (dielectric constant K) that fits exactly between the plates
of the oil is is held near the edge of the plates. It is pulled into the
(a) 4 (b) 0.4 U
capacitor by a force F = – where U is the energy of
(c) 2.5 (d) 1.0 x
the capacitor when dielectric is inside the capacitor up to
12. A parallel plate capacitor is made of two dielectric blocks
distance x (See figure). If the charge on the capacitor is
in series. One of the blocks has thickness d1 and dielectric then the force on the dielectric when it is near the edge is:
constant k1 and the other has thickness d 2 and dielectric
constant k2 as shown in the figure. This arrangement can
Q2 Q2
(a) (k 1) (b) k
2dl 2 0 2dl 2 0
13. A parallel plate capacitor with air between the plates has a
capacitance of 9 pF. The separation between its plates is d. Q2 d
(c) (k 1) (d) None
The space between the plate is now filled with two 2l 2 0
dielectrics. One of the dielectrics has dielectric constant k1
17. A parallel plate condenser with a dielectric of dielectric
= 3 and thickness d/3 while the other one has dielectric
constant K between the plates has a capacity C and is
constant k2 = 6 and thickness 2d/3. Capacitance of the
charged to a potential V volts. The dielectric slab is slowly
capacitor is now removed from between the plates and then reinserted. The
(a) 1.8 pF (b) 45 pF net work done by the system in this process is
(c) 40.5 pF (d) 20.25 pF 1
14. A parallel plate capacitor with a slab of dilectric constant 3 (a) K 1 CV 2 2
(b) CV (K – 1)/K
2
filling the whole space between the plates is charged to 2
(c) (K – 1) CV (d) zero
certain potential and isolated. Then the slab is drawn out
and another slab of equal thickness but dielectric constant 18. Two parallel plates of area 0.1 m2 each are at separation of
2 is introduced between the plates. The ratio of the energy 10 mm . One-fourth of the space between them is filled
stored in the capacitor later to that stored initially is with a dielectric of dielectric constant 4 as shown in the
(a) 2 : 3 (b) 3 : 2 figure. The equivalent capacitance between terminals A
(c) 4 : 9 (d) 9 : 4 and B is given by 0 (in SI units). Find .
15. An air capacitor of capacity C = 10 F is connected to a
constant voltage battery of 12 V. Now the space between
the plates is filled with a liquid of dielectric constant 5. The
charge that flows now from battery to the capacitor is
(a) 120 C
(b) 699 C
(c) 480 C
(d) 24 C
CAPACITANCE 108
19. A fully charged capacitor has a capacitance C. It is 23. In the figure, the equivalent capacitance between A and B is
discharged through a small coil of resistance wire embedded
in a thermally insulated block of specific heat capacity s
and mass m. If the temperature of the block is raised to T.
Then the potential difference V across the capacitor is
2mTC
(a)
s C1C 2 C 2C3 C3C1
(a) C1 + C2 + C3 (b)
C1 C 2 C3
2mTs
(b)
C C1C 2C3
(c) C C C C C C (d) none of these
1 2 2 3 3 1
3mTC
(c)
2s 24.
mTC
(d)
2s
Combination of Capacitors
Circuit laws & Miscellaneous problems in capacitors 33. A parallel plate capacitor of capacitance C is connected to
a battery and is charged to a potential difference V. Another
28. For the circuit shown, which of the following statements capacitor of capacitance 2C is connected to another battery
is true? and is charged to potential difference 2V. The charging
batteries are now disconnected and the capacitors are
connected in parallel to each other in such a way that the
positive terminal of one is connected to the negative terminal
of the other. The final energy of the configuration is–
38. In the figure shown the plates of a parallel plate capacitor 41. N identical capacitor are joined in parallel and the
have unequal charges. Its capacitance is 'C'. P is a point combination is charged to a potential V. Now if they are
outside the capacitor and close to the plate of charge- separated and then joined in series then energy of
Q. The distance between the plates is 'd' then which combination will :–
statement is wrong (A) remain same and potential difference will also remain
same
2Q -Q
(B) remain same and potential difference will become NV
(C) increase N times and potential difference will
P become NV
(D) increase N time and potential difference will remains
same
42. The value of equivalent capacitance of the combination
(A) A point charge at point 'P' will experience electric force shown in figure between the points P and Q is :–
due to capacitor
2C 2C P
3Q
(B) The potential difference between the plates will be 2C
2C C
C C Q
(C) The energy stored in the electric field in the region
(A) 3 C (B) 2 C
9 Q2
between the plates is (C) C (D) C/3
8C
43. In the given circuit if point C is connected to the earth and a
(D) The force on one plate due to the other plate is potential of +2000 V is given to point A, the potential at B is
Q2 10F 10F
2
2 0 d
5F
39. Five identical plates are connected across a battery as 10F C
A B
follows :
46. Two parallel plate capacitors whose capacities are C and 2 50. A fully charged capacitor has a capacitance C. It is
C respectively, are joined in parallel. These are charged by discharged through a small coil of resistance wire embedded
V potential difference. If the battery is now removed and in a thermally insulated block of specific heat capacity s
a dielectric of dielectric constant K is filled in between the and mass m. If the temperature of the block is raised by DT,
plates of the capacitor C, then what will be the potential the potential difference V across the capacitance is–
difference across each capacitor?
2mC T
V 2V (a)
(a) (b) s
K2 K2
mC T
3V 2 K (b)
(c) (d) s
K 2 3V
47. A parallel plate capacitor with air between the plates has a msT
(c)
capacitance of 9 pF. The separation between its plates is C
'd'. The space between the plates is now filled with two
dielectrics. One of the dielectric has dielectric constant 2msT
d (d)
K1= 3 and thickness while the other one has dielectric C
3
2d 51. The capacitance (C) for an isolated conducting sphere of
constant K2 = 6 and thickness . Capacitance of the radius (a) is given by 4a. This sphere is enclosed within
3
capacitor is now an earthed concentric sphere. The ratio of the radii of the
(A) 1.8 pF (B) 45 pF n
(C) 40.5 pF (D) 20.25 pF spheres being then the capacitance of such a
(n 1)
48. A parallel plate capacitor of capacitance C (without sphere will be increased by a factor-
dielectrics) is filled by dielectric slabs as shown in figure.
Then the new capacitance of the capacitor is (a) n
n
(b)
K=6
3
(n 1)
K=2 K=4 (n 1)
1 2
(c)
n
54. Four identical plates 1,2,3 and 4 are placed parallel to each
other at equal distance as shown in the figure. Plates 1
and 4 are joined together and the space between 2 and 3
is filled with a dielectric of dielectric constant k=2. The
capacitance of the system between 1 and 3 & 2 and 4 are
C1
C1 and C2 respectively. The ratio C is-
2
1
2
3
4
5
(a) (b) 1
3
3 5
(c) (d)
5 7
55. The equivalent capacitance across A & B is
23F 7F
13F 1F
A B
12F
10F 1F
28 15
(a) f (b) F
3 2
(c) 15 mF (d) None of these
CAPACITANCE 113
(a) (b)
31 32
(a) F (b) F
23 23
(c) (d)
33 34
(c) F (d) F
23 23
6. A capacitance of 2 F is required in an electrical circuit
across a potential difference of 1.0 kV. A large number of
1 F capacitors are available which can withstand a
3. A combination of capacitors is set up as shown in the
potential difference of not more than 300 V. The minimum
figure. The magnitude of the electric field, due to a point
number of capacitors required to achieve this is : (2017)
charge (having a charge equal to the sum of the charges
on the 4 F and 9 F capacitors), at a point distant 30 m (a) 32 (b) 2
from it, would equal: (2016) (c) 16 (d) 24
CAPACITANCE 114
K 0 a 2 K 0 a 2
(a) (b) ln K
2d K 1 d K 1
K 0 a 2 1 K 0 a 2
(c) ln K (d)
d 2 d
13. A parallel plate capacitor is of area 6 cm2 and a separation
(a) 2.4 F (b) 4.9 F 3 mm. The gap is filled with three dielectric materials of
(c) 3.6 F (d) 5.4 F equal thickness (see figure) with dielectric constants K1
= 10, K2 = 12 and K3 = 14. The dielectric constant of a
10. A parallel plate capacitor with area 200 cm2 and separation
single material which when fully inserted in the above
between the plates 1.5 cm, is connected across a battery capacitor gives the same capacitance would be: (2019)
of emf V. If the force of attraction between the plates is
25×10-6 N, the value of V is approximately:
-12 C
2
0 = 8.85×10 (2018)
N.m 2
(a) 250 V (b) 100 V (a) 4 (b) 14
(c) 300 V (d) 150 V (c) 12 (d) 36
11. A capacitor C1 = 1.0 F is charged up to a voltage V = 60 14. A parallel plate capacitor having capacitance 12 pF is
charged by a battery to a potential difference of 10 V
V by connecting it to battery B through switch (1). Now
between its plates. The charging battery is now
C1 is disconnected from battery and connected to a circuit
disconnected and a porcelain slab of dielectric constant
consisting of two uncharged capacitors C2 = 3.0 F and
6.5 is slipped between the plates. The work done by the
C3 = 6.0 F through switch (2), as shown in the figure. capacitor on the slab is: (2019)
The sum of final charges on C2 and C3 is : (a) 692 pJ (b) 508 pJ
(2018) (c) 560 pJ (d) 600 pJ
CAPACITANCE 115
15. Seven capacitors, each of capacitance 2 μF, are to be 17. Voltage rating of a parallel plate capacitor is 500 V. Its
connected in a configuration to obtain an effective dielectric can withstand a maximum electric field of
10 6 V/m. The plate area is 10–4 m2. What is the dielectric
capacitance of 6 F . Which of the combinations, constant if the capacitance is 15 pF? (given 0 = 8.86 ×
13 10–12 C2 m2)
shown in figures below, will achieve the desired value? (2019)
(2019) (a) 3.8 (b) 8.5
(a) (c) 4.5 (d) 6.2
18. A parallel plate capacitor has 1μF capacitance. One of its
two plates is given + 2μC charge and the other plate,
+4μC charge. The potential difference developed across
the capacitor is :
(2019)
(a) 3 V (b) 1 V
(b)
(c) 5 V (d) 2 V
19. A capacitor with capacitance 5 F is charged to 5 C .
If the plates are pulled apart to reduce the capacitance to
2 F how much work is done?
(2019)
(a) 6.25 10 J
6 (b) 3.75 10 6 J
(c)
(c) 2.16 106 J (d) 2.55 10 6 J
20. The parallel combination of two air filled parallel plate
capacitors of capacitance C and nC is connected to a
battery of voltage, V. When the capacitors are fully
charged, the battery is removed and after that a dielectric
material of dielectric constant K is placed between the
two plates of the first capacitor. The new potential
(d)
difference of the combined system is:
(2019)
nV
(a) (b) V
K n
16. In the given circuit, after the switch ‘S’ is turned from
position ‘A’ to position ‘B’, the energy dissipated in the V n 1 V
(c) (d)
circuit in terms of capacitance ‘C’ and the total charge ‘Q’ K n K n
is _____. (2019)
21. Figure shows charge (q) versus voltage (V) graph for
series and parallel combination of two given capacitors.
The capacitances are : (2019)
1 Q2 3 Q2
(a) (b)
8 C 8 C
22. A simple pendulum of length L is placed between the 24. In the given circuit, the charge on 4 F capacitor will be:
plates of a parallel plate capacitor having electric field E, (2019)
as shown in figure. Its bob has mass m and charge q. The
time period of the pendulum is given by:
(2019)
g g
m m
23. Two identical parallel plate capacitors, of capacitance C
each, have plates of area A, separated by a distance d.
The space between the plates of the two capacitor, is
filled with three dielectrics, of equal thickness and
dielectric constants K1, K2 and K3. The first capacitors is
filled as shown in Fig. I, and the second one is filled as A 0 k d
2
A 0 k 2 d
(c) d 1 2
Ak 0
(d) 1 d
d
E1 K1 K 2 K 3 26. Effective capacitance of parallel combination of two
(a) capacitors C1 and C2 is 10 μF. When these capacitor are
E2 K1 K 2 K 3 K 2 K 3 K 3 K1 K1 K 2
individually connected to a voltage source of 1 V, the
energy stored in the capacitor C2 is 4 times of that in C1. If
(b) E1 K1 K 2 K 3 K 2 K 3 K 3 K1 K1 K 2 these capacitors are connected in series, their effective
E2 K1 K 2 K 3
capacitance will be:
E1 9K1 K 2 K 3 (2020)
(c)
E2 K1 K 2 K 3 K 2 K3 K3 K1 K1 K 2 (a) 1.6μF (b) 3.2μF
(c) 4.2μF (d) 8.4μF
(d) E1 K1 K 2 K 3 K 2 K3 K 3 K1 K1 K 2
E2 9 K1 K 2 K 3
CAPACITANCE 117
27. A capacitor is made of two square plates each of side ‘a’ 31. A capacitor C is fully charged with voltage V0. After
making a very small angle between them, as shown in disconnecting the voltage source, it is connected in
figure. The capacitance will be close to parallel with another uncharged capacitor of
(2020) C
capacitance . The energy loss in the process after the
2
charge is distributed between the two capacitors is:
(2020)
1 1
(a) CV02 (b) CV02
2 4
1 1
(c) CV02 (d) CV02
3 6
0a2 a 0a2 3 a
(a) 1 (b) 1 32. Two capacitors of capacitances C and 2C are charged to
d 2d d 2d
potential differences V and 2V, respectively. These are
then connected in parallel in such a manner that the
0a2 a 0a 2 a
(c) 1 (d) 1
positive terminal of one is connected to the negative
d 2d d 4d terminal of the other. The final energy of this configuration
28. A 5F capacitor is charged fully by a 220V supply. It is is :
then disconnected from the supply and is connected in (2020)
series to another uncharged 2.5 F capacitor. If the energy 9
(a) zero (b) CV 2
X 2
change during the charge redistribution is J then
100
value of X to the nearest integer is ………….. . (2020) 25 3
(c) CV 2 (d) CV2
6 2
29. A 10 F capacitor is fully charged to a potential difference
of 50V. After removing the source voltage it is connected 33. In the circuit shown, charge on the 5 F capacitor is :
to an uncharged capacitor in parallel. Now the potential (2020)
difference across them becomes 20 V. The capacitance of
the second capacitor is :
(2020)
(a) 15 F (b) 20 F
(c) 10 F (d) 30 F
30. In the circuit shown in the figure, the total charge is
750 C and the voltage across capacitor C2 is 20 V. Then (a) 5.45 C (b) 18.00 C
the charge on capacitor C2 is :
(c) 10.90 C (d) 16.36 C
(2020)
34. A parallel plate capacitor has plate of length ‘l’, width ‘w’
and separation of plates is ‘d’. It is connected to a battery
of emf V.A dielectric slab of the same thickness ‘d’ and of
dielectric constant k = 4 is being inserted between the
plates of the capacitor. At what length of the slab inside
plates, will the energy stored in the capacitor be two times
the initial energy stored?
(2020)
(a) 650 C
21 1
(b) 450 C (a) (b)
3 2
(c) 590 C
1 1
(d) 160 C (c) (d)
4 3
CAPACITANCE 118
35. A parallel plate capacitor whose capacitance C is 14pF is 38. A parallel plate capacitor has plate area 100m2 and plate
charged by a battery to a potential difference V = 12V separation of 10m. The space between the plates is filled
between its plates. The charging battery is now up to a thickness 5 m with a material of dielectric constant
disconnected and a porcelain plate with k = 7 is inserted of 10. The resultant cpacitance of the system is pF. The
between the plates, then the plate would oscillate back value of 0 8.85 10 –12 F.m –1 The value of x to the
and forth between the plates with a constant mechanical
nearest integer is ................. .
energy of (in pJ). Find N. (Assume no friction)
(2021)
(2021)
39. For changing the capacitance of a given parallel plate
36. Four identical rectangular plates with length, l = 2m and
capacitor, a dielectric material of dielectric constant K is
3
breadth, b m are arranged as shown in figure. The used, which has the same area as the plates of the capaci-
2
x0 3
tor. The thickness of the dielectric slab is d , where ‘d’ is
equivalent capacitance between A and C is . The 4
d
the separation between the plates of parallel plate capaci-
value of x is (Round off to the Nearest Integer)
tor. The new capacitance (C’) in terms of orginal capaci-
(2021)
tance (C0) is given by the following relation.
(2021)
4k 3 K
(a) C ' C0 (b) C ' C0
K 3 4K
4 4k
(c) C ' C0 (d) C ' C0
3 K 3
3
(b) CV 2 (a) 4 F (b) 2 F
2
(c) 18 F (d) 6 F
25 5. A capacitor of capacitance C1 = 1 F can with stand maximum
(c) CV 2 voltage V1 = 6kV (kilo-volt) and another capacitor of
6
capacitance C2 = 3 F can withstand maximum voltage
9 V2 = 4 kV. When the two capacitors are connected in series,
(d) CV 2 the combined system can withstand a maximum voltage of
2
(a) 4 kV (b) 6 kV
2. A parallel plate capacitor of capacity is charged to a (c) 8 kV (d) 10 kV
potential 6. Five identical plates are connected across a battery as
follows. If the charge on plate 1 be +q, then the charges on
(i) The energy stored in the capacitor when the battery is
the plates 2, 3, 4 and 5 are
disconnected and the separation is doubled E1
(ii) The energy stored in the capacitor when the charging
battery is kept connected and the separation between
the capacitor plates is doubled is E2. Then E1/E2 value is
(a) 4 (b) 3/2
(c) 2 (d) ½
3. An infinite number of identical capacitors each of (a) – q, + q, – q, + q (b) – 2q, + 2q, – 2q, +q
capacitance 1F are connected as in adjoining figure. Then
(c) – q, + 2q, – 2q, + q (d) None of these
the equivalent capacitance between A and B is
7. Four plates, each of area A and each side are placed parallel
to each other at a distance d. A battery is connected between
the combinations 1 and 3 and 2 and 4. The modulus of
charge on plate 2 is
(a) 1 F (b) 2 F
2 0 A 3 0 A
1 (a) E (b) E
(c) 2 F (d) d d
2 0 A 0 A
(c) E (d) E
3d d
CAPACITANCE 120
8. Five capacitors are connected as shown in the diagram. If 13. Seven capacitors each of capacitance 2 F are connected
the p.d. between A and B is 22 V, the emf of the cell is in a configuration to obtain an effective capacitance
10/11 F. Which of the following combination will achieve
the desired result be ?
(a)
(a) 26 V (b) 42 V
(c) 38 V (d) 46 V
9. Find the equivalent capacitance between X and Y (Q.8
to Q.9)
(b)
(c)
12 A
(d) d d
2 1 1 2
CAPACITANCE 121
(b) C K1 K 2 ε0 A
t1k1 t2 k3
ε0 A
(c) C (c)
t1 t2
k1 k3
A
(d) (K1 – K2) 0 K K (t1 + t2)
1 2
17. A parallel plate capacitor with air in between the plates
has capacitance of 9 F .The separation between the
(a) Both C2 and C3 > C1
plates is d .The space between the plates is now filled
with two dielectrics .One of the dielectric has dielectric (b) C3 > C1 but C2 < C1
constant k1=3 with thickness d/3 and other one has (c) Both C2 and C3 < C1
dielectric constant k2=6 and thickness 2d/3.The new (d) C1 = C2 = C3
capacitance will be 22. The expression for the capacity of the capacitor formed
(a)1.8 F (b) 20.25 F by compound dielectric placed between the plates of a
(c) 40.5 F (d) 45 F parallel plate capacitor as shown in figure, will be (area of
18. Condenser A has a capacity of 15F when it is filled with a plate = A)
medium of dielectric constant 15. Another condenser B
has a capacity of 1F with air between the plates. Both are
charged separately by a battery of 100 V. After charging,
both are connected in parallel without the battery and the
dielectric medium being removed. The common potential
now is
(a) 400 V (b) 800 V
(c) 1200 V (d) 1600 V
19. An uncharged capacitor with a solid dielectric is connected
to a similar air capacitor charged to a potential of V0. If the 0 A
0 A
common potential after sharing of charges becomes V, then (a) (b)
the dielectric constant of the dielectric must be d1 d 2 d 3 d1 d 2 d 3
K1 K 2 K 3 K1 K 2 K 3
V0 V
(a) (b) V
V 0 0 A K1K 2 K 3 AK1 AK 2 AK 3
(c) (d) 0
V0 V V0 V d1d 2 d3 d1 d2 d3
(c) (d)
V V0
CAPACITANCE 122
23. The space between the plates of a parallel plate capacitor 29. A charge of 1 C is given to one plate of a parallel-plate
is filled completely with a dielectric substance having capacitor of capacitance 0.1 F and a charge of 2C is
dielectric constant 4 and thickness 3 mm. The distance given to the other plate. Find the potential difference, in
volts developed between the plates.
between the plates in now increased by inserting a second
sheet of thickness 5 mm and dielectric constant K. If the 30. A capacitor having a capacitance of 100F is charged to
a potential difference of 50 V. The charging battery is
capacitance of the capacitor so formed is one-half of the
disconnected and a dielectric slab of dielectric constant
original capacitance, the value of K is 2.5 is inserted. What charge in millicoloumb would have
(a) 10/3 (b) 20/3 produced this potential difference in absence of the
(c) 5/3 (d) 15/3 dielectric slab.
24. In a parallel–plate capacitor, the plates are kept vertical. Subjective Type Questions
The upper half of the space between the plates is filled
31. Two parallel plate capacitors of capacitances C and 2C are
with a dielectric with dielectric constant K and the lower
connected in parallel and charged to a potential difference
half with a dielectric with dielectric constant 2K. The ratio
V. The battery is then disconnected and the region between
of the charge density on the upper half of the plates to the the plates of capacitor C is completely filled with a material
charge density on the lower half of the plates will be equal to of dielectric constant K. The potential difference across
(a) 1 (b) 2 the capacitors now becomes ...........
(c) 1/2 (d) 3/2 32. The capacitance of a parallel plate capacitor with plate
area A and separation d, is C. The space between the plates
25. In a parallel–plate capacitor, the region between the plates
is filled with two wedges of dielectric constants K1 and K2
is filled by a dielectric also. The capacitor is connected to respectively (figure). Find the capacitance of the resulting
a cell and the slab is taken out. capacitor.
(a) Some charge is drawn from the cell
(b) Some charge is returned to the cell
(c) The potential difference across the capacitor is
reduced
(d) No work is done by an external agent in taking the
slab out
26. If we increase ‘d’ of a parallel plate condensor to ‘2d’ and
fill wax to the whole empty space between its two plate,
then capacitance increase from 1pF to 2pF. What is the
33. Three identical capacitors C1, C2 and C3 have a capacitance
dielectric constant of wax.
of 1.0 F each and they are uncharged initially. They are
(a) 2 (b) 4 connected in a circuit as shown in the figure and C1 is
(c) 6 (d) 8 then filled completely with a dielectric material of relative
27. A capacitor is connected across another charged capacitor. permittivity r. The cell electromotive force (emf) V0 = 8 V.
The energy in the two capacitors will: First the switch S1 is closed while the switch S2 is kept
(a) be equal to the energy in the initial capacitor open. When the capacitor C3 is fully charged, S1 is opened
(b) Be less than that in the initial capacitor and S2 is closed simultaneously. When all the capacitors
reach equilibrium, the charge on C3 is found to be 5C.
(c) Be more than that in the initial capacitor
The value of r = ________.
(d) Be more or less depending on the relative capacities of
the two capacitors
34. The figure shows two identical parallel plate capacitors 37. In the circuit shown in steady state
connected to a battery with the switch S closed. The switch
is now opened and the free space between the plates of
the capacitors is filled with a dielectric of dielectric constant
(or relative permittivity) 3. Find the ratio of the total
electrostatic energy stored in both capacitors before and
after the introduction of the dielectric
(a) charge across 4F capacitor is 20 C
(b) charge across 4F capacitor is 10 C
(c) potential difference across 4F capacitor is 5 Volt
(d) potential difference across 4F capacitor is 10 Volt
38. A dielectric slab is inserted between the plates of an
isolated charged capacitor. Which of the following
quantities will change.
35. Two parallel plate capacitors–4
A and B have the same (a) the electric field in the capacitor
separation d = 8.85 × 10 m 2between the2 plates. The
plate areas of A and B are 0.04 m and 0.02 m respectively. (b) the charge on the capacitor
A slab of dielectric constant (relative permittivity) K = 9 (c) the potential difference between the plates
has dimensions such that it can exactly fill the space
between the plates of capacitor B. (d) the stored energy in the capacitor
39. A parallel plate air capacitor is connected to a battery. The
quantities charge, voltage, electric field and energy
associated with this capacitor are given by Q0, V0, E0 and
U0 respectively. A dielectric slab is now introduced to fill
the space between the plates with the battery still in
connection. The corresponding quantities now given by
Q, V, E and U are related to the previous one as
(a) > Q0 (b) V > V0
(i) The dielectric slab is placed inside A as shown in figure (c) E > E0 (d) U > U0
(a). A is then charged to a potential difference of 110 V.
Calculate the capacitance of A and the energy stored in it. 40. A parallel plate capacitor is charged and the charging
battery is then disconnected. If the plates of the capacitor
(ii) The battery is disconnected and the dielectric slab is are moved farther apart by means of insulating handles
removed from A. Find the work done by the external agency
(a) the charge on the capacitor increases
in removing the slab from A.
(b) the voltage across the plates increases
(iii) The same dielectric slab is now placed inside B, filling
it completely. The two capacitors A and B are then (c) the capacitance increases
connected as shown in figure (c). Calculate the energy (d) the electrostatic energy stored in the capacitor
stored in the system. increases
More than one correct options 41. A parallel plate capacitor of plate area A and plate
separation d is charged to potential difference V and then
36. A dielectric slab of thickness d is inserted in a parallel the battery is disconnected. A slab of dielectric constant
plate capacitor whose negative plate is at x = 0 and K is then inserted between the plates of the capacitor so
positive plate is at x = 3d. The slab is equidistant from the as to fill the space between the plates. If Q, E and W
plates. The capacitor is given some charge. As x goes denote respectively, the magnitude of charge on each plate,
from 0 to 3d the electric field between the plates (after the slab is
(a) the magnitude of the electric field remains the same inserted), and work done on the system, in question, in
(b) the direction of the electric field remains the same the process of inserting the slab, then
(c) the electric potential increases continuously 0 AV 0 KAV
(a) Q (b) Q
(d) the electric potential increases at first then decreases d d
and again increases
V 0 AV 2 1
(c) E (d) W 1 K
Kd 2d
CAPACITANCE 124
Match the Column Type Questions 44. Five identical capacitor plates, each of area A, are arranged
42. Consider the situation shown below. The switch S is open such that adjacent plates are at a distance d apart, the
for a long time and then closed. Then match the columns plates are connected to a source of emf V as shown in
the figure.
Column-1 Column-2
1
(a) Charge flown through (P) CE 2
2
battery when switch is closed The charge on plate 1 is ......... and on plate 4 is ..........
1 2
(b) Change in energy stored (Q) CE
4
in capacitor
1
(c) Heat developed in system (R) CE 2
8
CE
(d) Work done by battery (S)
2
43. In case of an isolated parallel plate capacitor there is effect
on its capacity when a dielectric is introduced or plate
separation is changed. Match Column-1 with column-2
for the statements in Column-1.
Column-1 Column-2
(a) When the plates of (P) Work done by external
parallel plate capacitor agent
is negative are pulled apart
keeping charge constant
(b) When the plates of (Q) Work done by battery
parallel plate capacitor is
positiveare pulled apart
keeping it potential constant
(c) When a dielectric slab (R) Electric potential is
gradually inserted energy of
the system between the
plates of decreasesparallel
plate capacitor and its
potential is kept constant
(d) When a dielectric slab (S) Work done by external is
gradually inserted agent is
positive between the plates
of an isolated parallel plate
capacitor
CAPACITANCE 125
(a) + 32 C (b) + 40 C
(c) + 48 C (d) + 80 C
6. In the circuit shown in the figure, there are two parallel
plate capacitors each of capacitance C. The switch S1 and
(a) zero (b) q/2 pressed first to fully charge the capacitor C1 and then
released. The switched S2 is then pressed to charge the
(c) q (d) 2q
capacitor C2. After some time, S2 is released and then S3 is
3. A parallel plate capacitor C with plates of unit area and pressed. After some time. (2013)
separation d is filled with a liquid of dielectric constant
K = 2. The level of liquid is d/3 initially. Suppose the liquid
level decreases at a constant speed v, the time constant as
a function of time t is (2008)
7. A parallel plate capacitor has a dielectric slab of dielectric 9. A parallel plate capacitor of capacitance C has spacing d
constant K between its plates that covers 1/3 of the area between two plates having area A. The region between
of its plates, as shown in the figure. The total capacitance the plates is filled with N dielectric layers, parallel to its
of the capacitor is C while that of the portion with dielectric d
in between is C1. When the capacitor is charged, the plate plates, each with thickness . The dielectric constant
N
area covered by the dielectric gets charge Q1 and the rest
of the mth layer is K m K 1
of the area gets charge Q2. The electric field in the dielectric m
. For a very large N(>10 ),
3
is E1 and that in the other partion is E2. Choose the correct N
option/options, ignoring edge effects. (2014)
K 0 A
the capacitance C is .The value of α will be
d In2
Find Answer key and Detailed Solutions at the end of this book
CAPACITANCE
CURRENT ELECTRICITY
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Chapter 03 129
CURRENT ELECTRICITY
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130
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(iii) Gases: In gases current carriers are positive ions and (b) Current carriers in liquids:
free electrons. 1. In an electrolyte like CuSO4, NaCl etc., there are
(iv) Semiconductor: In semiconductors current carriers positively and negatively charged ions
are holes and free electrons. (like C u , SO 4 , Na , Cl ).
(v) The amount of charge flowing through a cross
2. These are forced to move in definite directions under
section of a conductor from t = ti to t = tf is given by :
the effect of an external electric field, causing electric
tf
current.
q ti
I dt
3. Thus, in liquids, the current carriers are positively and
Graphs
negatively charged ions.
(i) Slope of Q vs t graph gives instantaneous current.
(c) Current carriers in gases:
1. Ordinarily, the gases are insulators of electricity.
2. They can be ionized by applying a high potential
difference at low pressure
3. Thus, positive ions and electrons are the current
carriers in gases.
Fig. 3.5
(ii) Area under the I vs t graph gives net charge flown. 3. DRIFT VELOCITY
“If u1 , u2 , u3 , ...un are random thermal velocities of n free
electrons in the metal conductor, then the average thermal
u1 u2 u3 ... un
velocity of electrons is 0
n
As a result, there will be no net flow of electrons of charge
in one particular direction in a metal conductor, hence no
Fig. 3.6 current”.
“Drift velocity is defined as the average velocity with which
2. CURRENT CARRIERS the free electrons get drifted towards the positive end of the
conductor under the influence of an external electric field ”.
(a) Current carriers in solid conductors: 4 –1
1. The drift velocity of electrons is of the order of 10 ms
1. In solid conductors like metals, the valence electrons
2. If V is the potential difference applied across the ends
of the atoms do not remain attached to individual
of the conductor of length l, the magnitude of electric
atoms but are free to move throughout the volume of
Potential difference V
the conductor which are also known as free electrons. field set up is E
length
2. Under the effect of an external electric field, the
E
valence electrons move in a definite direction causing
electric current in the conductors.
– +
3. Thus, free electrons are the current carriers in solid
conductors.
Fig. 3.7
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131
CURRENT ELECTRICITY
3. Each free electron in the conductor experiences a 10. If cross-section is constant, I J i.e. for a given cross-
force, F e E. sectional area, greater the current density, larger will
eE be current.
4. The acceleration of each electron is a
m 11. The drift velocity of electrons is small because of the
5. At any instant of time, the velocity acquired by frequent collisions suffered by electrons.
12. The small value of drift velocity produces a large
electron having thermal velocity u 1 will be
amount of electric current, due to the presence of
v1 u1 a1
extremely large number of free electrons in a
where 1 is the time elapsed since it has suffered its
conductor. The propagation of current is almost at the
last collision with ion/atom of the conductor it is also
speed of light and involves electromagnetic process. It
known as relaxation time.
is due to this reason that the electric bulb glows
6. Similarly, the velocities acquired by other electrons in
immediately when switch is turned on.
the conductor will be
13. In the absence of electric field, the paths of electrons
v2 u 2 a2 , v3 u 3 a3 , ....., vn u n an . between successive collisions are straight line while in
7. The average velocity of all the free electrons in the presence of electric field the paths are generally
conductor under the effect of external electric field is curved.
the drift velocity v d of the free electrons. 14. Free electron density in a metal is given by
v1 v2 ... vn NA x d
Thus, vd n
n A
u1 a1 u2 a2 ... un an where N A = Avogadro number, x = number of
n free electrons per atom, d = density of metal
u u 2 ... u n 1 2 ... n 0 a a and A = Atomic weight of metal.
1 a
n n
3.1 Relaxation Time (τ)
1 2 ... n
where, = average time that has
n The time interval between two successive collisions of
elapsed since each electron suffered its last collision electrons with the positive ions in the metallic lattice is
with the ion/atom of conductor and is called average defined as relaxation time
relaxation time. mean free path
8. Its value is the order of 10
–14
second. r.m.s. velocity of electrons v rms
With rise in temperature v rms increases consequently
9. Putting the value of a in the above relation, we have
e E decreases.
vd
m
3.2 Mobility
eE
Average drift speed, vd Drift velocity per unit electric field is called mobility of
m
The negative sign show that is opposite to the direction vd m2
electron i.e. It’s unit is
E volt sec
of E .
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132
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conductor, q Ane eE
We know that vd
m
4. The electric field set up across the conductor is given
by E = V/l (in magnitude) eV
But E = V/l vd
5. Due to this field, the free electrons present in the
m
Also, I = A n e vd
conductor will begin to move with a drift velocity vd
eV Ane
2
towards the left hand side as shown in figure
I=Ane V
m m
V m
or = R = a constant for a given
I A n e2
conductor for a given value of n, l and at a given
temperature. It is known as the electrical resistance of the
conductor.
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133
CURRENT ELECTRICITY
3. The device or substances which don’t obey ohm’s nature of material and temperature of the conductor.
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134
CURRENT ELECTRICITY
(v) Resistivity depends on the temperature. For metals t Volume remains constant i.e., A1l1 = A2l2
= 0(1 + t) i.e. resistivity increases with After stretching length = l2, area of cross-section = A2,
l2
temperature.(where is the temperature coefficient radius = r2, diameter = d2 and resistance R 2
A2
of resistance.)
Ratio of resistances before and after stretching
(vi) Resistivity increases with impurity and mechanical 2 2 4 4
R1 l1 A2 l1 A2 r2 d 2
stress. = × = = = =
R 2 l2 A1 l2 A1 r1 d1
(vii) Magnetic field increases the resistivity of all metals
2
except iron, cobalt and nickel. R l
(1) If length is given then R l 1 1
2
R 2 l2
(viii) Resistivity of certain substances like selenium,
4
cadmium, sulphides is inversely proportional to 1 R r
intensity of light falling upon them. (2) If radius is given then R 4
1 2
r R 2 r1
V m m
3. We have, R 2
I Ane ne A
2
If a conducting wire stretches, it’s length increases, area of Conductor conduct charges very easily. When a small
cross-section decreases so resistance increases but volume potential difference is applied across the two ends of
remain constant.
Suppose for a conducting wire before stretching it’s length = conductor, a strong current flow through the
l1, area of cross–section = A1, radius = r1, diameter = d1, and conductor. For super-conductor, the value of electrical
l1 conductivity is infinite and electrical resistivity is zero.
resistance = R 1
A1 Examples of conductors are all metals like copper,
silver, aluminium, tungsten etc.
3. Semiconductors: These are those material whose
electrical conductivity lies in between that of
Fig. 3.12 insulators and conductors.
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135
CURRENT ELECTRICITY
semiconductor due to motion of electrons and holes. Colour Letter as No. Multi Colour Toler
Examples of semiconductors are germanium, silicon No. an plier ance
etc. Aid to
memory
The value of electrical resistance R increases with rise Black B 0 100 Gold 5%
of temperature. Brown B 1 101 Silver 10%
Red R 2 102 20%
R t R0 increase in resistance
Orange O 3 103 No
R0 t original resistance × rise of temp. Yellow Y 4 104 colour
Green G 5 105
Thus, temperature coefficient of resistance is defined
Blue B 6 106
as the increase in resistance per unit original resistance Violet V 7 107
per degree celsius or kelvin rise of temperature. Grey G 8 108
White W 9 109
1. For metals like silver, copper, etc., the value of is
Gold 10-1
positive, therefore, resistance of a metal increases with Silver 10-2
rise in temperature.
–1 –1 To remember the value of colour coding used for carbon
The unit of is K or °C .
resistor, the following sentences are found to be of great
2. For insulators and semiconductors is negative, help (where bold letters stand for colours).
therefore, the resistance decreases with rise in B B ROY Green, Britain Very Good Wife Gold Silver.
Way of finding the resistance of carbon resistor from its
temperature.
colour coding.
In the system of colour coding, Strips of different colours
6.2 Non-Ohmic Devices are given on the body of the resistor, figure. The colours on
strips are noted from left to right.
Those devices which do not obey Ohm’s law are
called non-ohmic devices. For example, vaccum tubes,
semiconductor diode, liquid electrolyte, transistor etc.
For all non-ohmic devices (where there will be failure
A B C R
of Ohm’s law), V–I graph has one or more of the
Fig. 3.14
following characteristics:
(i) Colour of the first strip A from the end indicates the
1. The relation I and V is non-linear, first significant figure of resistance in ohm.
(ii) Colour of the second strip B indicate the second
significant figure of resistance in ohm.
(iii) The colour of the third strip C indicates the
multiplier, i.e., the number of zeros that will follow
after the two significant figure.
(iv) The colour of fourth strip R indicates the tolerance
limit of the resistance value or percentage accuracy of
resistance.
Fig. 3.13
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8.1 Kirchhoff’s first law or Kirchhoff’s junction law or 1. The algebraic sum of changes in potential around any
Kirchhoff’s current law. closed path of electric circuit (or closed loop)
1. The algebraic sum of the currents meeting at a junction involving resistors and cells in the loop is zero, i.e.,
in a closed electric circuit is zero, i.e., I 0 V 0.
2. Consider a junction O in the electrical circuit at which 2. In a closed loop, the algebraic sum of the emfs and
the five conductors are meeting. Let I1, I2, I3, I4 and I5 algebraic sum of the products of current and resistance
be the currents in these conductors in directions, in the various arms of the loop is zero, i.e.,
shown in figure, IR 0.
3. Kirchhoff’s second law supports the law of
I3 I2 conservation of energy, i.e., the net change in the
energy of a charge, after the charge completes a closed
path must be zero.
I1 4. Kirchhoff’s second law follows from the fact that the
electrostatic force is a conservative force and work
I4
I5 done by it in any closed path is zero.
5. Consider a closed electrical circuit as shown in figure.
containing two cells of emfs. 1 and 2 and three
3.28
resistors of resistances R1, R2 and R3.
3. Let us adopt the following sign convention: the current
flowing in a conductor towards the junction is taken as
positive and the current flowing away from the
junction is taken as negative.
4. According to Kirchhoff’s first law, at junction O
(–I1) + (–I2) + I3 + (–I4) + (–I5)= 0
or –I1 – I2 + I3 – I4 – I5 = 0
or I 0
or I3 – I5 = I1 + I2 + I4
5. Total current flowing towards the junction is equal to Fig. 3.29
We adopt the following sign convention:
total current flowing out of the junction.
1 I3R2 I1R1 0 ......(1)
6. Current cannot be stored at a junction. It means, no
point/junction in a circuit can act as a source or sink of 2 I3R2 I2 R3 0 ......(2)
charge.
7. Kirchhoff’s first law supports law of conservation of 1 2 I2 R3 I1R1 0 ......(3)
charge. Traverse a closed path of a circuit once completely in
clockwise or anticlockwise direction.
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8.3 Difference between Kirchhoff’s I and II Laws 4. If Rs is the equivalent resistance of the given series
combination of resistances, then the potential
First Law Second Law
difference across A and B is, V = IRs. We have
1. This law supports the This law supports the law
IRs = I (R1 + R2 + R3) or R s R1 R 2 R 3
law of conservation of conservation of energy.
of charge. NOTE:
2. According to this law According to this law (i) If t In a series resistance circuit, it should be noted
I0 iR that the current is same in every resistor.
3. This law can be used This law can be used in (ii) The current in the circuit is independent of the
relative positions of the various resistors in the series.
in open and closed closed circuit only.
circuits. (iii) The voltage across any resistor is directly
proportional to the resistance of the resistor.
(iv) The total resistance of the circuit is equal to the sum
9. COMBINATION OF RESISTORS of the individual resistances, plus the internal resistance
of a cell if any.
9.1 Resistances in Series (v) When the total resistance in the series circuit is
obviously more than the greatest resistances in the
Resistors are said to be connected in series, if the same
circuit
current is flowing through each resistor when some potential
difference is applied across the combination. 9.2 Resistances in Parallel
Fig. 3.15
1. Let V be the potential difference applied across A and
B using the battery . In series combination, the same
current (say I) will be passing through each resistance.
2. Let V1, V2, V3 be the potential difference across R1, R2
and R3 respectively. According to Ohm’s law Fig. 3.17
V1 = IR1, V2 = IR2, V3 = IR3 1. Let V be the potential difference applied across A and
B with the help of a battery of e.m.f. .
3. Here, V = V1 + V2 + V3 = IR1 + IR2 + IR3 = I (R1 + R2 +
R3) 2. Let I be the main current in the circuit from battery. I
divides itself into three unequal parts because the
resistances of these branches are different and I1, I2, I3
be the current through the resistances R1, R2 and R3
respectively. Then,
Fig. 3.16 I = I1 + I2 + I3
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Fig. 3.19
In the figure shown, the resistances specified are in ohms.
We wish to determine the equivalent resistance between
Fig. 3.18 point A and D. Point B and C, E and F are the the same
V = IRp or I = V/Rp potential so the circuit can be redrawn as in figure shown.
1 1 1 1
V
V
V
V or
Rp R1 R 2 R3 R p R1 R 2 R3
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Fig. 3.26
Fig. 3.23 Find the effective resistance between A & B?
Example: In the circuit shown in figure (a) find the current Sol. Let the effective resistance between A & B be RE since
flowing through the 100 resistor connecting points U and the network is infinite long, removal of one cell from the
chain will not change the network. The effective resistance
S.
between points C & D would also be RE.
Sol. Figure (b) shows simplified circuit. The battery is
The equivalent network will be as shown below
directly attached to resistor 90 hence current in it is 2 A,
see figure (c), The total resistance of second branch is also
90, hence current divides equally. Now current through
45 resistor is 2 A and it is a combination of two equal 90
resistors. Once again current divides equally. 90 resistor is
a series combination of 40 and 50, hence current through Fig. 3.27
The original infinite chain is equivalent to R in series with R
them is equal, i.e.,
RR E
& RE in parallel R E = R +
R + RE
RER + RE2 = R2 + 2RRE RE2 – RRE – R2 = 0
RE =
R 1 5
2
Example: Find the equivalent resistance between A & B?
Fig. 3.24
Fig. 3.28
Sol. As moving rom one section to next one, resistance is
increasing
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The effective resistance between points C & D would be II Method : Folding Symmetry
kRE (where RE is the effective resistance)
∴ The potential difference in R between (B, C) and between
R kR E (B, D) is same VC = VD
∴ Effective R between A & B R E = R +
kR E + R
On solving we get
R – 2kR
2
2kR – R + + 4R 2
RE =
2k
9.3.2. Symmetrical Circuits : Fig. 3.32
Some circuits can be modified to have simpler solution by Hence the point C and D are same hence circuit can be
using symmetry if they are solved by traditional method of simplified as. This called folding.
KVL and KCL then it would take much time. Now , it is Balanced Wheatstone bridge
Example: Find the equivalent Resistance between A and B
Fig. 3.33
Fig. 3.30
2R × R 2R
Sol. I Method : Mirror Symmetry R eq = =
2R + R 3
The branches AC and AD are symmetrical
In II Method it is not necessary to know the currents in CA
∴ current through them will be same.
and DA
The circuit is also similar from left side and right side like
Example:
mirror images with a mirror placed alone CD therefore
Find the equivalent Resistance between A and B
current distribution while entering through B and an exiting
from A will be same. Using all these facts the currents are as
shown in the figure. It is clear that current in resistor
between C and E is 0 and also in ED is 0. It's equivalent is
shown in figure (b)
Fig. 3.34
In this case the circuit has symmetry in the two branches
AC and AD at the input
∴ current in them are same but from input and from exit the
Fig. 3.31 circuit is not similar (∵ on left R and on right 2R)
2R
R eq =
3
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Fig. 3.38
Here VA = VC and VB = VD
Here the circuit can be simplified as
Fig. 3.35
Here Vc = Vd
hence C and D are same point
So, the circuit can be simplified as
Now it is balanced wheat stone bridge.
Fig. 3.39
Sol.
Fig. 3.36
3R 9
3R × R Fig. 3.40
R eq = 2 = 2 = R
3R 9
3R +
2 2
Example: Find the equivalent Resistance between A and B
Fig. 3.41
Example:
Twelve equal resistors each R are connected to form the
edges of a cube. Find the equivalent resistances of the
network.
Fig. 3.37
Sol.
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Fig. 3.42
(a) When current enters at 1 & leaves at 6 (body diagonal)
Sol. Here 2, 4, 8 are equipotential points (if we move from
1 2, 4, 8 it comes along the edge & 62, 4, 8 it comes along
face diagonal). Similarly 3, 5, 7 are equipotential points.
Fig. 3.45
10. CELL
Fig. 3.43 The device which converts chemical energy into electrical
(b) When current enters at 1 and leaves at 2 energy is known as electric cell.
Sol. Here 3, 7 are equipotential surface (if we move from 1 Cell is a source of constant emf but not constant current.
Fig. 3.46
1. Emf of cell (E) : The potential difference across the
i.e. V = iR.
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R=0
E, r
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Fig. 3.51
1. Series grouping: In series grouping anode of one
cell is connected to cathode of other cell and so on. If
n identical cells are connected in series
10.2 Distinction between E.M.F. and Potential
Difference
E, r E, r E, r E, r
.R
the source of electric is measured between R nr
current. any two points of the (vii) Condition for maximum power R = nr and
circuit. E2
Pmax n
4r
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E, r V
E, r R
E, r
Fig. 3.54
i
R (i) Equivalent emf of the combination Eeq = nE
(ii) Equivalent internal resistance of the combination
Fig. 3.53 nr
req
(i) Equivalent emf Eeq = E m
(ii) Equivalent internal resistance Req = r/n
(iii) Main current flowing through the load
E
(iii) Main current i
Rr/ n nE mnE
i
(iv) Potential difference across external resistance = p.d. nr mR nr
R
across each cell = V = iR m
i (iv) Potential difference across load V = iR
(v) Current from each cell i '
n V
(v) Potential difference across each cell V '
(vi) Power dissipated in the circuit P E
2
n
.R
R r/n i
(vi) Current from each cell i '
(vii) Condition for max. power is n
nr
E2 (vii) Condition for maximum power R and
R r / n and Pmax n m
4r E2
Pmax (mn)
(viii) This type of combination is used when nr >> R 4r
(viii) Total number of cell = mn
Generalized Parallel Battery
E1 E 2 E NOTE:
... n
r r2 rn
E eq 1 and 1 1 1 ... 1 . (i) If the two cells connected in parallel are of the same
1 1 1 req r1 r2 rn
... emf and same internal resistance r, then
r1 r2 rn
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equivalent emf of all the cells is equal to the emf of one 11.1 Electric Power
cell. The rate at which electrical energy is dissipated into other
1 1 1 n or r = r/n forms of energy is called electrical power i.e.
... n term s eq
req r r r W V2
P= = Vi = i2 R =
t R
1. Units: It’s S.I. unit is Joule/sec or Watt. Bigger S.I.
11. HEATING EFFECT OF CURRENT units are KW, MW and HP, remember 1 HP = 746 Watt
2. Rated values: On electrical appliances (Bulbs, Heater
When some potential difference V is applied across a … etc.)
resistance R then the work done by the electric field on
charge q to flow through the circuit in time t will be
40 W
V2 t
W = qV = Vit = i2Rt . Joule 220 V
R
Fig. 3.56
Wattage, voltage, ……. etc. are printed called rated
3.55
values e.g. If suppose we have a bulb of 40 W, 220 V
This work appears as thermal energy in the resistor.
then rated power (PR) = 40 W while rated voltage (VR)
Heat produced by the resistance R is
= 220 V. It means that on operating the bulb at 220 volt,
2 2
W Vit i Rt V t the power dissipated will be 40 W or in other words 40
H Cal. This relation is called
J 4 2 4 2 4 2R
J of electrical energy will be converted into heat and
Joules heating. light per second.
Some important relations for solving objective questions are 3. Resistance of electrical appliance: If variation of
as follow:
resistance with temperature is neglected then resistance
Condition Graph
If R and t are constant of any electrical appliance can be calculated by rated
H i2 and H V2 2
power and rated voltage i.e. R = V R by using e.g.
PR
VA2 V R2
then Pconsumed also we have R
R PR
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If 100 W, 220 V bulb operates on 110 volt supply then 11.2 Maximum power transfer theorem
2
110
Pconsum ed
Power dissipated in external resistance (load)
10 0 25 W
220 2
V2 E
P Vi i2 R .R
R Rr
NOTE:
Power delivered will be maximum when
If VA < VR then % drop in output power
E2
(P P ) R = r so Pmax .
R consumed 100 4r
PR
This statement is called “maximum power transfer
For the series combination of bulbs, current through
theorem”.
them will be same so they will consume power in the
ratio of resistance i.e., P R {By P = i2R) while if they
are connected in parallel i.e. V is constant so power
consumed by them is in the reverse ratio of their
1
resistance i.e. P .
R
Fig. 3.57
5. Thickness of filament of bulb: We know that resistance
of filament of bulb is given by VR2 also
R , 11.3 Electric Energy
PR
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Note that the number of units of electricity consumed = No. 1 i.e. in parallel combination
V R
watt hour Prated bulb of greater wattage will
of kWh =
1000 give more bright light and
Electric energy VI t I 2 Rt V 2 t / R i.e. in series more current will pass
combination bulb of through it.
11.4 Electricity Consumption
lesser wattage will
1. The price of electricity consumed is calculated on the give more bright light
basis of electrical energy and not on the basis of and p.d. appeared
electrical power. across it will be
2. The unit Joule for energy is very small hence a big more.
practical unit is considered known as kilowatt hour
(KWH) or board of trade unit (B.T.U.) or simple
12. EXPERIMENTS
unit.
3. 1 KWH or 1 unit is the quantity of electrical energy 12.1 Galvanometer
which dissipates in one hour in an electrical circuit It is an instrument used to detect small current passing
when the electrical power in the circuit is 1 KWH through it by showing deflection. Galvanometers are of
thus 1 KWH = 1000 W 3600 sec = 3.6 10 J. 6
different types e.g. moving coil galvanometer, moving
3. Important formulae to calculate the no. of consumed magnet galvanometer, hot wire galvanometer. In dc circuit
units is usually moving coil galvanometer is used.
Total watt Total hours
n (i) It’s symbol: ; where G is the total
1000 internal resistance of the galvanometer.
(ii) Full scale deflection current: The current required
11.5 Combination of Bulbs (or Electrical Appliances) for full scale deflection in a galvanometer is called
Bulbs (Heater etc.) Bulbs (Heater etc.) full scale deflection current and is represented by ig.
are in series are in parallel (iii) Shunt: The small resistance connected in parallel to
(1) Total power Total power consumed galvanometer coil, in order to control current flowing
consumed Ptotal = P1 + P2 + P3 .... + Pn through the galvanometer is known as shunt.
1 1 1 P1 TABLE: Merits and demerits of shunt
....
Ptotal P1 P2 Merits of shunt Demerits of shunt
P2
P1 P2 To protect the galvanometer Shunt resistance
Supply
coil from burning. decreases the
Supply It can be used to convert any sensitivity of
galvanometer into ammeter galvanometer.
(2) In ‘n’ bulbs are If ‘n’ identical bulbs are in of desired range.
identical, parallel. Ptotal = nP
12.2 Ammeter
P Pconsumed Brightness It is a device used to measure current and is always
Ptotal
N connected in series with the ‘element’ through which current
1 is to be measured.
Pconsumed Brightness PR i
R
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R
V
i A R
+ – + –
V V
Fig. 3.59
Fig. 3.61
GS
1. Equivalent resistance of the combination (a) Equivalent resistance of the combination = G + R
G S
(b) According to ohm’s law Maximum reading of V
2. G and S are parallel to each other hence both will
which can be taken V = ig (G + R); which gives
have equal potential difference i.e. i g G (i i g ) S ;
V V
which gives Required shunt S = ig Required series resistance R = – G = –1 G
G
ig V
(i – i g ) g
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P R P R
In that case Dividing (3) by (4), we get
Q S Q S
arm.
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5. Across one gap, a resistance box R and in another gap length of the wire. The potentiometer is provided with
the unknown resistance S are connected. a jockey J with the help of which, the contact can be
6. The positive pole of the battery E is connected to made at any point on the wire, figure. A battery
terminal A and the negative pole of the battery to (called driving cell), connected across A and B sends
terminal C through one way key K. the current through the wire which is kept constant by
7. The circuit is now exactly the same as that of the using a rheostat Rh.
Wheatstone bridge figure.
Fig. 3.64
Adjust the position of jockey on the wire (say at B) the fact that the fall of potential across any portion of
where on pressing, galvanometer shows no deflection. the wire is directly proportional to the length of that
Note the length AB ( = l say) to the wire. Find the portion provided the wire is of uniform area of cross-
length BC ( = 100 – l) of the wire. section and a constant current is flowing through it.
P = resistance of the length l of the wire AB = lr Let V be the potential difference across the portion of
Q = resistance of the length (100 – l) of the wire BC = the wire of length l whose resistance is R.
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A battery of emf is connected between the end terminals A 3. If r is the resistance of potentiometer wire of length L,
and B of potentiometer wire with ammeter A1, resistance
then current through potentiometer wire is I
box R and key K in series. This circuit is called an auxillary Rr
circuit. The ends of resistance R1 are connected to terminals Potential drop across potentiometer wire. Potential
A and Jockey J through galvanometer G. A cell 1 and key gradient of potentiometer wire, i.e., fall of potential per
Fig. 3.66
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12.9 Precautions of Experiment potentiometer wire. Now emf of the cell, = potential
1. The current in the potentiometer wire from driving cell difference across the length l1 of the potentiometer wire.
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becomes infinite. (Here Q and V are the instantaneous values of charge and
4. Its sensitivity is Its sensitivity is low. potential difference while maximum charge on capacitor is
high. Q0 CV0 )
5. It is based on null It is based on deflection (ii) Discharging:
deflection method. method. VC + V R = 0
6. It can be used for It can be used only to Charge at any instant
Q = Q0e–t/RC
various purposes. measure emf or potential
At t = = RC = time constant
difference.
Q = Q0 e–1 = 0.368 Q0
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V V V0 1
( 0.37 ) or during discharging it is defined as the time
i e
Fig. 3.678
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SOLVED EXAMPLES
Example - 1 Example - 3
The charge flowing in a conductor varies with time as, A wire of resistance 5 is drawn out so that its length is
1 1
q at bt 2 ct 3
increased by twice its original length. Calculate its new
2 6
where a, b, c are positive constants.
resistance.
Then, f ind
Sol. Here, R1 = 5; l1 = l say ; A1 = A, say
(i) the initial current
(ii) the time after which the value of curent reaches a R2 = ? ; l2 = 2l + l = 3l; A2 = ?
maximum value
If V is the volumne of the wire of length l and area of cross-
(iii) the maximum or minimum value of current.
section A, then
Sol.
(i) Current, V = Al or A = V/l
dq d 1 1 1
i at bt 2 ct 3 a bt ct 2 ...(i) 2
dt dt 2 6 2 Now R R 2
A V/ V
When t = 0, i = a
di [ and V are constants]
(ii) b ct.
dt
or R 2 3 9
R 2 22 2
For i to be maximum or minimum, Hence,
R1 12 5 2
di b
0 b ct or t
dt c or R2 = 5 × 9 = 45 .
(iii) Putting this value of t in (i), we have
Example - 4
b 1 b2 b2 b2 b2
i = a – b × c a a .
c 2 c c 2c 2c A copper wire is stretched to make it 0.2% longer. What is
As this value of i is less than that at t = 0, it must be minimum. the percentage change in its resistance ?
Example - 5 Sol. We know that number for brown colour is 1, for violet colour
is 7 and that of yellow is 4. For gold, accuracy is 5%.
1 kg piece of copper is drawn into a wire 1 mm thick, and
another piece into a wire 2 mm thick. Compare the resistance Resistance of resistor shown in figure.
of these wires. 4
= 17 × 10 ± 5%
(a) 2 : 1 (b) 4 : 1
(c) 8 : 1 (d) 16 : 1 = 0.17 Megaohm ± 5%
Ans. (d) Example - 8
Sol. As volume will remains same in case of both plese.
The resistance of a tungsten filament at 150° C is 133.
A1 l1 A2 l2
What will be its resistance at 500°C ? The temperature
–1
l1 106 l2 4 106 coefficient of resistance of tungsten is 0.0045°C at 0°C.
–1
l1 4l2 Sol. Here; R150 = 133 , R500 = ?, = 0.0045°C
l We know, Rt = R0 (1 + t)
1
R1 A1 l A 4l 4 106 16
1 2 2 R150 = R0 (1 + × 150)
R2 l2 A1 l2 l2 10 6 1
A2
or 133 = R0 (1 + 0.0045 × 150) ...(i)
Example - 6
And R500 = R0 (1 + × 500)
A wire of resistance R is cut into n equal parts. These parts
or R500 = R0 (1 + 0.0045 × 500) ...(ii)
are then connected in parallel. The equivalent resistance of
the combination will be Dividing (ii) by (i) we get
(a) nR (b) R/n
2
(c) n/R (d) R/n . R 500 1 0.0045 500 3.25
Ans. (d)
133 1 0.0045 150 1.675
Sol. Resistance of wire=R
If the wire is cut into n equal parts, resistance of each wire is
3.25
R or R 500 133 258.1
and connected in parallel 1.675
n
R Example - 9
R eq
n2
A uniform wire or resistance R is shaped into a regular n
Example - 7 sided polygon where n is even. Find the equivalent
resistance between (i) opposite corners of polygon (ii)
The colour coded resistor is shown in figure. Compute adjacent corners of polygon.
the resistance in megaohm.
Sol. Let the polygon be as shown in figure. The resistance of
each side of polygon = R/n.
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R / n n 1 R / n n 1 Sol.
equivalent resistance is R
R / n n 1 R / n n2 (a) For maximum effective resistance, the n resistors must be
connected in series.
Example - 10
Maximum effective resistance, in Rs = nR.
Six equal resistances each of 4 ohm are connected to form
a network as shown in figure. What is the resistance For minimum effective resistance, the n resistors must be
between A and B ? connectedd in parallel
Rs nR
n2
Rp R / n
2
Equivalent resistance of and 3 in series
3
2 11
3
3 3
Sol.
E = E2 – E1 = 40 V – 10V = 30 V
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Example - 21
R1 = 100 Ω , R2 = R3 = 50 Ω ,
R4 =75 Ω , ε = 4.75 V..
Work out the equivalent resistance of the circuit and the
current in each resistor.
1 1 1 1 1 1 1
R p R 2 R 3 R 4 50 50 75
emf
current through the cell will be = i =
net resistance
6 6 4 16 4
300 300 75
50 25
i= A.
or Rp = 75/4 . 20×5 7
10+
20+5
Total resistance of circuit = R1 + Rp = 100 + 75/4 = 475/4
25 100
Voltage across 5 = 50 – × 10 =
4.75 7 7
Current, i1 0.04 A;
475 / 4
100 20
I2 A
Pot drop across R1 = i1R1 = 0.04 × 100 = 4V 75 7
Example - 24 Example - 27
Ans. (a)
A voltmeter has a resistance of 20000. When connected Sol. While the slide is in the middle of the potetiometer only half
in series with a large resistance R across 110 V line, the of its resistance R0/2) will be between the points A and B.
meter reads 5 V. Find the resistance R. Hence, the total resistance between A and B, say, R1, will be
Sol. p.d. across voltmeter is 5 V. given by the following expression :
1 1 1
R1 R R 0 / 2
R0R
R1
R 0 2R
The total resistance betwen A and C will be sum of resistance
between A and B and B and C, i.e., R1 + R0/2
The current flowing through the potentiometer will be
Rv 20000 V 2V
5 110 110 I
R Rv R 20000 R1 R 0 / 2 2R1 R 0
R = 420000 = 420 k The voltage V1 taken from the potentiometer will be the
product of current I and resistance R1.
Example - 26
2V
V1 IR1 R1
Why do we prefer a potentiometer to measure emf of a cell
2R1 R 0
rather than a voltmeter ?
Substituting for R1, we have a
Sol. A potentiometer does not draw any current from the cell
2V R0 R
whose emf is to be determined, wheereas a voltmeter always V1
draws some little current. Therefore, emf measured by voltmeter R R R 0 2R
2 0 R0
is slightly less then actual value of emf of the cell. R 0 2R
2VR
V1
2R R 0 2R
or V1 2VR
R 0 4R
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Example - 28 Example - 29
When two known resistance R and S are connected in the Name the physical quantity which has its unit joule
left and right gaps of a metre bridge, the balance point is –1
coulomb . Is it a scalar or vector quantity ?
found at a distance l1 from the zero end of the metre bridge –1
wire. An unknown resistance X is now connected in Sol. A physical quantity with unit, Jc = workdone/charge =
parallel to the resistance S and the balance point is now potential difference. It is a scalar quantity.
found at a distance l2 from the zero end of the metre bridge
wire figure. Obtain a formula for X in terms of l1, l2 and S. Example - 30
Two bulbs are marked 220 V – 100 watt and 220 V – 50 watt
respectively. They are connected in series to 220 V mains.
Find the ratio of heats generated in them.
Sol. Here, P1 = 100 W, P2 = 50 W, V = 220 V
Sol. When resistance R and S are connected to the left and right 2 2
R1 = (200) /100 = 484 and R2 = (220) /50 = 968 .
gaps of metre bridge and bridge is balanced at length l1 from
zero end, then
When bulbs are connected in series to mains, there will be
R 1 same current I in each bulb. Then
...(i)
S 100 1
2 2 –1 2 2 –1
H1 = I R1 = I × 484 Js and H2 = I R2 = I × 968 Js
When unknown resistance X is connected in parallel to S,
R 2
Example - 31
S 100 2
In figure, the emf of the cell is 1.8 V and internal resistance is
Putting the value of S’, we have
2/3 , calculate the current in the 3 resistance and the
R S X 2 power dissiplated in the whole circuit.
...(iii)
SX 2
100
Dividing (iii) by (i), we get
S X 2 100 1 or S 100 1
1 2
X 100 2 1 X 1 100 2
1 100 2
or X S
100 2 1
165
CURRENT ELECTRICITY
2
42 2 1 0.2 106
P = I (R + r) = (0.89) = 1.6 watt
2
n 6.25 1011 / s .
31 3 q 2 1.6 1019
Example - 32 No. of –particles striking a surface in time 4 second
11 12
To reduce the brightness of a light bulb, should an = nt = 6.25 × 10 × 4 = 2.5 × 10 .
auxiliary resistance be connected in series with it or in Note :
parallel ? There is no significance of angle (= 60°) for finding the
number of –particles striking the surface.
Sol. To reduce the brightness of a light bulb, we should decreases
the current flowing through the bulb, which is possible when (ii) If v is the velocity of –particle while travelling towards a
an auxiliary resistance is connected in series with the bulb. surface, then
1 2E 2 83.5 1.6 1016
Example - 33 E mv 2 or v
6
= 2 × 10 m/s
2 m 6.68 1027
Three resistances are connected in series across 12 volt 6
It means a beam of length v = 2 × 10 m crosses a section in
battery. The first resistor has value of 1 ohm second has one second. But number of –particle passing through a
voltage drop of 4 volt and third has power dissipation of 11
section in one second is, n = 6.25 × 10
12 watt. What is circuit current.
(a) 1A (b) 2A n
No. of –particles in until length of the beam
(c) 3A (d) 4A v
Ans. (b) No. of –particles in length l of the beam
Sol. V V1 V2 V3
n 6.25 1011
0.20 = 6.25 × 104
12 v 2 106
12 I 1 4 { Since, P=VI}
I
166
CURRENT ELECTRICITY
(a) zero (b) 2A
(c) 5A (d) 1 A
(a) zero (b) 20 V
Resistance & Variation with Temperature
(c) 60 V (d) 120 V
4. In figure, if the potential at point P is 100 V, what is the 8. Thermistors are usually made of
magnitude of potential(in V) at point Q ? (a) metals with low temperature coefficient of resistivity
(b) metals with high temperature coefficient of resistivity
(c) metal oxides with high temperature coefficient of
resistivity
(d) semiconductors with low temperature coefficient of
resistivity
9. A conductor behaves as a super conductor
(a) above critical temperature
(b) at critical temperature
(c) at 40 C°
(d) at boiling point of that metal
167
CURRENT ELECTRICITY
2
10. Which of the following material has negative temperature 18. A nichrome wire 1 m long and 1 mm in cross-section area
coefficient of resistance. draws 4 A current at 2 volt. The resistivity of nichrome is
–7 –7
(a) brass (b) copper (a) 1 × 10 –m (b) 2 × 10 –m
–7 –7
(c) porcelain (d) carbon (c) 4 × 10 –m (d) 5 × 10 –m
11. Which of the following material has nearly zero temperature 19. The temperature coefficient of resistance for a wire is
–1
0.00125°C . At 300 K its resistance is 1 ohm. The
coefficient of resistance.
temperature at which the resistance becomes 1.5 ohm is
(a) carbon (b) porcelain
(a) 450 K (b) 727 K
(c) copper (d) manganin
(c) 454 K (d) 900 K
12. A piece of copper and silicon are cooled from room
20. You need to produce a set of cylindrical copper wires 3.50
temperature to 100 K. The resistance of
m long that will have a resistance of 0.125 each. What
(a) each of them increases
will be the mass (in gram) of each of these wires? Given
(b) each of them decreases
that resistivity of copper is 1.72 108 m, density of
(c) copper increases and that of silicon decreases
(d) copper decreases and that of silicon increases copper 8.9 103 kg / m3 .
13. On increasing the temperature of a conductor, its resistance 21. Two plates of same material R and S are in the form of a
increases because square and have the same thickness. A side of S is twice
(a) relaxation time decreases the side of R. Compare their resistances. The direction of
current is shown by an arrow head.
(b) mass of the electron increases
(c) electron density decreases
(d) none of the above
14. A resistance of 2 is to be made from a copper wire
–8 (a) The resistance of R is twice that of S
(specific resistance 1.7 × 10 m) using a wire of length
(b) Both have the same resistance
50 cm. The radius of the wire is
(c) The resistance of S is four times that of R
(a) 0.0116 mm (b) 0.0367 mm
(d) The resistance of R is half to that of S.
(c) 0.116 mm (d) 0.367 mm
3 22. The resistance of a 10 m long wire is 10 . Its length is
15. The specific resistance of a wire is , its volume is 3 m and
increased by 25% by stretching the wire uniformly. The
its resistance is 3 ohm, then its length will be
resistance of wire will change to (approximately)
(a) 1/ (b) 3 / (a) 12.5 (b) 14.5
(c) 15.6 (d) 16.6
(c) 3 / (d) / 3
23. The length of conductor is doubled and its radius is halved,
16. The heating element of an electric heater should be made its specific resistance is
with a material which should have (a) unchanged (b) halved
(a) High specific resistance and high melting point (c) doubled (d) uadrupled
(b) High specific resistance and low melting point 24. A coil has resistance of 18 when its mean temperature
(c) Low specific resistance and high melting point is 20oC and of 20 when its mean temperature is 50 oC.
(d) Low specific resistance and low melting point What will be mean temperature rise when its resistance is
17. As the temperature of a conductor increases, its resistivity 21 and the surrounding temperature is 15oC.
and conductivity changes .The ratio of resistivity to (a) 40oC (b) 50oC
conductivity (c) 60oC (d) 70oC
(a) Increases 25. All the edges of a block with parallel faces are unequal. Its
(b) decreases longest edge is twice its shortest edge. The ratio of the
(c) remains constant maximum to minimum resistance between parallel faces is
(d) may increase or decrease depending on material (a) 2 (b) 4
(c) 8 (d) 16
CURRENT ELECTRICITY 168
Combination of Resistors 31. A wire of resistance 6 is cut into three equal pieces,
which are joined to form a triangle. The equivalent
26. The approx effective resistance between A and B in figure resistance between any two corners of the triangle is
is (a) (3/4) (b) (4/3)
(c) 2 (d) 4
32. Two resistances are joined in parallel whose resistance is
3/5 . One of the resistance wire is broken and the effective
resistance becomes 3 . The resistance in ohm of the wire
that got broken was
(a) 10 (b) 12 (a) 4/3 (b) 2
(c) 9.81 (d) 10.85 (c) 6/5 (d) 3/4
33. You are given three eual resistors. How many resistances
can be obtained by joining them in series and parallel
27. What is the total resistance of the circuit ?
grouping ?
(a) 6 (b) 4
(c) 3 (d) 2
34. What is the equivalent resistance between A and B in the
circuit of figure, if R = 3 ,
(a) 6 (b) 7
(c) 8 (d) 9
28. What will be the resistance between P and Q in the
following circuit?
(a) 8 (b) 9
(c) 12 (d) 15
35. One end of copper wire of length 2L and cross sectional
area A is attached to one end of another copper wire of
length L and cross section area 2A. If the free end of the
(a) (1/3) (b) (2/3) longer wire is at electric potential of 8 volt and free end of
(c) 2 (d) 5 longer wire is at potential of 1 volt ,what is the potential of
29. What is the current I in the circuit as shown in figure junction of two wires.
(a) 1.2 V (b) 2.1 V
(c) 2.4 V (d) 3.6 V
36. Consider the following statements regarding the network
shown in the figure
37. Calculate battery current (in amp) of the network shown 42. In a Wheatstone’s bridge, three resistances P, Q and R are
in figure. connected in the three arms and the fourth arm is formed by
two resistances S1 & S2 connected in parallel. The condition
for the bridge to be balanced will be
P 2R P R S1 S2
(a) (b)
Q S1 S2 Q S1S2
P R S1 S2 P R
(c) (d) Q S S
Q 2S1S2 1 2
38. Total current supplied to the circuit by the battery is
43. The Wheatstone’s bridge shown in figure is balanced. If
the position of the cell E and galvanometer G are now
interchanged, G will show zero deflection
(a) 4A (b) 6A
(c) 2A (d) 1A
39. A uniform wire of resistance 36 is bent in the form of a
circle. The effective resistance across the points A and B
is (a) only if all resistor are equal
(b) only if R1 = R3 and R2 = R4
R1 R
(c) only if = 2
R4 R3
(d) in all cases
44. An electric circuit is shown in figure. Calculate the potential
(a) 36 (b) 18
difference (in V) across the resistor of 400 , as it will be
(c) 9 (d) 2.75
measured by the voltmeter V of resistance 400 .
40. The figure shows currents in a part of an electric circuit,
then current I is
2A
1A
1.3A
2A
I
(a) 1.7 A (b) 3.7 A
(c) 1.3 A (d) 1 A
41. In the figure shown below, the electric current flowing
through 2R resistor is 45. The figure shows a network of currents. The magnitude of
currents is shown here. The current I will be
B
A D
R 2R C R
46. An infinite ladder network of resistances is constructed 52. For a cell terminal potential difference is 2.2 volt when
with 1 and 2 resistances, as shown in figure.The 6 V circuit is open and reduces to 1.8 volt when cell is
battery between A and B has negligible internal resistance : connected to an external resistance of 5 ohm.What is the
internal resistance of the cell ?
10 10
(a) (b)
9 19
1 5
(i) Show that the effective resistance between A & B is 2 . (c) (d)
9 9
(ii) What is the current that passes through the 2
resistance nearest to the battery ? 53. Two identical cells connected in series send 1.0 amp of
current through a 5 ohm resistance.When they are
47. In the circuit shown, some potential difference is applied connected in parallel, they send current of 0.8 amp through
between A & B. The euivalent resistance between A & B is the same resistor. What is the internal resistance of the cell ?
(a) 0.5 (b) 1
(c) 2.5 (d) 4
A B
54. A d.c. main supply of e.m.f. 220 V is connected across a
storage battery of e.m.f 200 V through a resistance of 1.
The battery terminals are connected to an external
(a) 14 (b) 12.5 resistance ‘R’. The minimum value of ‘R’, so that a current
(c) 3.6 (d) 2.1 passes through the battery to charge it is:
(a) 7 (b) 9
Cells
(c) 11 (d) Zero
48. Two real batteries are connected in series. Consider the 55. A battery of EMF 5V and internal resistance 20 is
following statements (i) The euivalent e.m.f. is larger than
connected with a resistance R1 50 and a resistance
either of two emf’s (ii) The euivalent internal resistance is
smaller than either of two internal resistances. R2 40. A voltmeter of resistance 1000 is used to
(a) Both statements are correct measure the potential difference across R1 . What
(b) Statement (i) is correct but statement (ii) is wrong percentage error is made in the reading.
(c) Statement (ii) is correct but statement (i) is wrong
(d) Both are wrong statements Heating Effects of Current
49. How will you connect 24 cells each of internal resistance 56. In a large building, there are 15 bulbs of 40 W, 5 bulbs of
of 1so as to get maximum power output across a load of 100 W, 5 fans of 80 W and 1 heater of 1 kW. The voltage of
1.50 the electric mains is 220 V. The minimum capacity of the
(a) 4 rows, 6 cells in a row main fuse of the building will be :
(b) 12 rows, two cells in a row (a) 10 A (b) 12 A
(c) 2 rows, 12 cells in a row (c) 14 A (d) 8 A
(d) 3 rows, 8 cells in a row 57. If two bulbs of wattage 25 and 30 watt each rated at 220
50. An idealized voltmeter is connected across the terminals volt are connected in series with 440 volt supply. Which
bulb will fuse ?
of a 15 volt battery and a 7.5 ohm appliance is also
connected across its terminals .If the voltmeter reads 12.5 (a) 25 watt bulb (b) 30 watt bulb
volt .What is the internal resistance of the battery. (c) neither of them (d) both of them
(a) 1.0 (b) 1.2 58. A 200 volt,1000 watt bulb is connected across 100 volt
main supply.What will be power consumed.
(c) 1.4 (d) 1.5
(a) 200 watt (b) 250 watt
51. Five cells each of e.m.f. E and internal resistance r are
connected in series. If due to oversight one cell gets (c) 500 watt (d) 750 watt
connected wrongly ,then the euivalent e.m.f. and internal 59. When three identical bulbs of 60 watt, 200 volt rating are
resistance of this combination is connected in series to a 200 volt supply, the power drawn
by them
(a) 5E,5r (b) 4E,4r
(a) 60 watt (b)180 watt
(c) 3E,4r (d) 3E,5r
(c) 10 watt (d) 20 watt
171
CURRENT ELECTRICITY
61. A 100 watt,110 volt and a 50 watt,110 volt lamps are (a) it does not get overheated
connected in series across 220 volt d.c. source. If the (b) it does not draw excessive current
resistances of two lamps are assumed to remain constant, (c) it does not appreciably change the current to measure
the voltage across 100 watt lamps is (d) it can measure large current
(a) 100 volt (b) 143.3 volt 69. In a potentiometer experiment, there is no current at the
(c) 73.3 volt (d) 200 volt balance point in
62. Two electric bulbs marked 25 W–220 V and 100 W-220 V (a) main battery circuit
are connected in series to a 440 V supply. Which of the
(b) galvanometer circuit
bulbs will fuse ?
(c) potentiometer circuit
(a) Both (b) 100 W
(d) both main and galvanometer circuit
(c) 25 W (d) Neither
70. The infinity resistance plug in a post office box (resistance
63. 50 electric bulbs are connected in series across 200 volt
box) has
supply and illumination produced is I1. 5 bulbs get fused.
If the remaining bulbs are again connected across the (a) an air gap only
source in series, the illumination produced is I2.Which of (b) a resistance coil of infinite resistance
the following is true. (c) largest resistance value in box
(a) I1= I2 (b) I1> I2 (d) resistance of value –100 ohm
(c) I1< I2 (d) anything is possible 71. In meter bridge or wheatstone bridge for measurement of
64. A coil takes 15 minutes to boil certain amount of water, resistance, the known and the unknown resistances are
another coil takes 20 minutes for the same process .Time interchanged. The error so removed is
taken to boil the same amount of water when both the (a) end correction (b) index error
coils are connected in series across same source is
(c) due to temperature effect (d) random error
(a) 5 min (b) 8.6 min
72. In the meter bridge circuit shown in figure, find the length
(c) 35 min (d)12 min (in cm) AC at null deflection in galvanometer.
65. A battery has an open circuit potential difference of 6 V
between its terminals. When a load resistance of 60 is
connected across the battery, the total power dissipated
by the battery is 0.4 W. What should be the load resistance
R(in ohm), so that maximum power will be dissipated in R.
66. The supply voltage in a room is 120 V. The resistance of
the lead wires is 6. A 60W bulb is already switched on.
What is the decrease of voltage across the bulb, when a
240 W heater is switched on in parallel to the bulb ?
(a) zero (b) 2.9 V
(c) 13.3 V (d) 10.04 V
172
CURRENT ELECTRICITY
73. Figure shows a 200 cm potentiometer wire AB with 78. The current in the primary circuit of a potentiometer wire is
–7
resistance 15 , find the balancing length (in cm) for a 3V 0.5 A, specific resistance of wire is 4 × 10 –m and area of
–6 2
cell. cross-section of wire is 8 × 10 m . The potential gradient
in the wire would be
(a) 2.5 mV/metre (b) 25 mV/metre
(c) 25 V/metre (d) 10 V/metre
79. A candidate connects a moving coil voltmeter V, a moving
coil ammeter A and a resistor R as shown in figure. If the
voltmeter reads 20 volt and the ammeter reads 4 A, the
value of R is
(a) equal to 5 ohm
(b) greater than 5 ohm
(c) less than 5 ohm
(d) may be greater or less than 5 ohm
80. A resistance of 2 is connected across one gap of a
meter bridge, the length of wire is 1 meter, and an unknown
resistance, greater than 2 is connected across the other
gap. When these resistances are interchanged, the balance
75. The length of a potentiometer wire is 5 metres. An electron
–19 point shifts by 20 cm. Neglecting any correction, find the
in this wire experiences a force of 4.8 × 10 newton, e.m.f.
unknown resistance.
of the main cell used in potentiometer is
(a) 3 (b) 4
(a) 3 volt (b) 15 volt
(c) 5 (d) 6
(c) 1.5 volt (d) 5 volt
81. A wire connected in the left gap of a meter bridge balance
76. In a potentiometer, experiment, the balancing length is at
a 10 in the right gap at a point which divides the bridge
240 cm with a cell. On shunting the cell with a resistance
wire in the ratio of 3:2.What is the resistance of the wire
of 2 the balancing length becomes 120 cm .What is the
internal resistance of the cell. (a) 10 (b) 1.2
(a) 0.5 (b) 1 (c) 15 (d) 1
(c) 2 (d) 4 82. What shunt resistance (in ohm) is required to make the
77. In the following circuit figure, the resistance of wire AB 1.00 mA, 20 galvanometer into an ammeter with a range
is 10 and its length is 1m. Rest of the quantities are shown of 0 to 50.0 mA ?
in the figure. The potential gradient on potentiometer wire will 83. A galvanometer having a coil of resistance 100 gives a
be
full scale deflection when a current of one milli- ampere is
passed through it. What is the value of resistance (in ohm)
which can convert this galvanometer into ammeter giving
a full scale deflection for a current of 10A ?
84. A galvanometer of coil resistance 20 gives a full scale
deflection with a current of 5 mA. What arrangement should
be made in order to measure current upto 1.0 A.
(a) add a series resistance of 2
(b) add a parallel resistance of 2
(a) 0.8 V/m (b) 0.08 V/m (c) add a series resistance of 0
(c) 0.008 V/m (d) none of the above (d) add a parallel resistance of 0.1
173
CURRENT ELECTRICITY
R-C Circuits
85. A resistor R and 2F capacitor in series are connected 88. The heat produced in the capacitors on closing the switch
through a switch to 200 V direct supply. Across the S is
capacitor is a neon bulb that lights up at 120V. Calculate
4F 5F
the value of R to make the bulb light up 5 s after the switch
has been closed (log10 2.5 =0.4)
20V 2
(a) 1.7×105 (b) 2.7×106 R 2
(c) 3.3×107 (d) 1.3×104 S
a resistor R. Suppose t1 is the time taken for the energy (a) 0.0002 J (b) 0.0005 J
stored in the capacitor to reduce to half its initial value and
(c) 0.00075 (d) Zero
t2 is the time taken for the charge to reduce to one-fourth
89. Four identical capacitors are connected in series with a
t1 battery of emf 10V. The point X is earthed. Than the
its initial value. Then the ratio will be
t2 potential of point A is–
1 10V
+ -
(a) 1 (b)
2
1
(c) (d) 2 C C C X C
4
A B
87. The time constant of the shown circuit for charging
is
(a) 10 V (b) 7.5 V
R C (c) –7.5 V (d) 0 V
90. A conducting solid sphere is joined in an electrical circuit
V1
V2 R as shown in figure. Two imaginary points A and B are taken
2R inside the sphere. For given conditions-
R
5 5
(a) RC (b) RC
3 2
7 7
(c) RC (d) RC
4 3
1 1
(a) (b)
4 2
3
(c) (d) 1
4
7. A 50 resistance is connected to a battery of 5 V. A
(a) 0.13 A, from Q to P (b) 0.13 A, from P to Q
galvanometer of resistance 100 is to be used as an
(c) 1.3 A, from P to Q (d) 0A ammeter to measure current through the resistance, for
3. In the electric network shown, when no current flows this a resistance rs is connected to the galvanometer. Which
through the 4 resistor in the arm EB, the potential of the following connections should be employed if the
difference between the points A and D will be : measured current is within 1% of the current without the
(2015) ammeter in the circuit? (2016)
(a) rs = 0.5 in parallel with the galvanometer
(b) rs = 0.5 in series with the galvanometer
(c) rs = 1 in series with galvanometer
(d) rs = 1 in parallel with galvanometer
8. To know the resistance G of a galvanometer by half
deflection method, a battery of emf VE and resistance R is
used to deflect the galvanometer by angle θ. If a shunt of
resistance S is needed to get half deflection then G, R and
(a) 6 V (b) 3 V S are related by the equation: (2016)
(c) 5 V (d) 4 V (a) 2S (R + G) = RG (b) S (R + G) = RG
4. A galvanometer having a coil resistance of 100 gives a (c) 2S = G (d) 2G = S
full scale deflection, when a current of 1 mA is passed 9. The resistance of an electrical toaster has a temperature
through it. The value of the resistance, which can convert
dependence given by R(T) = R0 [1 + (T T0 ) ] in its range
this galvanometer into ammeter giving a full scale deflection
for a current of 10 A, is : (2016) of operation. At T0 300K, R 100 and at T = 500 K,
(a) 2 (b) 0.1 R = 120 . The toaster is connected to a voltage source at
(c) 3 (d) 0.01 200 V and its temperature is raised at a constant rate from
5. The temperature dependence of resistance of Cu and 300 to 500 K in 30 s. The total work done in raising the
undoped Si in the temperature range 300–400 K, is best temperature is : (2016)
described by : (2016)
2
(a) 60000 l n J
6
(a) Linear increase for Cu, exponential increase for Si. (b) 200 ln J
5 3
(b) Linear increase for Cu, exponential decrease for Si.
5
(c) Linear decrease for Cu, linear decrease for Si. (c) 400 ln J (d) 300 J
6
(d) Linear increase for Cu, linear increase for Si
CURRENT ELECTRICITY 175
10. A galvanometer having a coil resistance of 100 gives a 15. A 9 V battery with internal resistance of 0.5 is connected
full scale deflection when a current of 1 mA is passed across an infinite network as shown in the figure. All
through it. The value of the resistance which can convert ammeters A1, A2, A3 and voltmeter V are ideal.
this galvanometer into ammeter giving full scale deflection (2017)
for a current of 10 A is: (2016)
(a) 0.1 (b) 0.01
(c) 100 (d) 0.001
11. In a meter bridge experiment resistances are connected as
shown in the figure. Initially resistance P = 4 and the
neutral point N is at 60 cm from A. Now an unknown
resistance R is connected in series to P and the new
position of the neutral point is at 80 cm from A. The value Choose correct statement.
of unknown resistance R is : (2017)
(a) Reading of A1 is 2A (b) Reading of A2 is 18 A
(c) Reading of V is 9 V (d) Reading of V is 7V
16. A potentiometer PQ is set up to compare two resistances
as shown in the figure. The ammeter A in the circuit reads
1.0 A when two way key K3 is open. The balance point is
at a length l1 cm from P when two way key K3 is plugged in
between 2 and 1, while the balance point is at a length l2
33 cm from P when key K3 is plugged in between 3 & 1. The
(a) (b) 6
5 R1
ratio of two resistances , is found to be : (2017)
20 R2
(c) 7 (d)
3
12. In the below circuit, the current in each resistance is:
(2017)
18. The figure shows three circuits I, II and III which are 22. On interchanging the resistances, the balance point of a
connected to a 3V battery. If the powers dissipated by the meter bridge shifts to the left by 10 cm. The resistance of
configurations I, II and III and P1, P2 and P3 respectively, their series combination is 1k. How much was the
then : (2017) resistance on the left slot before interchanging the
resistances? (2018)
(a) 500 (b) 910
(c) 990 (d) 505
23. In the given circuit all resistances are of value R ohm
each. The equivalent resistance between A and B is :
(2018)
(a) P1 P2 P3 (b) P1 P3 P2
(c) P2 P1 P3 (d) P3 P2 P1
19. In the given circuit diagram when the current reaches
steady state in the circuit, the charge on the capacitor of
capacitance C will be : (2017) (a) 2R (b) 3R
5R 5R
(c) (d)
3 2
24. In a meter bridge, as shown in the figure, it is given that
resistance Y =12.5 and that the balance is obtained at a
distance 39.5 cm from end A (by Jockey J). After
interchanging the resistances X and Y, a new balance point
is found at a distance 2 from end A. What are the values
r1
(a) CE (b) CE
r1 r of X and 2 ? (2018)
r1 r2
(c) CE (d) CE
r2 r r r2
20. Two batteries with e.m.f 12V and 13V are connected in
parallel across a load resistor of 10. The internal
resistance of the two batteries are 1 and 2 respectively.
The voltage across the load lie between: (2018)
(a) 11.4 V and11.5 V
(b) 11.7 V and11.8 V
(a) 8.16 and 60.5 cm (b) 19.15 and 39.5 cm
(c) 11.6 V and 11.7 V (c) 8.16 and 39.5 cm (d) 19.15 and 60.5 cm
(d) 11.5 V and 11.6 V 25. A copper rod of cross-sectional area A carries a uniform
21. In a potentiometer experiment, it is found that no current current I through it. At temperature T, if the volume charge
passes through the galvanometer when the terminals of density of the rod is , how long will the charges take to
the cell are connected across 52cm of the potentiometer travel a distance d ? (2018)
wire. If the cell is shunted by resistance of 5, a balance is
2 d A 2 d A
found when the cell is connected across 40 cm of the wire. (1) (2)
I IT
Find the internal resistance of the cell. (2018)
(a) 2 (b) 2.5 d A d A
(c) 1 (d) 1.5 (3) (4) I T
I
CURRENT ELECTRICITY 177
26. A constant voltage is applied between two ends of a 31. Determine the charge on the capacitor (in C ) in the
metallic wire. If the length is halved and the radius of the following circuit:
wire is doubled, the rate of heat developed in the wire will (2019)
be : (2018)
(a) Doubled (b) Halved
(c) Unchanged (d) Increased 8 times
27. A heating element has a resistance of 100 at room
temperature. When it is connected to a supply of 220 V, a
steady current of 2 A passes in it and temperature is 5000C 32. A resistance is shown in the figure. Its value and tolerance
more than room temperature. What is the temperature are given respectively by: (2019)
coefficient of resistance of the heating element? (2018)
(a) 0.5×10-4 0 C -1 (b) 5×10-4 0 C -1
35. A uniform metallic wire has a resistance of 18 and is 39. The resistance of the meter bridge AB in given figure is
bent and ends joined into an equilateral triangle. Then, 4. With a cell of emf ε = 0.5 V and rheostat resistance Rh
the resistance between any two vertices of the triangle = 2 the null point is obtained at some point J. When the
(in ) is: cell is replaced by another one of emf ε = ε2 the same null
(2019) point J is found for Rh = 6. The emf ε2 is: (2019)
36. The actual value of resistance R, shown in the figure
is 30 U. This is measured in an experiment as shown using
V
the standard formula R , where V and I are the reading
I
of the voltmeter and ammeter, respectively. If the measured
value of R is 5% less, then the internal resistance of the
voltmeter is:
(2019)
(a) 0.4 V (b) 0.3 V
(c) 0.6 V (d) 0.5 V
40. Two equal resistances when connected in series to a
battery, consume electric power of 60 W. If these
resistances are now connected in parallel combination to
the same battery, the electric power consumed will be:
(a) 600 U (b) 570 U
(2019)
(c) 35 U (d) 350 U
(a) 60 W (b) 240 W
37. A current of 2 mA was passed through an unknown
(c) 120 W (d) 30 W
resistor which dissipated a power of 4.4 W. Dissipated
power when an ideal power supply of 11. V is connected 41. In a Wheatstone bridge (see fig.), Resistances P and Q
across it is: (2019) are approximately equal. When R = 400 , the bridge is
balanced.On interchanging P and Q, the value of R, for
(a) 11105 W (b) 11103 W balance is 405 . The value of X is close to: (2019)
51. For the circuit shown, with R1 = 1.0 , R2 = 2.0 , E1 = 2 V 55. A wire of resistance R is bent to form a square ABCD as
and E2 = E3 = 4 V, the potential difference between the shown in the figure. The effective resistance between E
points ‘a’ and ‘b’ is approximately (in V) : (2019) and C is: (E is mid-point of arm CD) (2019)
12
(a)
5
5
(b)
2
59. In a meter bridge experiment, the circuit diagram and the 63. To verify Ohm’s law, a student connects the voltmeter
corresponding observation table are shown in figure. across the battery as, shown in the figure. The measured
voltage is plotted as a function of the current, and the
following graph is obtained :
1 1 1 1
(a) (b)
4 a b 2 a b
(d)
1 1 1 1
(c) (d)
2 a b 4 a b
CURRENT ELECTRICITY 182
65. The resistive network shown below is connected to a D.C. 70. A galvanometer having a coil resistance 100 gives a full
source of 16 V. The power consumed by the networkis 4 scale deflection when a current of 1 mA is passed through
W. The value of R (in ) is (2019) it. What is the value of the resistance which can convert
this galvanometer into voltmeter giving full scale deflection
for a potential difference of 10 V?In full scale deflection,
current in galvanometer of resistance is 1mA. Resistance
required in series to convert it into voltmeter of range 10V.
(2020)
(a) 7.9 K (b) 9.9 K
(c) 8.9 K (d) 10K
66. A moving coil galvanometer, having a resistance G, 71. The series combination of two batteries both of the same
produces full scale deflection when a current Ig flows emf 10V, but different internal resistance of 20Ω and 5Ω, is
through it. This galvanometer can be converted into (i) an connected to the parallel combination of two resistors 30Ω
ammeter of range 0 to I0 (I0 > Ig) by connecting a shunt and RΩ. The voltage difference across the battery of
resistance RA to it and (ii) into a voltmeter of range 0 to V internal resistance 20Ω is zero, the value of R(in Ω) is
(V = GI0) by connecting a series resistance Rv to it. Then, (2020)
(2019) 72. In the given circuit diagram, a wire is joining point B & C.
2 Find the current in this wire (2020)
I0 I g RA I g
(a) RA RV G
2
and
Ig RV I 0 I g
(b) RA RV G 2 and
2
Ig RA I 0 I g
(c) R R G 2
A V I0 I g and R I
(a) 0.4 A (b) 2 A
V g
67. The current i1 (in A) flowing through 1 resistor int A respectively. Then : (2020)
75. Model a torch battery of length to be made up of a thin 78. Two resistors 400 and 800 are connected in series
cylindrical bar of radius ‘a’ and a concentric thin cylindrical across a 6 V battery. The potential difference measured by
shell of radius ‘b’ filled in between with an electrolyte of a voltmeter of 10 k across 400 resistor is close to :
resistivity (see figure). If the battery is connected to a (2020)
resistance of value R, the maximum joule heating in R will (a) 2.05 V (b) 2 V
take place for : (2020) (c) 1.95 V (d) 1.8 V
79. A battery of 3.0 V is connected to a resistor dissipating 0.5
W of power. If the terminal voltage of the battery is 2.5V,
the power dissipated within the internal resistance is:
(2020)
(a) 0.072 w (b) 0.10 w
(c) 0.125 w (d) 0.50 w
80. The value of current i1 flowing from A to C in the circuit
diagram is: (2020)
b
(a) R (b) R 2 ln b
2l a
l a
b b
(c) R ln (d) ln
l a 2l a
76. Which of the following will NOT be observed when a
multimeter (operating in resistance measuring mode)
probes connected across a component, are just reversed?
(2020)
(a) 4A (b) 5A
(a) Multimeter shows NO deflection in both cases i.e.
(c) 2A (d) 1A
before and after reversing the probes if the chosen
component is metal wire. 81. Four resistances 40 , 60 , 90 and 110 make the
(b) Multimeter shows a deflection, accompanied by a arms of a quadrilateral ABCD. Across AC is a battery of
splash of light out of connected component in one emf 40 V and internal resistance negligible.The potential
direction and NO deflection on reversing the probes if the differenceacross BD in V is _______. (2020)
chosen component is LED.
(c) Multimeter shows an equal deflection in both cases
i.e. before and after reversing the probes if the chosen
component is resistor.
(d) Multimeter shows NO deflection in both cases i.e.
before and after reversing the probes if the chosen
component is capacitor.
77. An ideal cell of emf 10 V is connected in circuit shown in
figure. Each resistance is 2. The potential difference (in V)
82. An electrical power line, having a total resistance of 2
across the capacitor when it is fully charged is ……(2020) delivers 1 kW at 220 V. The efficiency of the transmission
line is approximately : (2020)
(a) 72% (b) 91%
(c) 85% (d) 96%
83. A galvanometer of resistance G is converted into a
voltmeter of range 0 – 1V by connecting a resistance R1 in
series with it. The additional resistance that should be
connected in series with R1 to increase the range of t h e
voltmeter to 0 – 2V will be : (2020)
(a) G (b) R1
(c) R1 – G (d) R1 + G
CURRENT ELECTRICITY 184
84. In the circuit, given in the figure currents in different 89. When the switch S, in the circuit shown, is closed then
branches and value of one resistor are shown. Then the valued of current i (in A) will be: (2021)
potential at point B with respect to the point A is : (2020)
(a) +2 V (b) –2 V 90. A carbon resistance has following colour code. What is
the value of the resistance? (2021)
(c) +1 V (d) –1 V
85. A galvanometer is used in laboratory for detecting the
null point in electrical experiments. If, on passing a current
of 6 mA it produces a deflection of 2º, its figure of merit is (a) 53× 104 ± 5% (b) 5.3 × 104 ± 5%
close to : (2020)
(c) 6.4 × 104 ± 5% (d) 64 × 104 ± 10%
(a) 6 103 A / div. (b) 3 103 A / div. 91. In the given circuit the internal resistance of the 18 V cell
is negligible. If R1 = 400, R3 = 100 and R4 = 500 and
(c) 666º A / div. (d) 333º A / div.
the reading of an ideal voltmeter across R4 is 5 V, then the
86. A circuit to verify Ohm’s law uses ammeter and voltmeter value of R2 will be: (2021)
in series or parallel connected correctly to the resistor.In
the circuit: (2020)
(a) Ammeter is always connected in series and voltmeter
in parallel
(b) Both, ammeter and voltmeter must be connected in series
(c) Both ammeter and voltmeter must be connected in
parallel
(a) 300 (b) 450
(d) ammeter is always used in parallel and voltmeter is series
(c) 550 (d) 230
87. In the figure shown, the current in the 10V battery is close
92. A moving coil galvanometer allows a full scale current of
to : (2020)
10–4 A. A series resistance of 2 MO is required to convert
the above galvanometer into a voltmeter of range 0 – 5 V.
Therefore the value of shunt resistance required to convert
the above galvanometer into an ammeter of range 0 – 10
mA is: (2021)
(a) 100 b)10
(a) 0.21 A from positive to negative terminal (c)None of these (d) 200
(b) 0.36 A from negative to positive terminal 93. In the figure given, the electric current flowing through
(c) 0.42 A from positive to negative terminal
the 5 k resistor is ‘x’ mA.
(d) 0.71 A from position to negative terminal
88. An ideal cell of emf 10 V is connected in circuit shown in
figure. Each resistance is 2. The potential difference (in
V) across the capacitor when it is fully charged is ……. .
(2020)
94. A cylindrical wire of radius 0.5 mm and conductivity 100. Two wires of same length and thickness having specific
resistances 6 and 3 respectively are connected in
5 107 S / m is subjected to an electric field of 10 m V/m. parallel. The effective resistivity is cm. The value of
The expected value of current in the wire will be x3 mA . , to the nearest integer, is ___________.
The value of x is (2021)
101. A current of 6 A enters one corner P of an equilateral
(2021)
triangle PQR having 3 wires of resistance 2 each and
95. The energy dissipated by a resistor is 10 mJ in 1 s when an
electric current of 2 mA flows through it. The resistance is leaves by the corner R. The currents i1 in ampere is
________ . (Round off to the Nearest Integer)
(2021)
96. The equivalent resistance of series combination of two
resistors is ‘s’ When they are connected in parallel, the
equivalent resistance is ‘p’. If s = np, then the minimum
value for n is (Round off to the Nearest Integer)
(2021)
97. The circuit shown in the figure consists of a charged
capacitor of capacity 3 µF and a charge of 3 µC. At time t
= 0, when the key is closed, the value of current flowing
through the 5 M resistor is ‘x’ µA. (2021)
The value of ‘x’ to the nearest integer is ___________. 102. In the given circuit of potentiometer, the potential
difference E across AB (10 m length) is larger than E1 and
E2 as well. For key K1 (closed), the jockey is adjusted to
touch the wire at point J1 so that there is no deflection in
103. A cell E1 of emf 6V and internal resistance 2 is connected 108. A current of 10 A exists in a wire of cross-sectional area of
5 mm2 with a drift velocity of 2 103 m / s . The number of
with another cell E2 of emf 4 V and internal resistance 8 free electrons in each cubic meter of the wire is ____ .
(as shown in the figure). The potential difference across (2021)
points X and Y is:
(a) 2 106 (b) 625 1025
(2021)
(c) 2 1025 (d) 1 1023
109. Two cells of emf 2E and E with internal resistance r1 and r2
respectively are connected in series to an external resistor
R (see figure). The value of R, at which the potential
difference across the terminals of the first cell becomes
zero is: (2021)
(a) 10.0 V (b) 5.6 V
(c) 3.6 V (d) 2.0 V
104. A current through a wire depends on time as i = 0t + t2
where 0 = 20 A/s and = 8 As–2. Find the charge crossed
through a section of the wire in 15 s. (2021)
(a) 11250 C (b) 2250 C r1
(a) r1 r2 (b) r2
(c) 260 C (d) 2100 C 2
105. Five equal resistances are connected in a network as
r1
shown in figure. The net resistance between the points A (c) r2 (d) r1 r2
and B is (2021) 2
110. The four arms of a Wheatstone bridge have resistances
as shown in the figure. A galvanometer of 15 resistance
is connected across BD. Calculate the current through the
galvanometer when a potential difference of 10 V is
maintained across AC. (2021)
3R R
(a) (b)
2 2
(c) R (d) 2R
106. A wire of 1 has a length of 1 m. It is stretched till its
length increase by 25%. The percentage change in
resistance to the nearest integer is (2021)
(a) 2.44 A (b) 2.44 mA
(a) 76% (b) 25%
(c) 56% (d) 12.5% (c) 4.87 mA (d) 4.87 A
107. A resistor develops 500 J of thermal energy in 20 s when a 111. In the experiment of Ohm’s law, a potential difference of
current of 1.5 A is passed through it. If the current is 5.0 V is applied across the end of a conductor of length
increased from 1.5 A to 3 A, what will be the energy 10.0 cm and diameter of 5.00 mm. The measured current in
developed (2021) the conductor is 2.00 A. The maximum permissible
(a) 2000 J (b) 500 J percentage error in the resistivity of the conductor is
(c) 1000 J (d) 1500 J (2021)
(a) 8.4 (b) 7.5
(c) 3.9 (d) 3.0
CURRENT ELECTRICITY 187
(a) 1 A (b) 2 A
(a) 7R/13 (b) 7R/6 (c) 4 A (d) 6 A
(c) 14R/8 (d) 15R/7 8. The electrical resistance between points A and B of the
figure shown is
4. What is the equivalent resistance of the network across
points A and B shown in figure below.
2
(a) (b) 2
3
9. In the given circuit RB = 6, RC = 3, RD = 1. The ratio 14. This question has Statement-I and Statement-II. Of the
of current through resistors C, B and D will be in the ratio four choices given after the statements, choose the one
that best describes the two Statements.
Statement-I : Higher the range, greater is the resistance
of ammeter.
Statement-II : To increase the range of ammeter, additional
shunt needs to be used across it.
(a) If Statement-I is true, Statement-II is true; Statement-II
is the correct explanation for Statement-I.
(b) If Statement-I is true, Statement-II is true; Statement-II
is not a correct explanation for Statement-I.
(a) 2 : 1 : 6 (b) 3 : 2 : 1 (c) If Statement-I is true; Statement-II is false.
(c) 2 : 3 : 6 (d) 1 : 2 : 3 (d) If Statement-I if false; Statement-II is true.
15. Statement–1 : At cryogenic temperatures, the electrical
10. In the circuit, the galvanometer G shows zero deflection. If
resistivity in metallic conductors diminishes.
the batteries A and B have negligible internal resistance,
the value of the resistor R will be Statement–2 : Thermal oscillations of atoms which hinder
motion of free electrons under the influence of an external
field become insignificant.
(a) A (b) B
(c) C (d) D
16. Statement– 1 : In the circuit shown assume both cell are
ideal and of fixed e.m.f., the resistor R1 is fixed, the resistor
R2 is non zero variable .Then the electric power delivered
to the resistor R1 is independent of value of resistance R2
20. Statement–1 : Two unequal resistors are connected in 28. Statement–1 : Voltmeter always gives e.m.f. of a cell if it is
series across a cell. Then the potential drop across the connected across the terminals of a cell.
larger resistor is more. Statement–2 : Terminal potential difference of a cell is
Statement–2 : Current will be same in both the resistors. given by V= E-Ir.
(a) A (b) B (a) A (b) B
(c) C (d) D (c) C (d) D
21. Statement–1 : If the current in the lamp decreases by 29. Statement–1 : Potential measured by voltmeter across a
20%, the percentage decrease in its illumination is 40%. wire is always less than the actual potential difference
across it.
Statement–2 : Illumination is directly proportional to the
square of current flowing through lamp. Statement–2 : Finite resistance of voltmeter changes
current flowing through the resistance across which
(a) A (b) B
potential difference is to be measured.
(c) C (d) D
(a) A (b) B
22. Statement–1 : In a simple battery circuit the point of lowest (c) C (d) D
potential is negative terminal of battery.
30. Statement–1 : When two conducting wires of different
Statement–2 : Current flows towards the point of higher resistivity and having same cross sectional area are joined
potential as it does in such a circuit from negative to in series ,the electric field in them would be equal when
positive terminal. they carry current.
(a) A (b) B Statement–2 : When wires are in series they carry equal
(c) C (d) D current.
23. Statement–1 : Fuse wire must have high resistance and (a) A (b) B
low melting point. (c) C (d) D
Statement–2 : Fuse is used for small current flow only.
Comprehensive type questions
(a) A (b) B
(c) C (d) D Passage - 1
24. Statement–1 : A 200 watt bulb glows with more brightness Using the following passage, solve Q. 31 to 34
than 100 watt bulb. A potential difference is applied across a copper wire of
Statement–2 : 100 watt bulb has more resistance than 200 radius 0.5 mm. It results in a uniform electric field of 1.5 V/
watt. m along the length of the wire. Consequently there is a
(a) A (b) B current in the wire. Temperature of the wire is
(c) C (d) D 60oC.Assuming each copper atom contributes one free
electron .Given that density of copper is 8.9 gm/cm3 and
25. Statement–1 : It is advantageous to transmit electric power
resistivity of copper at 20oC is 1.7×10–8 ohm.m and its
at high voltage.
temperature coefficient of resistivity at 20oC is 3.9×10–3/
Statement–2 : High voltage implies high current. O
C . Atomic mass of copper is 63.5. Answer the following
(a) A (b) B questions.
(c) C (d) D 31. Resistivity of copper under experimental condition is
26. Statement–1 : Current density at any point in ohmic (a) 4.5 × 106 m (b) 1.96 × 10–8 m
resistor is in the direction of electric field at that point. (c) 3.2 × 10–8 m (d) 5.1 × 10–8 m
Statement–2 : A point charge when released from rest in 32. Current density in the wire is
a region having only electrostatic field always move along
(a) 11.24 × 106 A/m2 (b) 5.4 × 107 A/m2
electric field lines.
(c) 7.65 × 107 A/m2 (d) 52.5 × 106 A/m2
(a) A (b) B
33. Drift speed of the electron is
(c) C (d) D
(a) 2.3 × 10–2 m/s (b) 4.5 × 10–3 m/s
27. Statement–1 : The drift velocity of electrons in metallic
wire will decrease if the temperature of the wire is increased. (c) 7.5 × 103 m/s (d) 5.7 × 103 m/s
Statement–2 : On increasing temperature, conductivity 34. Potential difference that need to be applied between the
of metallic wire decreases. ends of a 4 m long wire to produce the given field is
(a) A (b) B (a) 3 volt (b) 4.5 volt
(c) C (d) D (c) 6 volt (d) 1.5 volt
190
CURRENT ELECTRICITY
46. What resistance must be connected in parallel to the 52. A battery of e.m.f. E and internal resistance r is connected
galvanometer to turn it into an ammeter capable of reading across a resistance R. Resistance R can be adjusted to
electric current up to 10.01 A. any value greater than equal to zero. A graph is plotted
(a) 0.1 ohm (b) 1 ohm between the current passing through the resistance and
(c) 1.01 ohm (d) 10.01 ohm potential difference across it. Select the correct alternative.
Passage - 5
Using the following passage, solve Q. 47 to 50
Bohr and Biot purchased two electric kettles A and B of
the same size and thickness and same volume of 0.4 litre.
They studied the specification of kettles as under
Kettle A : specific heat capacity;1680 J/kg–k, mass = 200
gm cost = Rs. 400 (a) internal resistance of the battery is 5 ohm
Kettle B : specific heat capacity = 2450 J/kg–K mass = 400 (b) e.m.f. of the battery is 10 volt
gm cost = Rs 400. When kettle A is switched on with
(c) maximum current which can be taken from battery is 2
constant potential source, the tea begins to boil in 6
ampere
minutes. When kettle B is switched on with same source
separately, tea begins to boil in 8 minutes. They made (d) V versus I graph can never be straight line as shown
discussion on specification and efficiency of kettles and here.
prepared a list of questions to draw a conclusion. Some 53. In the circuit shown in the figure
of these are as under (take specific heat of tea liquid =
4200 J/kg–k, and density = 1000 kg/m3)
47. Efficiency of kettle A is
(a) 93.34% (b) 83.34%
(c) 73.34% (d) 100%
48. Efficiency of kettle B is
(a) 93.5% (b) 83.5% (a) voltage V=10 volt
(c) 73.5% (d) 62.5% (b) resistance R1=10 ohm
49. If the resistances of coil of kettle A and B are RB and RB (c) resistance R2=20 ohm
respectively, then we can say (d) equivalent resistance of the circuit is 10 ohm
(a) RA > RB
(b) RA > RB Numeric Type Questions
(c) RA = RB 54. In the circuit shown below , calculate the current flowing
(d) no relation can be predicted through 3 ohm resistance.
50. If both the kettles are joined with the same source in series
one after the other .The boiling starts in kettle A & B after
(a) four times of their original time
(b) two times of their original time
(c) equal to their original time
(d) can not be ascertained from this data.
56. The potential difference between the points A and B 63. Calculate the steady state current in the 2 resistor shown
shown in the circuit will be …..Volt. in the circuit (see figure). The internal resistance of the
battery is negligible and the capacitance of the condenser
C is 0.2 F.
2 R
64. A leaky parallel plane capacitor is filled completely with a
4
10V 4V material having dielectric constant k = 5 and electrical
–12 –1 –1
conductivity = 7.4 × 10 m . If the charge on the
6V
plane at instant t = 0 is = 8.85 mC, then calculate the leakage
D current at the instant t = 12 s.
58. Three 2 ohm resistors are connected as shown in figure 65. Find the emf (V) and internal resistance (r) of a single
below. Each can withstand a maximum power of 32 watt battery which is equivalent to a parallel combination of
without becoming excessive hot .What maximum power two batteries of emfs V1 and V2 and internal resistance r1
in watt that can be delivered to the combination of and r2 respectively, with polarities as shown in figure
resistors.
Subjective Type Questions 74. A heating coil of 2000 watt is immersed in an electric kettle.
How much time will it take in raising the temperature of 1
68. It is desired to make a 20.0 coil of wire whose temperature litre of water from 4°C to 100°C ? Only 80% part of the
coefficient of resistance is zero. To do this, carbon resistor therml energy produced is used in raising the temperature
of resistance R1 is placed in series with an iron resistor of of water.
resistance R2. The proportion of iron and carbon are so 75. A house is fitted with 20 lamps of 60 watt each, 10 fans
chosen that R1 + R2 = 20 for all temperatures near 20°C. consuming 0.5 A each and and electric kettle of resistance
Find the values of R1 and R2. Temperature coefficient of 110 . If the energy is supplied at 220 V and costs 50 paise
–3 per k Wh, calculate bill for November while running these
resistance for carbon, C = – 0.5 × 10 /°C and that of
–3 appliances for 6 hours a day.
iron is Fe = 5 × 10 /°C.
76. We have 30 watt, 6 volt bulb which we want to glow by a
69. (a) Estimate the average drift speed of conduction electrons
–7 2 supply of 120 V. What will have to be done for it ?
in a copper wire of cross-sectional area 1.0 × 10 m carrying
77. A heater is designed to operate with a power of 1000 W in
a current to 1.5 A. Assume that each copper contributes
a 100 V line. It is connected to two resistance of 10 and
roughly one conduction electron. The density of copper is
3 3 R , Figure. If the heater is now operating with a power of
9.0 × 10 kg/m , and its atomic mass is 63.5 u.
62.5 W, calculate the value R.
(b) Compare the drift speed obtained above with,
(i) thermal speeds of copper atoms at ordinary
temperatures,
(ii) speed of propagation of electric field along the
conductor which causes the drift motion.
70. A wire carries a current of 0.5 A, when a potential difference
of 1.5 V is applied across it. What is its conductance ? If
2 78. Five equal resistance each of R are connected in a
the wire is of length 3 m and area of cross-section 5.4 mm , network as shown in figure. Calculate the equivalent
calculate its conductivity. resistance between the points A and B.
71. Write the mathematical relation between mobility and drift
velocity of charge carriers in a conductor. Name the mobile
charge carriers responsible for conduction of electric
current in (a) an electrolyte (b) an ionised gas.
72. Find the current in each branch of the given circuit. What
is the energy supplied by the battery to the circuit in one
minute?
73. The resistance of the galvanometer G in the circuit is 25. 79. The given network is part of another larger circuit.
The meter deflects full scale for a current of 10 mA. Calculate the potential of point D.
The meter behaves as an ammeter of three different
ranges. The range is 0–10 A, if the terminals O and P
are taken; range is 0 – 1 A between O and Q; range is 0
– 0.1 A between O and R. Calculate the resistance R 1,
R 2 and R3.
CURRENT ELECTRICITY 194
Match the Column Type Questions 83. Consider two identical cells each of e.m.f. E and internal
resistance of r connected to a load resistance R. Match
80. For an electric conductor, Match column I with Column II column-1 with column-2.
Column-I Column-II Column-1 Column-2
(a) Electric conductivity of a (p) Electric field
conductor depends on strength E2
(a) For parallel combination (P)
(b) Conductance of a (q) Temperature 4r
conductor depends on of cells
(c) For a given conductor (r) Nature of the
r
at a given temperature conductor (b) For series combination (Q) Ee= E, req
current density depends on 2
(d) For a given potential (s) Dimensions of of cells
difference applied across a conductor (area/ (c) For maximum power (R) Ee= 2E, re= 2q
conductor, current in it will length) transfer to load when
depends on. cells in series
81. In an electric circuit with two batteries in series with E2
opposite polarities are shown in the figure, match the (d) For maximum power (S)
2r
following quantities in column I with column-II. transfer when cells
are in parallel
Column-I Column-II
(a) Potential difference across (p) B1
B1 battery
(b) Potential difference across (q) 9 volt
B2 battery
(c) power is supplied by battery (r) B2
(d) Powerr is consumed by (s) 14 volt
battery (t) none
82. Referring to a circuit shown below, match the column I
with Column II
Column-I Column-II
(a) Equivalent resistance (p) The circuit is a balanced
between A and C wheatstone bridge plus a
resistance parallel to bridge
(b) Equivalent resistance (q) R/2
between B and D
(c) Equivalent resistance (r) On removing the
A and B between B and D
equivalent resistance
becomes R/2
(d) Equivalent resistance (s) On removing
between C and D resistance between
B and D equivalent
resistance becomes 5R/8
CURRENT ELECTRICITY 195
2 Rr 8 R R r
(a) (b)
Rr 3R r
5R
(c) 2r + 4R (d) 2r
2
4. Two identical capacitors, have the same capacitance C.
One of them is charged to potential V1 and the other to V2. (a) W1 > W2 = W3 (b) W1 > W2 > W3
The negative ends are also connected, the decrease in
(c) W1 < W2 = W3 (d) W1 < W2 < W3
energy of the combined system is (2002)
CURRENT ELECTRICITY 196
13. Find out the value of current through 2 resistance for Multiple Choice Questions
the given circuit. (JEE 2005)
17. For the circuit shown in the figure (JEE 2009)
(a) 5 A (b) 2 A
(c) zero (d) 4 A
14. Two bars of radius r and 2r are kept in contact as shown.
An electric current I is passed through the bars. Which
(a) the current I through the battery is 7.5 mA
one of following is correct ? (JEE 2006)
(b) the potential difference across RL is 18 V
(c) ratio of powers dissipated in R1 and R2 is 3
(d) if R1 and R2 are interchanged, magnitude of the power
dissipated in RL will decrease by a factor of 9
18. Consider a thin square sheet of side L and thickness t,
made of a material of resistivity . The resistance between
(a) Heat produced in bar BC is 4 times the heat produced two opposite faces, shown by the shaded areas in the
in bar AB figure is (JEE 2010)
(b) Electric field in both halves is equal
(c) Current density across AB is double that of across BC
(d) Potential difference across AB is 4 times that of across BC
15. A resistance of 2 is connected across one gap of a
metre-bridge (the length of the wire is 100 cm) and an
unknown resistance, greater than 2 , is connected across
the other gap. When these resistances are interchanged,
the balance point shifts by 20 cm. Neglecting any
(a) directly proportional to L
corrections, the unknown resistance is (JEE 2007)
(a) 3 (b) 4 (b) directly propotional to t
(c) 5 (d) 6 (c) independent of L
16. Figure shows three resistor configurations R1, R2 and R3 (d) independent of t
connected to 3 V battery. If the power dissipated by the
19. Incandescent bulbs are designed by keeping in mind that
configuration R1, R2 and R3 is P1, P2 and P3, respectively,
then (JEE 2008) the resistance of their filament increases with the increase
in temperature. If at room temperature, 100W, 60W and 40W
bulbs have filament resistance R100, R60 and R40, respectively,
the relation between these resistances is (JEE 2010)
1 1 1
(a) R R R
100 40 60
20. To verify Ohm’s law, a student is provided with a test 23. A meter bridge is set-up as shown in figure, to determine
resistor RT, a high resistance R1, a small resistance R2, two an unknown resistance X using a standard 10 resistor.
identical galvanometers G1 and G2, and a variable voltage The galvanometer shows null point when tapping-key is
source V. The correct circuit to carry out the experiment is at 52 cm mark. The end-corrections are 1 cm and 2 cm
(JEE 2010) respectively for the ends A and B. The determined value
(a) of X is (JEE 2011)
(c)
26. Two ideal batteries of emf V1 and V2 and three resistances 2475 1875
R1, R2 and R3 are connected as shown in the figure. The (a) (b)
64 64
current in resistance R2 would be zero if (JEE 2014)
1875 2475
(c) (d)
49 132
30. Consider two identical galvanometers and two identical
resistors with resistance R. If the internal resistance of
the galvanometers RC < R/2, which of the following
statement(s) about any one of the galvanometers is(are)
true? (JEE 2016)
(a) The maximum voltage range is obtained when all the
components are connected in series
(a) V1 = V2 and R1 = R2 = R3
(b) The maximum voltage range is obtained when the two
(b) V1 = V2 and R1 = 2R2 = R3 resistors and one galvanometer are connected in series,
(c) V1 = 2V2 and 2R1 = 2R2 = R3 and the second galvanometer is connected in parallel to
(d) 2V1 = V2 and 2R1 = R2 = R3 the first galvanometer
27. A galvanometer gives full scale deflection with 0.006 A (c) The maximum current range is obtained when all the
current. By connecting it to a 4990 resistance, it can be components are connected in parallel
converted into a voltmeter of range 0-30 V. If connected to (d) The maximum current range is obtained when the two
2n galvanometers are connected in series and the combination
a resistance, it becomes an ammeter of range 0-1.5
249 is connected in parallel with both the resistors
A. The value of n is (JEE 2014) 31. In process 1, the energy stored in the capacitor EC and
28. In the following circuit, the current through the resistor R heat dissipated across resistance ED are related by :
(=2) is I amperes. The value of I is (JEE 2015) (JEE 2017)
(a) EC = ED ln2 (b) EC = ED
1
(c) EC = 2ED (d) EC E D
2
32. In process 2, total energy dissipated across the resistance
ED is : (JEE 2017)
V0 t R 2
(a) I ln
R1
(b) the outer surface is at a higher voltage than the inner
surface
(c) the outer surface is at a lower voltage than the inner
surface
(a) The key S1 is kept closed for long time such that
(d) V I 2
capacitors are fully charged. Now key S2 is closed, at this
time the instantaneous current across 30 resistor 37. In the balanced condition, the values of the resistances
(between points P & Q) will be 0.2A (round off to 1st of the four arms of a Wheatstone bridge are shown in
decimal place). the figure below. The resistance R3 has temperature
(b) If key S1 is kept closed for long time such that coefficient 0.0004° C–1. If the temperature of R3 is
increased by 100 ºC, the voltage developed between S
capacitors are fully charged, the voltage across the
and T will be__________ volt. (JEE 2020)
capacitor C1 will be 4V.
(c) At time t = 0, the key S1 is closed, the instantaneous
current in the closed circuit will be 25 mA
(d) if key S1 is kept closed for long time such that
capacitors are fully charged, the voltage difference
between points P and will be 10V.
35. Two identical moving coil galvanometers have 10
resistance and full scale deflection at 2A current. One
of them is converted into a voltmeter of 100 mV full scale 38. In the circuit show below, the switch 5 is connected to
reading and the other into an Ammeter of 1mA full scale position P for a long time so that the charge on the
current using appropriate resistors. These are then used capacitor becomes q1μC . Then 5 is switched to position
to measure the voltage and current in the Ohm’s law Q. After a long time, the charge on the capacitor is q 2μC .
experiment with R = 1000 resistor by using an ideal cell. [JEE 2021]
Which of the following statement(s) is/are correct ?
(JEE 2019)
(a) The resistance of the Voltmeter will be 100
(b) The resistance of the Ammeter will be 0.02(round
off to 2nd decimal place)
(c) If the ideal cell is replaced by a cell having internal The magnitude of q2 is
resistance of 5 then the measured value of R will be 39. In the circuit show below, the switch 5 is connected to
more than 1000 position P for a long time so that the charge on the
(d) The measured value of R will be 5 capacitor becomes q1μC . Then 5 is switched to position
36. Shown in the figure is a semicircular metallic strip that Q. After a long time, the charge on the capacitor is q 2μC .
has thickness t and resistivity Its inner radius is R1 and [JEE 2021]
outer radius is R2. If a voltage V0 is applied between its
two ends, a current I flows in it. In addition, it is observed
that a transverse voltage V develops between its inner
and outer surfaces due to purely kineticeffects of moving
electrons (ignore any role of the magnetic field due to the
current). Then (figure is schematic and not drawn to
scale)– (JEE 2020) The magnitude of q1 is
CURRENT ELECTRICITY 201
Find Answer key and Detailed Solutions at the end of this book
CURRENT ELECTRICITY
Answer Key
CHAPTER -1 ELECTROSTATICS
EXERCISE - 1 : EXERCISE - 2 :
BASIC OBJECTIVE QUESTIONS PREVIOUS YEARS JEE MAIN QUESTIONS
DIRECTION TO USE -
DIRECTION TO USE - Scan the QR code and check detailed solutions.
Scan the QR code and check detailed solutions.
EXERCISE - 3 :
ADVANCED OBJECTIVE QUESTION 1 Qq
(c) 40 2ma
47. (a) 2and 8at ends, at 3 cm from 2(b) zero
5
48.
6
49. (b)
2
Q2 R R
n
U
50. (a) n 1
8 0 r 2 R r
DIRECTION TO USE -
Scan the QR code and check detailed solutions.
Q2 R
(b) U
1. (d) 2. (d) 3. (c) 4. (b) 8 0 r 2
5. (a) 6. (b) 7. (a) 8. (c) 51. (b, d) 52. (a,b,c)
9. (b) 10. (b) 11. (a) 12. (d) 53. (b,c) 54. (c,d)
13. (d) 14. (c) 15. (d) 16. (c) 55. (b,d) 56. (a,b,cd)
17. (a) 18. (a) 19. (c) 20. (a) 57. (a,b) 58. (a,b,c,d)
21. (a) 22. (b) 23. (c) 24. (d) 59. (c,d) 60. (a,d)
25. (a) 26. (b) 27. (b) 28. (b) 61. (a,b,c) 62. (c,d)
29. (c) 30. (d) 31. (d) 63. (a,c) 64. (a,b)
65. (A–R, P; B–Q; C–P; D–P,S)
32. (True) 33. (False) 34. (0.2) 35. (0.9)
66. (–Ea)
36.(0) 37. (0.5) 38. (3.16 × 10–9C)
39. (8.48 m) 40. (0.628 sec.) 67. ( 8iˆ )
4
41. (a) H a (b) H 3
a
9 q2
3 68. 9 10
L2
a2 69. (c)
42. (i) VA a b c , VB b c ,
0 0 b
1 Q2
70. 180º , , 2
a 2 b2 40 4L
VC c , (ii) a + b = c
0
c c
3 Q2 3 GM 2
43. (a) U , (b) U E 1.5 1032 J ,
20 0 R 5 R
(c) U Q2
80 R
q
44. 2 0 m
QR r
45. 4 R 2 r 2
0
1
46. (a) 4a, (5a, 0); (b) KQ
2
;
3a x 3a x
204
ANSWER KEY
EXERCISE - 4 :
PREVIOUS YEARS JEE ADVANCED QUESTIONS
DIRECTION TO USE -
Scan the QR code and check detailed solutions.
1 q2 4
5. (b) 6.
40 a
.
6
3 3 3 6 2
qp pq ˆ
7. (c) 8. (a) KE 2
, (b) F i
40 d 20 d3
9. (c) 10.
1 2 qa 11. (c)
2 0
1/ 3
a
12. V ' V
3t
13. (b) 14. (a,b,c,d) 15. (c) 16. (d)
17. (b) 18. (a) 19. (d) 20. (c)
21. (b) 22. (a) 23. (b) 24. (c)
25. (b) 26. (a) 27. (a) 28. (a,d)
29. (d) 30. (a)
31. (a,b,c,d)
32. (c)
33. (a,b,c,d)
34. (c,d) 35. (c) 36. (a,c,d)
37. (6) 38. (b,d) 39. (c,d) 40. (c)
41. (c) 42. (a) 43. (c) 44. (d)
45. (6) 46. (c) 47. (b,d)
48. (P–5; Q–3; R–1, 4; S–2)
49. (a,b) 50. (a) 51. (c,d)
52. (b,c,d)
53. (b,c) 54. (6.00) 55. (2.00) 56. (a,c)
57. (3.00) 58. (6.40)
205
ANSWER KEY
Answer Key
CHAPTER -2 CAPACITANCE
EXERCISE - 1 : EXERCISE - 2 :
BASIC OBJECTIVE QUESTIONS PREVIOUS YEARS JEE MAIN QUESTIONS
EXERCISE - 3 : EXERCISE - 4 :
ADVANCED OBJECTIVE QUESTION PREVIOUS YEARS JEE ADVANCED QUESTIONS
DIRECTION TO USE -
DIRECTION TO USE -
Scan the QR code and check detailed solutions.
Scan the QR code and check detailed solutions.
1. (b) 2. (a) 3. (b) 4. (a) 1. (b) 2. (a) 3. (a) 4. (d)
5. (c) 6. (b) 7. (a) 8. (d) 5. (c) 6. (b,d) 7. (a,d) 8. (d)
9. (c) 10. (d) 11. (d) 12. (a) 9. (1.00) 10. (1.30)
13. (a) 14. (c) 15. (d) 16. (c)
17. (c) 18. (b) 19. (c) 20. (c)
21. (a) 22. (a) 23. (b) 24. (c)
25. (b) 26. (b) 27. (b) 28. (5)
29. (5) 30. (2)
31.
3
V
K2
CK1K 2 K A
32. CR n 2 where C 0
K 2 K1 K1 d
33. (1.50)
34. (3/5)
–9 –5
35. (i) CA = 2 × 10 F, UA = 1.21 × 10 J ;
–5 –5
(ii) W = 4.84 × 10 J ; (iii) U = 1.1 × 10 J
36. (b,c)
37. (a,c)
38. (a,c,d)
39. (a,d)
40. (b,d)
41. (a,c,d)
42. (A–S; B–Q; C–Q; D–P)
43. (A–S; B–S,R; C–P,Q; D–R,P)
0 AV 2AV
44. ,
d d
207
ANSWER KEY
Answer Key
CHAPTER -3 CURRENT ELECTRICITY
EXERCISE - 1 : EXERCISE - 2 :
BASIC OBJECTIVE QUESTIONS PREVIOUS YEARS JEE MAIN QUESTIONS
EXERCISE - 3 : EXERCISE - 4 :
ADVANCED OBJECTIVE QUESTION PREVIOUS YEARS JEE ADVANCED QUESTIONS
DIRECTION TO USE -
DIRECTION TO USE -
Scan the QR code and check detailed solutions.
Scan the QR code and check detailed solutions.
1. (b) 2. (c) 3. (b) 4. (b) 1. (c) 2. (b) 3. (a) 4. (c)
5. (b) 6. (c) 7. (c) 8. (a) 5. (d)
9. (a) 10. (b) 11. (b) 12. (c) 6. (a) No ; (c) 8 (bridge balanced)
13. (d) 14. (d) 15. (a) 16. (a) 7. (a) 8. (a) 9. (d) 10. (a)
17. (c) 18. (c) 19. (a) 20. (a) 11. (a) 12. (d) 13. (c) 14. (a)
21. (d) 22. (c) 23. (c) 24. (b) 15. (a) 16. (c) 17. (a, d) 18. (c)
25. (c) 26. (c) 27. (a) 28. (d) 19. (a) 20. (c) 21. (4) 22. (5)
29. (a) 30. (d) 31. (b) 32. (c) 23. (b)
33. (d) 34. (c) 35. (a) 36. (b) 24. (a, b, c, d) 25. (c)
37. (b) 38. (c) 39. (b) 40. (c) 26. (a, b, d) 27. (5)
41. (b) 42. (c) 43. (c) 44. (d) 28. (1 A) 29. (b)
45. (a) 46. (a) 47. (b) 48. (d) 30. (b, c) 31. (b)
49. (c) 50. (a) 51. (b, c) 32. (a) 33. (5.55)
52. (a, b, c) 34. (b, c) 35. (b, d)
53. (b, c, d) 54. (4) 55. (1) 36. (a, c, d) 37. (0.26 to 0.27)
56. (6) 38. (0.67) 39. (1.33)
57. (1) 58. (48) 59. (2)
60. (5)
61. (2 V, 1 A, 0, – 1A ; (ii) 1 A, 2 A, – 1A, 2 A)
62. (R/2) 63. (0.9 A) 64. (0.198 A)
V1r2 V2 r1 rr
65. ( V , r 12 )
r1 r2 r1 r2
66. (4.96 mA, 1.58 V)
2 21 19
67. ( V ; (ii) V, V)
13 13 13
68. (*)
69. (*) 70. (*) 71. (*) 72. (*)
73. (*) 74. (*) 75. (*) 76. (*)
77. (*) 78. (*) 79. (*)
80. (a–q, r; b–q, r, s; c – p; d – q, r, s)
81. (a–s; b – t; c – p; d – r)
82. (a – p, q, r; b – p, q; c – s; d – s)
83. (a–q; b–r; c–s; d–s)
MASTER INDEX .
VOLUME 1:
Electrostatics
Capacitance
Current Electricity
VOLUME 2:
Magnetic Effect of Current
Magnetism and Matter
Electromagnetic Induction
Alternating Current & Electromagnetic Waves
VOLUME 3:
Ray Optics and Optical Instruments
Wave Optics
VOLUME 4:
Modern Physics
Semiconductors and Communication system
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