Electric Charges Fields

www.sakshieducation.
com
ELECTRIC CHARGES AND FIELDS

Important Points:
1.
Coulombs Law:
The force of attraction (or) repulsion between two charges at rest is directly proportional to the
product of magnitude of two charges and inversely proportional to the square of the distance
between them.
2.
Electric Field:
Electric field exists in a region in which a charge experiences a force.
3.
Intensity of Electric Field:

The force experienced by a unit positive charge kept in an electric field is called the intensity of
the electric field (E). It is a vector quantity.
4.
Electric Lines Of Force:

The path traced by a unit positive charge placed in a uniform electric field is called electric line
of force.
5.
Gausss Law:
The total flux linked with a closed surface is (1/0) times the charge enclosed by the closed
surface. = E.ds =
6.
1
q.
0
The electric intensity at a point due to an infinitely long charged wire is, E =
.
2 0 r
Here is the uniform linear charge density, r is the radial distance of the point from the
axis of the wire.
7.
The electric intensity due to an infinite plane sheet of charge is, E =

Here is the uniform surface charge density.
www.sakshieducation.com
2 0
Very Short Answer Questions
1.
A.
What is meant by the statement that charge is quantized?

Charge always exists as an integer multiple of electronic charge (1.610-19C). Fractions of
electronic charge are not possible. i.e., Q = ne where n is an integer. Hence charge is said to
be quantized.
2.
A.
Repulsion is the sure test of electrification than attraction. Why?

Positively charged body can attract both negatively charged and neutral bodies, but positively
charged body can only repel another positively charged body. Hence repulsion is sure test of
electrification.
3.
How many electrons constitute 1C of charge?
A:
Charge q = ne
Given q = 1C
1 = n 1.6 10 19
n =
1
= 6.25 1018
19
1.6 10
4.
What happens to the weight of a body when it is charged positively?
A:
Every body acquires positive charge due to the loss of electrons. Hence the weight of a body
decreases when it is positively charged.
5.
What happens to the force between two charges if the distance between them is
a) Halved b) Doubled?
A.
1
d2
2
F d
F d
F2 = 4F1
a) 2 = 1 2 =
F1 d 2
F1 d / 2
The force between the charges increases by four times
2
F d d
F
b) 2 = 1 = F2 = 1
F1 d 2 2d
4
The force is reduced to of its original value.
6.
A.
The electric lines of force do not intersect. Why?

The tangent drawn to electric line of force gives the direction of electric field at the point. If
the electric lines of force intersect then at the point of intersection electric field will act in
two different directions, which is not possible. Hence electric lines of force do not intersect.
7.
Consider two charges +q and -q placed at B and C of an equilateral triangle ABC. For
this system, the total charge is zero. But the electric field (intensity) at A which is
equidistant from B and C is not zero. Why?
A:
Charge is a scalar, Hence it is added algebraically. But field is a vector, it should be added
vectorially. The fields due to the given charges are not in the same direction. Therefore the net
field is not zero.
8.
Electrostatic field lines of force do not form closed loops. If they form closed loops then
the work done in moving a charge along a closed path will not be zero. From the above
two statements can you guess the nature of electrostatic force?
A:
Electric force is conservative force. If field lines form closed loop, then work done by the
conservative force along the closed path is not be zero.
9.
State Gauss's law in electrostatics?
A:
The total electric flux through any closed Gaussian surface is equal to
times the algebraic
sum of charge enclosed by the surface. Here 0 is the permittivity of free space.
Mathematically the Gauss law can be stated as E.ds =
10. When will be the electric flux negative and when is it positive?
A.
For a closed surface, inward flux is taken to be negative and outward flux is taken to be
positive.
11. Write the expression for electric intensity due to an infinite long charged wire at a radial
distance 'r' from the wire?
A:
The electric intensity due to an infinitely long charged wire E =
. Where is the linear

2 0 r
charge density and r is the perpendicular distance from wire.
12. Write the expression for electric intensity due to an infinite plane sheet of charge?
A:
Electric field intensity due to an infinite non-conducting plane sheet of charge E =
, where
2 0
is surface charge density.
13. Write the expression for electric intensity due to charged conducting spherical shell at
points outside and inside the shell?
A. Outside the shell ( r > R ): If ' ' is the surface charge density E =
Inside the shell ( r < R ) : If ' ' is the surface charge density E = 0.
R2
o r2
Short Answer Questions

1.
State and explain Coulombs inverse square law in electricity?
A.
Statement:
The force of attraction (or) repulsion between two charges at rest is directly proportional to the
product of magnitude of two charges and inversely proportional to the square of the distance
between them and acts along the line joining the charges.
Explanation:
Let two point charges q1 and q2 at rest are separated by a distance r. The force F between them
is given by
F q1q2
1
d2
(Or) F
q1q2
r2
(Or) F =
q1q2
4 r 2
Where is called to absolute permittivity of the medium.

=0r Where 0 is called the permittivity of free space and r is the relative permittivity of
the medium
r = 1 For air or vacuum
Hence, for air (or) vacuum Fair =
Fair =
1 q1q2
4 0 r 2
1 q1q2
4 0 r 2
(Or) o = 8.854 1012 Farad / m.
2.
Define intensity of electric field at a point. Derive an expression for the intensity due to a
point charge?
A.
Intensity of Electric Field (E):

The force acting on unit positive charge placed at a point in the electric field is called the
Intensity of electric field at that point.
Expression:
Let a point charge Q be placed at point O in air (or) vacuum. Let P be a point at a distance
r from the charge Q. Let a unit positive charge is placed at P.
The force acting on this unit charge is given by,
F=
1 Q 1
4 0 r 2
Hence from definition, E =
1 Q
4 0 r 2
Intensity of electric field in vector form can be given by, E = 1 Q3 r

4 0 r
3.
Derive the equation for the couple acting on an electric dipole in a uniform electric field?
A.
Let an electric dipole of dipole moment P placed at an angle to the direction of a uniform
electric field of intensity E. Each charge of the dipole experiences a force of magnitude qE.
Hence the dipole experiences two equal, unlike and non collinear forces which form a couple.
The torque acting on the dipole is given by
= qE 2a sin Or = pE sin
In vector form = p E
+q
qE
2l sin
qE q
The torque on the dipole tends to align the dipole in the direction of electric field.
4.
Derive an expression for the electric intensity of electric field at a point on the axial line of
an electric dipole?
A. Electric Field on Axial Line:

Consider an electric dipole consisting of two point charges q and +q separated by a distance
2a. Let P be a point on the axial line at a distance r from centre of o the dipole.
Electric field intensity at P due to charge +q is given by,
EA =
q
4 0 ( r + a ) 2
Electric field intensity at P due to charge - q is given by,

EB =
q
4 0 ( r a ) 2
Eaxial = EB E A
Net intensity,
Eaxial =
q
q
2
4 0 (r a ) 4 0 (r + a )2
Eaxial =
1
q
2
4 0 (r a 2 ) 2
( EB > EA )
Since p = q(2a)
Eaxial =
1
2 pr
2
4 0 (r a 2 ) 2
If, r>>>a then Eaxial =
1 2p
4 0 r 3
5. Derive an expression for the electric intensity of electric field at a point on the equatorial
line of an electric dipole.
A. Electric Field on Equatorial Line:
Consider an electric dipole consisting of two point charges +q and q separated by a distance
2a. Let P be a point on the equatorial line at a distance r from centre of o the dipole.
Electric field intensity at P due to charge +q is given by,
EA =
q
1
q
(Along PD )
=
2
2
4 0 AP
4 0 ( r + a 2 )
Electric field intensity at P due to charge - q is given by,

EB =
q
4 0 ( r + a 2 )
2
(Along PC )
Clearly E A = EB .Let us resolve and EB into two components in two mutually perpendicular
directions components of E A and EB along the equatorial line cancel each other. But the
components perpendicular to equatorial line get added up because they act in same direction.
So magnitude of resultant intensity E at P
E = 2 E A cos = 2
Eequitorial =
q
a
2
2
2
4 0 ( r + a ) r + a 2
p
4 0 ( r + a 2 )3/ 2
If r >>> a, then Eequitorial =
p
. This intensity is parallel to AB from P.
4 0 r 3
1
6.
State Gauss's law in electrostatics and explain its importance?
A:
The total electric flux through any closed surface is equal to
times the net charge enclosed
by the surface.
Here 0 is the permittivity of free space, Mathematically, Gauss law can be stated as,
E = E.ds =
Here q is the total charge enclosed by the surface.

Importance:
1) Gausss theorem holds good for any closed surface.
2) It is useful in calculating the electric field and potentials due to Continuous symmetric
charge distribution.
3) It is valid for stationary charges as well as for rapidly moving charges.
4) Gauss theorem can help us to calculate electric flux that radiated outwards from one
coulomb of charge.
Long Answer Questions

1.
Define Electric Flux. Applying Gausss law derives the expression for electric intensity
due to an infinite long straight charged wire. (Assume that the electric field is everywhere
radial and depends only on the radial distance r of the point from the wire.)
A.
Electric Flux:
The total number of electric lines of force passing through a normal plane inside an electric
field is called Electric flux ( ) . It is a scalar quantity.
d = E . d s .
Gauss Theorem:
The Electric flux ( ) through any closed surface is equal to
times the net charge
enclosed by the surface.

= E.ds =
1
q
0
Here 0 is the permittivity of free space.

Electric field due to a line charge:
Consider an infinitely long thin straight wire having a uniform charge density . Imagine a
cylindrical Gaussian surface ABCD of length l and radial distance r with its axis along the
wire.
For the top and bottom flat surfaces AB and CD, E and d s are perpendicular. Hence electric
flux through them is zero.
Along the curved surface ADCB, E ds =
Or
E (2 rl ) =
E =
q
2 o r
2 o r
=
l
This gives electric intensity due to infinitely long charged wire.
2.
State Gausss law in electrostatics. Applying Gausss laws derive the expression for
electric intensity due to an infinite plane sheet of charge?
A.
Gauss Theorem:
by the surface.
=
E.ds =
This is the integral form of Gauss law. Here 0 is the permittivity of free space.
Electric field due to an infinite plane sheet of charge:
Consider an infinite plane sheet of charge OXPY of uniform surface charge density ' ' .
Imagine a Gaussian surface in the form of horizontal cylinder ABCD, passing through the
vertical plane sheet and having length 2r perpendicular to the plane sheet and symmetrical on
either side of the sheet. The areas ds3 and ds4 are perpendicular to E. Hence electric flux
through them is zero.
For the areas ds3 and ds4 ,
E.ds = E ds = E (S + S) = E (2S)
Here S is the area of each flat surface of the cylinder.

From the Gausss law we have,
E.ds =
But q = s
E (2S) =
, or
E=
.
2 0
Thus the electric intensity due to an infinite plane sheet of charge E =
.
2 0
If is positive, E will be along the outward drawn normal

If is negative, E will be along the inward drawn normal
3.
Applying Gausss law derive the expression for electric intensity due to charged
conducting spherical shell at (i) a point outside the shell (ii) a point on the surface of the
shell and (iii) a point inside the shell ?
A.
Gauss Theorem:
by the surface.
=
E.ds =
This is the integral form of Gauss law. Here 0 is the permittivity of free space
Electric field due to a spherical shell:
Consider a spherical shell of radius R and a charge q. Let us find the electric field at a point
P at a distance r from the centre O of the shell.
Surface charge density =
q
q = 4 R 2
2
4 R
(a) When P is outside the shell:

Consider a spherical Gaussian surface of radius r .The total flux through the surface is given
by,
= E ds = E (4 r 2 )
E (4 r 2 ) =
qE=
1
4 0
q
r2
4 R 2
R2
E =
E=
4 o r 2
0 r2
(b) When the point is on the surface (r = R):

Consider a spherical Gaussian surface of radius R .The total flux through the surface is given
by,
E (4 R 2 ) =
E=
q
0
R2 E =
4 0
0
(c)When P is inside the shell (r < R):
Consider a spherical Gaussian surface of radius r where (r < R). In this case Gaussian
surface does not enclose any charge and hence according to Gauss law.
E (4 r 2 ) =
E = 0
PROBLEMS
1. Two small identical balls, each of mass 0.20 g. carry identical charges and are suspended
by two threads of equal lengths. The balls position themselves at equilibrium such that the
angle between the threads is 600. If the distance between the balls is 0.5 m, find the charge
on each ball?
Sol: m = 0.2 = gm = 2 104 kg
= 300 , x = 0.5 m = 5 10 1 m
1
q2
= mg tan .
4 0 x 2
9 109 q 2
= 2 104 9.8 tan 300
25 10 2
q = 1.77 107 col.
2. An infinite number of charges each of magnitude q are placed on x- axis at distances of

1, 2, 4, 8 ... Meter from the origin respectively. Find intensity of the electric field at origin.
Sol: Eres =
q
4 0
1
1 1 1 8
12 + 22 + 42 + 82 + 162 + .....
q 1
=
=
4 0 1 1
4
3.
q
towards origin
3 o
A clock face has negative charges q, 2q, 3q.......12q fixed at the position of the
corresponding numerals on the dial. The clock hands do not disturb the net field due to
the point charges. At what time does the hour hand point in the direction of the electric
field at the centre of the dial?
.
-12q
-11q
E12
-q
E1
E11
-10q
E2
-2q
E10
E3
-9q
-3q
E9
E4
-8q
-4q
E8
-7q
E5
E7
E6
-5q
-6q
The resultant electrified due to all charges lies in between 9 and 10 i.e. at the time 9:30
4. Consider a uniform electric field E = 3103 N/C. (a) What is the flux of this field through
a square of 10 cm on a side whose plane is parallel to the yz plane? (b) What is the flux
through the same square if the normal to its plane makes a 600 angle with the x-axis?
Sol: E = 3 103 NC 1
A = (10 1 ) = 10 2 m 2
2
a) = EA = 3 103 102 = 30
b) = EA cos
= 3 103 10 2
5.
1
= 15. Here = 600
2
There are four charges, each with a magnitude q. Two are positive and two are negative.
The charges are fixed to the corners of a square of side L, one to each corner, in such a
way that the force on any charge is directed toward the center of the square. Find the
magnitude of the net electric force experienced by any charge?
A.
The attractive force experienced by charge at C due to the charges at B and D is same
F1 = F2 = F =
1 qq
.
4 0 L2
......... (1)
+q
B
-q
A
2F
C
D
q
F
L
-q
The resultant of two equal forces is

F ' = F2 + F2 = 2F = 2
1 q2
.
4 0 L2
........... (2)
The repulsive force experienced by the charge at C due to the charge at A is
1
qq
.
2
4 0
2L
F =
......... (3)
The net force experienced by the charge at C towards the centre is
FR = 2F F = 2
FR =
1 q2
1
q2
.
.
4 0 L2 4 0 2L2
1
q2
. 2 2 2 1
4 0 2L
6. The electric field in a region is given by E = ai + b j . Here a and b are constants. Find the
net flux passing through a square area of side L parallel to y-z plane.
Sol: A = L2 i; E = ai + b j
( )
= E. A = ( ai + bj ) . L2 i
= aL2
7. A hollow spherical shell of radius r has a uniform charge density . It is kept in a cube of
edge 3r such that the centre of the cube coincides with the centre of the shell. Calculate the
electric flux that comes out of a face of the cube?
Sol: For the spherical shell
=
q
q = .4r 2
4r 2
Electric flux through one face of the cube is given by Eo =

=
q
4r 2
=
6 Eo
6 Eo
2r 2
3Eo
8.
An electric dipole consists of two equal and opposite point charges +Q and Q, separated
by a distance 2l. P is a point collinear with the charges such that its distance from the
positive charge is half of its distance from the negative charge. The electric intensity at P
is-
A.
Given r l =
1
(r + l )
2
o
-q
+q
(r-l)
2l
r
(r+l)
2r 2l = r + l r = 3l
The electric intensity at P is E =
E=
1
2pr
.
2
4 0 ( r l 2 ) 2
2P ( 3l )
1
1 2P ( 3l )
=
2
4 0 ( 3l )2 l 2
4 0 ( 8l 2 )2
P = ( 2l ) Q
1 2 ( 2l Q ) 3l
1 3Q
.
=
4
4 0
64l
4 0 16l 2
9. Two infinitely long thin straight wires having uniform linear charge densities and 2
are arranged parallel to each other at a distance r apart. The intensity of the electric field
at a point midway between them isSol: 1 = , 2 = 2, r1 = r2 =
E1 =
1
=
2 0 r1
E2 =
2
=
2 0 r2
Enet = E2 E1 =
r
2
r
2 0
2
2
r
2 0
2
0 r
2
0 r
=
0 r 0 r 0 r
e and 2e
10. Two infinitely long thin straight wires having uniform linear charge densities
are arranged parallel to each other at a distance r apart. The intensity of the electric field
at a point midway between them isSol: 1 = , 2 = 2, r1 = r2 =
E1 =
1
=
2 0 r1
E2 =
2
=
2 0 r2
r
2
r
2 0
2
Enet = E2 E1 =
2
r
2 0
2
0 r
2
0 r
=
0 r 0 r 0 r
11. An electron of mass m and charge e is fired perpendicular to a uniform electric field
of intensity E with an initial velocity u. If the electron traverses a distance x in the
field in the direction of firing, find the transverse displacement y it suffers?
A.
The transverse distance of electron in the direction of firing x = ut....... (1)

+
y
m
e
1
2
The transverse distance of the electron in the perpendicular direction is y = at 2

Since acceleration acquired by the electron by the field is a =
Deflection of the electron
2
1 eE x
eEx 2
y = =
2 m u 2mu 2
eE
m

Electric Charges Fields

Uploaded by

Copyright:

Available Formats

Electric Charges Fields

Uploaded by

Document Information

Original Description:

Copyright

Available Formats

Share this document

Share or Embed Document

Sharing Options

Did you find this document useful?

Is this content inappropriate?

Copyright:

Available Formats

Electric Charges Fields

Uploaded by

Copyright:

Available Formats

www.sakshieducation.

ELECTRIC CHARGES AND FIELDS

Intensity of Electric Field:

Electric Lines Of Force:

The electric intensity due to an infinite plane sheet of charge is, E =

Very Short Answer Questions

What is meant by the statement that charge is quantized?

Repulsion is the sure test of electrification than attraction. Why?

How many electrons constitute 1C of charge?

What happens to the weight of a body when it is charged positively?

The force is reduced to of its original value.

The electric lines of force do not intersect. Why?

State Gauss's law in electrostatics?

times the algebraic

The electric intensity due to an infinitely long charged wire E =

. Where is the linear

charge density and r is the perpendicular distance from wire.

Electric field intensity due to an infinite non-conducting plane sheet of charge E =

is surface charge density.

Short Answer Questions

State and explain Coulombs inverse square law in electricity?

Where is called to absolute permittivity of the medium.

Hence, for air (or) vacuum Fair =

(Or) o = 8.854 1012 Farad / m.

Intensity of Electric Field (E):

Hence from definition, E =

Intensity of electric field in vector form can be given by, E = 1 Q3 r

A. Electric Field on Axial Line:

Electric field intensity at P due to charge +q is given by,

Electric field intensity at P due to charge - q is given by,

If, r>>>a then Eaxial =

Electric field intensity at P due to charge - q is given by,

If r >>> a, then Eequitorial =

State Gauss's law in electrostatics and explain its importance?

The total electric flux through any closed surface is equal to

times the net charge enclosed

Here q is the total charge enclosed by the surface.

Long Answer Questions

times the net charge

enclosed by the surface.

Here 0 is the permittivity of free space.

This gives electric intensity due to infinitely long charged wire.

times the net charge enclosed

Here S is the area of each flat surface of the cylinder.

Thus the electric intensity due to an infinite plane sheet of charge E =

If is positive, E will be along the outward drawn normal

times the net charge enclosed

(a) When P is outside the shell:

(b) When the point is on the surface (r = R):

2. An infinite number of charges each of magnitude q are placed on x- axis at distances of

The resultant of two equal forces is

The repulsive force experienced by the charge at C due to the charge at A is

The net force experienced by the charge at C towards the centre is

Electric flux through one face of the cube is given by Eo =

The electric intensity at P is E =

The transverse distance of electron in the direction of firing x = ut....... (1)

The transverse distance of the electron in the perpendicular direction is y = at 2

You might also like