Electron, charge body

Rest

Electrostatics
Coulomb’s law

q1 q2
r

Where K is a proportionality constant and its value is

Is called permittivity of free space or vacuum.

Electric Field
Electric field intensity (E)
Force experienced by unit positive charge placed at that point is called electric field
intensity or simply electric field. For a given charge q the electric field at distance r is
given by
P
1

q
It is a vector quantity direct outward for positive charge
and inward for negative charge.
 Electric Potential (V)
Electric potential of a given charge at any point is defined as the total work done
in bringing unit positive charge from infinity to that point.

Consider a unit positive charge is at point P at distance r from the given

charge q .
1
We have, small work done on moving small distance dr is
P
dw = -F.dr r

q
Total work done or potential can be obtained by integrating equation (i) from limit  to r
 1
P
r

q
Point Charge
Numericals :

1. A spherical drop of water carrying a charge of 30 pC has a potential of 500 V at its

surface (with V = 0 at infinity). (a) What is the radius of the drop? (b) If two such
drops of the same charge and radius combine to form a single spherical drop, what is
the potential at the surface of the new drop?

q= 30pC= 30 10-12 C R

V =500V

(a) 0.54mm (b) 790V

2. Two similar charges +q and mass m are suspended from
a thread of length l as shown in figure. Show that the θ
separation between them at equilibrium is TCosθ
l
Assuming θ is very small. T
θ

Tcosθ=mg ……(i) FC TSinθ

q q
Tsinθ= FC ………(ii) x

mg
3. Four charges 12nC,-24nC, 31nC and 17nC are placed at the corners of a square of
side 1.3m. Find the electric potential at the centre.

V= V1 + V2 + V3 + V4

q1 q4

q2 q3

(352V)
4. Two equal positive charges +q are placed at two corners of an equilateral triangle of
side a. Show that the electric field produced at third corner is

E
E1
E2
300

E= E1Cos30 + E2Cos30

q q
a
q1 q2
 Electric Dipole
An electric dipole consists of two equal and opposite charges separated by very small
distance. For a short dipole of length ‘2d’ having charges +q and –q dipole moment is
defined as the product of charge and distance of separation between charges
Dipole moment (p) = q × 2d
it is a vector quantity directed from negative to positive charge.

+q -q
2d
 Electric field due to a short dipole
(i) Along the axial line
r
O E- E+
Consider a short dipole of length 2d +q -q P
having charges +q and –q. P be a point
at distance r from the centre of dipole 2d
where electric field has to be determine.
We have electric field due to positive and negative charges are,

The resultant electric field at point P is

E = E- - E+
 For short dipole r >> d and p= 2qd is the dipole moment.

……………(iii)

This is the required expression for electric field along the axial line of an electric dipole.
 (ii) Along the equatorial line E+
θ
P E
Consider a short dipole of length 2d θ
having charges +q and –q. P be a point
at distance r from the centre of dipole along E-
the equatorial line where electric field has to
be determine. r

θ
We have electric field due to positive and negative
+q -q
point charges at point P is, O
2d

Along the direction as shown in figure.

Along the direction as shown in figure.

Since two electric field vectors are equal the resultant vector bisects the angle between
two vectors. Magnitude of resultant vector can be obtained by resolving the components
of equation (i) and (ii)
 Hence , the resultant Electric field at point P is E= E+Cosθ+E-Cosθ

For short dipole r >> d and p= 2qd is the dipole moment

Which shows that electric field along equatorial line is half of electric field along axial line
of an electric dipole.

……………..(iv)
 Electric Potential Due to a Dipole
P

Consider a short electric dipole AB of length

r2
2d having charges +q and –q with centre
at O. P be a point at distance r from the r1
centre of dipole and θ is angle made with r
axis by position vector r. M
-q +q
θ
We know electric potential at point P d O d θ B
A
due to positive and negative charges
are
N

Where r1 and r2 be the position of point P from

Charges +q and –q

The resultant electric potential at point P is V= V+ + V-

 For short dipole (r >> d) we can use the approximation MP ~ OP~NP~ r
And to find the value of r and r let us construct MN perpendicular to OP.
P
Now , in triangle BON Cosθ= NB/BO

Therefore , NB= dCosθ r2

Similarly , AM = dCosθ
r1
r
Hence equation (iii) becomes M
-q +q
θ
d O d θ B
A

For short dipole r >> d and p= 2qd is the

dipole moment

OR In vector form
Case I:Along the axial line θ= 00 and the potential is
maximum.

Case II: Along the equatorial line θ= 900 and

the potential is minimum.

V =0
V =0

- + Vmax
Electric Dipole in an Electric Field

Consider an electric dipole of length 2d is

placed in an electric field of intensity E by
making an angle θ with the axis. The 2
electric force experienced by each charge
in opposite direction is given by

F = qE……..(i)
Rotational torque experienced by the dipole is

torque (τ)= force x perpendicular distance

or, τ = qE Sinθ. 2d

= pE Sinθ
The arrangement of four charges in different geometry is called
an electric quadrupole. an electric quadrupole is an extension of
electric dipole consisting of two dipoles.

Quadrupole moment is define by the relation Q=2qd2 which is a Tensor quantity

Types of Quadrupole
d
+q -q +q -q

d
+q -q -q +q d d
d
2d +q -q -q +q
d
Linear Quadrupole Crossed Quadrupole
Square Quadrupole
 Electric Field Due to a Linear Quadrupole

Consider a linear quadrupole

AOB of length 2d having A O B
charges +q at point A and B , -2q EA
+q -q -q +q EO P
at point O. P be a point at
EB
distance r from the centre of
quadrupole where electric field 2d
r
has to be determine.
We have electric field at
point P due to charges at point
A, O and B are

direction is outward from Quadrupole

direction is outward from Quadrupole

direction is towards the Quadrupole

 The resultant electric field at point P is E= EA +EB – EO ………….(iv)

For short electric quadrupole r >> d and Q= 2qd2 is the Quadrupole moment

This is the required expression for electric field due to a linear Quadrupole.
Electric Potential Due to a Quadrupole

Consider a linear quadrupole

AOB of length 2d having A O B
charges +q at point A and B , -
+q -q -q +q P
2q at point O. P be a point at
distance r from the centre of 2d
quadrupole where electric r
potential has to be determine.
We have electric potential
at point P due to charges at
point A, O and B are
 The resultant potential at point P is V=VA + VB + VO ……………(iv)

For short electric quadrupole r >> d and Q= 2qd2 is the Quadrupole moment

Which is the required expression for electric potential along the axis of
a linear Quadrupole.
 Electric field due a continuous charge distribution

dECosθ
Electric Field Due to a Charged Rod dE

Consider a uniformly charged rod of plastic

having linear charge density λ (charge per θ
unit length). The rod where P be a point dESinθ
P
at distance z from the centre of the rod where
electric field has to be determine. θ
let dx be a segment of the rod having z
small charge dq at distance x from the centre
and produces a small electric field dE at point
P along the direction as shown in figure. dq
+ + + + + + + + + + + + + +q
x
We know the small electric field at point P is dx
This electric field makes an angle  with vertical. So, it can be resolved into two
components dEcos and dEsin. Taking similar charge element on the other side of the
rod, we notice that electric field set up by this element also has magnitude dE, the
components dE sin  is cancelled out being equal and opposite. So, resultant electric field
due to charged rod is

Case I : for a rod of infinite length

For even function

Let x = zTanθ
Then, dx= zSec2θdθ

When x = 0, 𝜃 = 0
𝜋
When 𝑥 = ∞, 𝜃 =
2

…………..(iii)
Case II : for a rod of finite length l

For even function

Let x = zTanθ
Then, dx= zSec2θdθ

When x = 0 , θ = 0

(say)
 l
θ0
2z

θ0 =Tan-1 (l /2z)

Where q =λl is total charge on rod

Which is the required expression for electric field due to a charged rod of
finite length.

HW; prove that: Which is the electric field

due to infinite length rod.
 dECosθ
Electric Field Due to a Charged Ring
dE

Consider a ring of radius a, charge q and linear

charge density . Electric field at point P and at dESinθ
distance z from centre of the ring can be calculated P
by integrating the electric field due to small θ
segment of ring of length dx. Horizontal
components dE sin will be cancelled out being z
equal and opposite round the ring, so the field is
only due to vertical components dEcos.
+ + + +
+
We know the small charge on small segment dx is
dq + +
dq = λdx ………….(i) dx R O +q
The small electric field at point P is + +
+ + +
 The total electric field can be obtained by integrating the vertical component of
small electric field from limit 0 to 2πR

Where q = λ .2πR is total

charge on ring.

This is the required expression for electric field due to a charged ring along its axis.
 At the centre of ring z =0, and E = 0

As the distance from centre increase value if electric field also increases, and again
decreases for large distance from the centre. So the maximum electric field can be
obtained by differentiating equation (iii) with respect to z.

And the maximum field is

 If a negative charge –q’ of mass m is placed along the axis of ring, it attracted towards the
centre of ring by an electric force
F = -q’E ………..(i)

+ +
+ +
R +q
O
+ +
+
+ +

where

Which is the differential equation of Simple

Harmonic Motion so the particle execute SHM
along the axis of the ring with time period
Near the centre of ring R >> z
 Electric Potential Due to a Charged Ring P dV

Consider a ring of radius a, charge q and linear charge density

. Electric potential at point P and at distance z from centre of
the ring can be calculated by integrating the small electric
potential due to small segment of ring of length dx. z

We know small electric potential at point P is

+ + + +
+
dq P
+ +
[ dq = λdx ] dx R rO +q
+ +
+ + +
Total electric potential can be obtained by integrating
small potential from limit 0 to 2πR

Where q = λ.2πR is total charge in ring.

 Electric Field Due to a Charged Disc dE

Consider the disc of radius R has uniform

surface charge density . To find the electric P
field at point P at distance ‘z’ from the centre of
the disc, we consider the disc as a set of
concentric rings and calculate electric field at P
due to one of the ring. The vector sum of all
those fields gives the field due to whole disc. z

Let a thin ring of radius x and width dx having small dx

positive charge dq lies inside the disc, and produces
a small electric field dE at point P along the axis of +
+
disc as shown in figure. + +
R
+ +
We have small charge on ring O x + +q
(dq)= surface charge density  Area +
+
+
dq =  2π xdx ………(i)
dx
2πx
 Now, the small electric field at point P due to the small charged ring of radius x is

The total electric field of the disc can be obtained by integrating small electric
field from limit 0 to R .
 R
θ0
z
Which is the expression for electric field of uniformly charged ring.

For a disc of large radius R>> z. And the electric field is

Electric Potential Due to a Charged Disc
Consider the disc of radius R has uniform surface charge
density . To find the electric potential at point P at distance dV
P
‘z’ from the centre of the disc, we consider the disc as a set
of concentric rings and calculate electric potential at P due
to one of the ring. The total electric potential can be
obtained by integrating potential due to such ring.
Let a thin ring of radius x and width dx having small z
positive charge dq lies inside the disc, and produces
a small electric potential dV at point P as shown in figure.
dx
We have small charge on ring
(dq)= surface charge density  Area + +
+ +
R
dq =  2π xdx ………(i) + +
O x + +q
We know the small potential at point P due to a thin +
+
ring of radius x with charge dq is +
 ……………(ii)

The total electric potential of the disc can be obtained by

integrating small electric potential from limit 0 to R

Let y2 = z2 + x2 When x=0, y= z

then ydy = x dx
When x=R, 𝑦 = 𝑧 2 + 𝑅2
 Relation between electric field and Potential
Consider a test charge q0 is moving
along given charge q. On moving a
dV
small distance dr; small work done
against the field at distance r is dw, and
change in potential is dV. +q q0
r dr
We know small work done against
P
the field is
dw = –F.dr ……(i)
Or, q0dV = –Eq0dr
……….(ii)

In three dimension,

And electric potential is ……….(iii)

Charges Electric Field (E) Electric Potential (V)

1. Point Charge

2. Electric Dipole

3. Electric
Quadrupole

4. Ring

5.Disc

6.Sphere
 Numericals:
1. What is the magnitude of the electric field at the point (2, 3) m if the electric
potential is given by V = 2x + 5xy + 3y2 volts? What acceleration does acceleration
experience in the x-direction? (IOE)
x-component of electric field

y-component of electric field

Resultant Electric field = 32.75V/m

Now, Force experienced by electron along x- axis (Fx) =eEx

or, ma =eEx

= 1.761011 17 = 29.92 1011 m/s2

 2. A particle of charge –q and a mass m is placed midway between two equal positive
charges qo of separation d. If the negative charge –q is displaced in perpendicular
direction to the line joining them and released, show that the particle describe a
SHM with a period
-q

Let the negative charge is placed at displacement y from

mid-point F1 θ F2

Force exerted by charge q0 from coulomb’s law y

F
- q0 q0
d

From figure resultant force makes an angle θ with individual force , given by

- -
F= 2F1Cosθ

-
Near the centre y 0

And, = 2π/T
 ds

It states that the total electric flux trough a

closed surface enclosing a charge is equal to 1/ε0
times the total charge enclosed by the surface.
Mathematically;
q
Total electric flux (φ)

Gaussian surface
Which is Gauss law in Electrostatics

In differential form
 Application of Gauss Law
P
Electric field due to a charged sphere
(i) Outside the Sphere r +
Consider a uniformly charged sphere of radius R with total +
charge +q. P be a point outside the sphere at distance r R
from the center where electric field has to be determine. + O +q
Let us construct a Gaussian sphere of radius r +
+
through the point P.

We have from Gauss law ………(i)

Gaussian Sphere

………(ii)

Which is same as electric field due to a point charge. Hence a charged sphere behaves
as if all the charge of the sphere were concentrated in a point charge at its centre.
(ii) Inside the Sphere
Consider a uniformly charged sphere of radius R with total
charge +q. P be a point inside the sphere at distance r +
from the center where electric field has to be determine. P
+ r
Let us construct a Gaussian sphere of radius r R
through the point P. + O +q
+
We have from Gauss law ………(i) +

Where q’ is charge enclosed by Gaussian sphere

Gaussian Sphere

………(i)

Now, volume charge density of sphere is

At the centre of sphere r = 0 and E=0

At the surface of sphere r = R

O R
r
1. Calculate the electric field due to a uniformly charged rod of length L at
distance a from its nearest end along its axis (IOE)
Let dx be the small segment of the rod with small charge dq which produce
small electric field dE at point P given by
x dx a
+q + + + + + + + P
dq dE
L
 2. A neutral water molecule in its vapor state has an electric dipole moment of
magnitude 7.1×10–30 Cm. If the molecule is placed in an electric field of
(IOE) 2.5×104N/C, (i) What maximum torque can the field exert on it? (ii) How much
work must an external agent do to turn this molecule end for end in this field?

Here, Electric dipole moment (p) = 7.1×10–30 Cm

Electric field (E) = 2.5×104N/C
We know the torque acting on dipole (τ) = pE = pE Sinθ

Maximum torque (τmax ) = pE Sin90

= 17.7510-26 Nm

Let dw be the small work done on rotating the dipole

through small angle dθ

or , small work done (dw) = τ dθ = pE Sinθdθ

dθ
-q +q
Now, work done by external agent

= 2pE
W= 3.5510-27 J
 3. A plastic rod contains a uniformly distributed -q charge. The rod has been bent in
120 circular arc of radius ‘r’ as shown in figure below. Prove that the electric field
intensity at the center of bent rod
(IOE)
dq
Let dx be the small segment of the rod at an angle θ with dx
dθ
small charge dq which produce small electric field dE at
centre; given by -q

dE
Direction is
radially outward dECosθ
θ
Resultant electric field directed outward bisecting the 1200
circular arc, which can be obtained by integrating horizontal
component of small electric field from limit -600 t0 +600
 4. A particle of charge –q and a mass m is placed midway between two equal positive
(IOE) charges qo of separation d. If the negative charge –q is displaced in perpendicular
direction to the line joining them and released, show that the particle describe a
SHM with a period
-q

Let the negative charge is placed at displacement y from

F1 θ F2
mid-point
Force exerted by charge q0 from coulomb’s law y
F
- q0 q0
d

From figure resultant force makes an angle θ with individual force , given by

- -
F= 2F1Cosθ

-
Near the centre y 0

And, = 2π/T
 2075 Ashwin
Numericals:
9. Charges of uniform volume charge density 3.2 mC/m3 fill a non- conducting solid
sphere of radius 5 cm. What is the magnitude of the electric field at (a) 3.5 cm (b) 8
cm from the centre of the sphere? (IOE)

We know, total charge on sphere (q) = .V = 1.675μC

+
(a) Electric field inside the sphere P
+

+ +q
+
+
= 4221 N/C
8cm
(b) Electric field outside the sphere P

= 2355.47 N/C
9. Two large parallel plates are separated by a distance of 5cm. The plates have equal but
opposite charges that create an electric field in the region between the plates. An alpha
particle (q = 3.2×10–19C,m = 6.68×10–27kg) is released from the positively charged plate, and
it strikes the negatively charged plate 2×10–6 sec later. Assuming that the electric field
between plates is uniform and perpendicular to the plates, what is the strength of electric
field? (IOE)

+ _
Force experienced by an - particle (F) = Eq
+ -
or, ma = Eq …..(i)
+ -

+ -

+ -
t = 2μs
+ -

5cm

521.8V/m
 10. Calculate the potential at a point due to a uniform line of charge of length L at a
distance D from its one end which lies in the perpendicular line. (IOE)

Let dx be the small segment of the rod at distance x from

the left end having small charge dq and produces a small P dV
potential dV at point P ,  be the linear charge density.
We have small potential at point p is
D

dq
+q + + + + + + +
x dx
L

And the total potential

 11. Charge is distributed uniformly throughout the volume of an infinitely long
cylinder of radius a. Show that electric field at a distance r from the cylinder
axis r < a is given by where  is volume charge density

Let us construct a Gaussian cylinder of length l and radius r enclosing

a charge q as shown in figure.

a
We have from Gauss law, + +
+ +
+ + +

+ + r+
P
+ + +
[ q= V ] Gaussian cylinder
 12. A spherical charge distribution is given by
Find (a) Total amount of charge (b) Electric field outside the sphere (c) inside the sphere (d)
Position for which Electric field is maximum. (a is radius of the sphere)

Soln: Consider a thin spherical shell of radius r and thickness dr having dr

small charge dq given by
dq = .dV………….(i)
+
Total charge (q) 4πr2dr + +
+
+ r a +
+ +++ ++ +
(b) Using Gauss law E.4πr2 = q/ε0 + +++++
++
+
+ +
(c) Again using Gauss law P
+
+ +
P
or, E.4πr2

(d) For maximum electric field

2074 Bhadra
13. If a copper coin has mass 3.11gm, what is the total charge on the nucleus of
the atom in the coin? Also find the number of protons in the nucleus. Molar
mass (M) = 63.5gm/mole, Avogadro number (NA) = 6.02×1023 atom/mole.

We know , 63.5gm of copper contains NA atoms of copper

1gm of copper contains (NA/M) atoms of copper

3.11gm of copper contains (NA/M)3.11 atoms of copper

Total charge on nucleus of copper coin (Q)

3.11gm
= 13.68104C

Total number of proton (n) = Q/e

= 8.551023 protons
8. Charges of uniform volume charge density 3.2 μC/m3 fill a non- conducting solid
sphere of radius 5 cm. What is the magnitude of the electric field at (a) 3.5 cm (b) 8
cm from the centre of the sphere?

We know, total charge on sphere (q) = .V = 1.675nC

+
(a) Electric field inside the sphere P
+

+ O +q
+
+
= 4410 N/C

8cm
(b) Electric field outside the sphere P

= 307.62N/C
 10. Calculate the potential at a point due to a uniform line of charge of length L at a
distance D from its one end which lies in the perpendicular line.

Let dx be the small segment of the rod at distance x from

the left end having small charge dq and produces a small P dV
potential dV at point P ,  be the linear charge density.
We have small potential at point p is
D

dq
+q + + + + + + +
x dx
L

And the total potential

11. Given the electric field due to a charged disc of radius R at distance z from its centre is

Find the potential at that point

12. Given the electric potential due to a charged ring of radius R at distance z from its
centre is
Find the electric field at that point