ELECTRIC FIELD

Example:

A ring-shaped conductor with radius a carries a total positive charge Q

uniformly distributed on it. Find the electric field at a point P that lies

on the axis of the ring at a distance x from its centre.

Solution:

The figure shows the electric field dE at the point P due to an

infinitesimal segment at the top of the ring. The x-and y-components of

dE is dE cos Ө and dE sin Ө. If we take another segment of the bottom,

we see that its contribution to the field at P has the same x-component

but opposite y-component. Hence total y-component becomes zero.

Hence the field at P will be along x-axis. Therefore, the net field at P is

E = dEcos

=

2 2 2 20

1 dQ x. .

4 x a x a

=

3

2 2 20

xdQ

4 x a

=

3

2 2 20

Qx

4 x a

or,

3

2 2 20

QxE i

4 x a

(1) At the centre of the ring, x = 0 , E = 0

(2) If x >> a, 2

0

1 QE i

4 x

This field is same as that of a point charge at the centre.

(3) E will be maximum at the points where dE

dx = 0

or, x = a

,2

and Emax = 2

0

1 2Q.

4 3 3a

The graph of the variation of E with x is as shown in figure.

The negative value of E indicates that the direction of the field is

opposite to the positive value.

(4) If a >> x,

E = 3

0

Qx

4 a

=

2

0

x,2 a

where is the linear charge density.

The force on a negative point charge –q is

F =

2

0

qx

2 a

= – kx,

where k =

2

0

q

2 a

On releasing the point charge (–q) of mass m, it will execute SHM (if x <<

a) with the time-period.

T = m

2k

=

02 m2 a

q

It should be noted that if x is not negligible relative to a, the negative

charged particle will oscillate about the centre but the motion will not be

simple harmonic.

Calculations of electric field due to continuous charge distribution.

(1) Linear charge distribution

Infinite line of charge:

x = d tan Ө

dx = d sec2 Ө dӨ

dEP = 2

0

dq1

4 r

=

2 20

1 dx

4 x d

dEP =

2

2 2 20

1 dsec d

4 d tan d

dEP =

0

1d

4 d

E = dEcos

=

0

d cos4 d

=

/2

/20

sin4 d

=

0

sin sin4 d 2 2

E =

02 d

Semi infinite wire

E1 =

04 d

E11 =

04 d

Enet =

0

2

4 d

E1 = E11

=

04 d

Enet =

0

2

4 d

For finite wire

E1 =

0

sin sin4 d

E11 =

0

cos cos4 d

Electric field for uniformly charged arc

dl = R dӨ

dq = l dl

= mRdӨ

dE = 2

0

1 dq

4 R

=

2

0

1 Rd

4 R

dE =

0

d4 R

E = dEcos

=

/2

0 /2

cos d4 R

E =

0

sin2 R 2

ф in radians.

ELECTROSTATIC POTENTIAL

Electrostatic potential is the scalar way of representing the region

around any electric charge configuration (as the electric field was the

vector way of representation) which is useful in calculating the work

involved when a charge moves around other charges or field.

When a charge is moved slowly around another charge, some electrical

work may be done and the work done by the electric field is equal and

opposite to the work done by external agent during the same

movement.

Physically you know that high potential means high capacity to do work.

Closer you are to a positive charge, higher is the potential experienced

by you (means the positive charge will do large work in pushing a unit

charge from that point to infinity). Closer you are to a negative charge,

lesser is the potential experienced by you.

SI unit for electric potential that follows from its definition is the "joule

per coulomb". This combination is given a special name, called "volt" in

honour of the Italian scientist Alessandro Volta (1745-1827).

1 volt = 1 V = 1 J/C = 1 joule/coulomb

(i) Test charge is an infinitesimal positive charge.

(ii) Potential is a scalar quantity.

(iii) Potential may be positive, negative or zero.

(iv) The potential at a point in the electric field due to an isolated positive

charge will be positive and that due to an isolated negative charge will

be negative relative to zero potential at infinity.

(v) Potential differences are fundamental since they do not depend on

defining an arbitrary reference point and assigning an arbitrary potential

energy to it. But the potential at a point depends on defining an arbitrary

reference point and assigning an arbitrary potential energy to it.

(vi) The potential difference Va – Vb is called potential of a with respect to b.

Sometimes it is abbreviated as Vab i.e., Vab = Va – Vb.

(vii) When a charge q moves from i to f in an electric field, the word done by

the electric field on the charge q is Wele = q(Vi – Vf). For a positive charge,

W > 0 if Vi > Vf. Hence a positive charge moves from high potential to low

potential if it is free to move. For a negative charge, W > 0 if Vi < Vf.

Hence a negative charge moves from low potential to high potential if it

is free to move.

(viii) The work done by the external force is Wext = q(Vf – Vi), (note the order

of subscripts), if the charge q moves from i to f in a electric field very

slowly.

Potential due to a Point Charge

It is the work done by an external agent in moving unit positive charge

slowly from infinity to the point where potential is to be calculated.

Figure shows an isolated positive point charge of magnitude q. We want

to find potential V due to q at a point P, at a radial distance r from that

charge.

This potential will be determined relative to zero potential at infinity.

Let us suppose that a test charge (positive) q0 moves from infinity to

point P. Since the work done by the electric field is independent of the

path followed by the test charge, we take the simplest path along the

radial line from infinity to q passing through P.

Consider an arbitrary position P' at a distance r' from q. The electrostatic

force on q at P' is

02

0

qq1F ,

4 r ' directed radially away. For a further differential

displacement ds, the work done by the force F is

dW = – F.ds

Since the direction of ds is opposite to the direction of increasing r'.

Hence ds dr '

So, dW = Fdrc

=

02

0

qq1dr '

4 r '

or, W =

r

0 2

0

1 dr 'qq

4 r '

= W

=

r

0

0

1 1qq

4 r '

=

0

0

1 1 1qq

4 r

=

0

0

qq

4 r

0 0

W qV

q 4 r

…[potential due to a point charge]

If the point charge is negative the electrostatic force points towards q,

and hence W will be positive, giving a negative potential V.

Thus the sign of V is same as the sign of q, when we assign zero potential

at infinity.

Potential due to an Electric Dipole

Figure shows a dipole of dipole moment p = qd. We have to find the

electric potential at the point P, distant r from the centre O of the dipole.

The line OP makes an angle Ө with the dipole axis. The potential at P due

to the positive charge (+ q) is

V = 0 1

1 q

4 r

and that due to the negative charge (– q) is

V2 =

0 2

1 q

4 r

The net potential at P is

V = V1 + V2

=

0 1 2

1 q q

4 r r

=

2 1

0 1 2

r rq

4 r r …(i)

For a small dipole, we have usually, r >> d, where "d" is the separation

between the two charges. Under these conditions, We may write, r2 – r1

- d cos Ө and r1 r2 - r2 using these quantities in equation (i), we get

V =

2

0

q dcos.

4 r

Example:

Infinite number of same charge q are placed at

x = 1, 2, 4, 8 ...... What is the potential at x = 0 ?

Solution:

V =

0

1 q q q q....

4 1 2 4 8

=

0

q 1

141

2

= 0

2q

4

= 0

q

2

Example:

If the alternative charges are unlike, then what will be the potential?

Solution:

Then,

V =

0

1 q q q q.........

4 1 2 4 8

=

0

1 q q q q..... .....

4 1 2 2 8

=

0

1 1 1 1

1 14 21 1

4 4

= 0

1 2q

4 3

Potential difference:

The work done in taking a charge from one point to the other in an

electric field is called the potential difference between two points.

Thus, if w be work done in moving a charge q0 from B to A then the

potential difference is given by–

VA – VB = 0

W

q

(i) Unit of potential difference is volt.

(ii) This is a scalar quantity.

(iii) Potential difference does not depend upon Co-ordinate system.

(iv) Potential difference does not depend upon the path followed.

This is, because electric field is a conservative force field and work

done is conservative force field does not depend upon path

followed.

Example:

In the following fig. Along which path the work done will be maximum

in carrying a charge from A to B in the presence of any another charge.

Solution:

Same for all the path

[Because the work done doesn't depend upon the path]

Example:

A charge 20 µC is situated at the origin of X-Y plane. What will be

potential difference between points (5a, 0) and (–3a, 4a)

Solution:

Distance between (0, 0) & (5a, a),

r1 = 225a 0

= 5a

1

kqV

5a

Distance between

(0, 0) & (–3a, 4a)

r2 = 2 29a 16a

= 5a

V2 = kq

5a

V1 – V2 = 0

Relationship between electric potential and intensity of electric field

(i) VA =

A

E.dr,

VA = electric potential at point A.

(ii) Potential difference between two points in an electric field is given by

negative value of line integral of electric field i.e.,

VB – VA = B

A

E.dr

(iii) E V

= – grad

= (gradient)

=

i j k

x y z

Ex =

V,

x

Ey =

V,

y

Ez =

V

z

(iv) If v is a function of r only, then E = dV

dr

(v) For a uniform electric field, E =

V

r and it's direction is along the

decrease in the value of V.

Example:

Electric potential for a point (x, y, z) is given by V = 4x2 volt. Electric

field at point (1, 0, 2) is –

Solution:

E = dV

dx

= – 8x

E at (1, 0, 2)

= – 8 V/m

Magnitude of E

= 8V/m direction along –x axis.

Example:

Electric field is given by E = 2

100

x potential difference between x = 10

and x = 20 m will be

Solution:

E = dV

dx

dV Edx

B B

A A

dV E.dx

20

B A 2

10

100V V

x

= – 5 volts

Potential difference = 5 volt.

Example:

The potential at a point (x, 0, 0) is given as V =

2 3

1000 1500 500.

x x x What will be electric field intensity at x =

1m ?

Solution:

E = – V

=

V V Vi j k

x y z

or iEx + jEy + kEz

=

V V Vi j k

x y z

=

V

x

V V0

y z

Comparing both sides

Ex =

V

x

=

2 3

1000 1500 5000

x x x x

=

2 3 4

1000 2 1500 3 5000

x x x

For x = 1,

(Ex) = 5500 V/m

Example:

What will be the electric field intensity at r = 3.

Solution:

For 2 < r < 4,

V = 5 volts

dV

E 0dr

In the above problem, what will value of E at r = 6 ?

at r2 = 7m

V2 = 2 volt

at r1 = 5m

V1 = 4 volt

E =

2 1

2 1

V V

r r

=

2 4

7 5

= 1 volt/metre.

Example:

An oil drop 'B' has charge 1.6 x 10–19 C and mass 1.6 × 10–14 kg. If the

drop is in equilibrium position, then what will be the potential diff.

between the plates.

[The distance between the plates is 10mm]

Solution:

For equilibrium, electric force = weight of drop

qE = mg

V

q. mgd

mgd

Vq

=

14 3

19

1.6 10 9.8 10 10

1.6 10

V = 104 volt

[When a charged particle is in equilibrium in electric field, the following

formula is often used qE = mg]

Equipotential Surface –

(i) These are the imaginary surface (drawn in an electric field) where

the potential at any point on the surface has the same value.

(ii) No two equipotential surfaces ever intersects.

(iii) Equipotential surfaces are perpendicular to the electric field lines.

(iv) Work done in moving a charge from a one point to the other on

an equipotential surface is zero irrespective of the path followed

and hence there is no change in kinetic energy of the charge.

(v) Component of electric field parallel to equipotential surface is

zero.

(iv) Nearer the equipotential surfaces, stronger the electric field

intensity.

Example:

Some equipotential surfaces are shown in fig.

What is the correct order of electric field intensity?

Solution:

EB > EC > EA, because potential gradient at B is maximum.

POTENTIAL ENERGY OF CHARGED PARTICLE IN ELECTRIC FIELD

(i) Work done in bringing a charge from infinity to a point against the

electric field is equal to the potential energy of that charge.

(ii) Potential energy of a charge of a point is equal to the product of

magnitude of charge and electric potential at that point i.e. P.E. =

qV.

(iii) Work done in moving a charge from one point to other in an

electric field is equal to change in it's potential energy i.e. work

done in moving Q from A to B = qVB – qVA

= UB – UA

(iv) Work done in moving a unit charge from one point to other is

equal to potential difference between two points.

NOTE:

Circumference of the circle in above example can be considered as

equipotential surface and hence work done will be zero.

Potential Energy of System:

(i) The electric potential energy of a system of charges is the work

that has been done in bringing those charges from infinity to near

each other to form the system.

(ii) If a system is given negative of it's potential energy, then all

charges will move to infinity. This negative value of total energy is

called the binding energy.

(iii) Energy of a system of two charges

PE = 1 2

0

q q1

4 d

(iv) Energy of a system of three charges

PE = 2 3 3 11 2

0 12 23 31

q q q qq q1

4 r r r

(v) Energy of a system of n charges.

PE = n n

j

ii 1 j 10 ij

i j

q1 1. q

2 4 r

Electric Field

Example:

Two symmetrical rings of radius R each are placed coaxially at a

distance R meter. These rings are given the charges Q1 & Q2

respectively, uniformly. What will be the work done in moving a charge

q from center of one ring to centre of the other. i

Solution:

Work done = q (V2 – V1) potential at the centre of first ring

V1 = 1 2

2 20 0

Q Q1 1

4 R 4 R R

= 21

0

Q1Q

4 R 2

potential at the centre of second ring

V2 = 2 1

2 20 0

Q Q1 1

4 R R 4 R R

= 12

0

Q1Q

4 R 2

work done = q (V2 – V1 )

= 2 11 2

0

Q QqQ Q

4 R 2 2

1 2 1 2

0

q 1W Q Q 1

4 R 2

NOTE:

Work done in moving the same charge from second to the first ring will

be negative of the work done calculated above i.e.

2 1 1 2

0

q 1W Q Q 1

4 R 2

ELECTRIC DIPOLE

(i) A system consisting of two equal and opposite charges separated

by a small distance is termed an electric dipole.

Example:

Na+ Cl–, H+ Cl– etc.

(ii) An isolated atom is not a dipole because centre of positive charge

coincides with centre of negative charge. But if atom is placed in

an electric field, then the positive and negative centres are

displaced relative to each other and atom become a dipole.

(iii) DIPOLE MOMENT:

The product of the magnitude of charges and distance between

them is called the dipole moment.

(a) This is a vector quantity which is directed from negative to

positive charge.

(b) Unit:

Coulomb – metre (C-M)

(c) Dimension:

[M0 L1 T1 A1]

(d) It is denoted by p that is p qd

Electric field due to a dipole

(i) There are two components of electric field at any point

(a) Er in the direction of r

(b) Er in the direction perpendicular to r

Er = 3

0

1 2Pcos.

4 r

E = 3

0

q Psin.

4 r

(ii) Resultant

E = 2 2

rE E

= 2

3

0

P1 3cos

4 r

(iii) Angle between the resultant E and r, is given by

Ө= tan–1 r

E

E

= tan–1 1

tan2

(iv) If Ө = 0, i.e. point is on the axis –

Eaxis = 3

0

1 P.

4 r

Ө = 0, i.e. along the axis.

(v) If Ө = 90°, i.e. point is on the line bisecting the dipole

perpendicularly

Eequator = 3

0

1 P.

4 r

(vi) So, Eaxis = 2Eequator (for same r)

(vii) Eaxis =

2

2 20

1 2Pr.

4 r

Eequator =

3/2

2 20

1 2P.

4 r

where P = q . 2

(viii) V =

2

0

q 2 cos1.

4 r

= 2

0

1 Pcos.

4 r

= 2

0

1 P.r.

4 r

= 3

0

1 P.r.

4 r

where Ө is the angle between P and r.

(ix) If Ө = 90°, Vequator = 0

(x) Here we see that V = 0 but E 0 for points at equator.

NOTE :

(i) This is not essential that at a point, where E = 0, V will also be

zero there eg. inside a uniformly charged sphere, E = 0 but V ≠ 0

(ii) Also if V = 0, it is not essential for E to be zero eg. in equatorial

position of dipole V = 0, but E ≠ 0

Electric Dipole In an Electric Field – Uniform Electric Field

(i) When an electric dipole is placed in an uniform electric field, A

torque acts on it which subjects the dipole to rotatory motion.

This Ʈ is given by Ʈ = PE sin Ө

or P E

(ii) Potential energy of the dipole

U = – PE cosӨ

= P.E

or, 2

0

1 pcosV .

4 r

…*potential due to an electric dipole+

where p = qd is the magnitude of the electric dipole moment.

Potential due to a System of Charges

Suppose a system contains charges q1, q2, q3 ....... qn, at distances r1, r2, r3

....... rn, from a point P.

The potential at P,

Due to q1 is V1 = 1

0 1

q1

4 r

Due to q2 is V2 = 2

0 2

q1

4 r

Due to q3 is V3 = 3

0 3

q1

4 r

and so on.

By the superposition principle, the potential V at P due to all the charges

is the algebraic sum of potentials due to the individual charges.

Therefore, net potential at P is

V = V1 + V2 + V3 + …… + Vn

or, 31 2 n

0 1 2 3 n

qq q q1V .............

4 r r r r

Example :

In the given figure, there are four point charges placed at the vertices

of a square of side a = 1.4 m. If q1 = + 18 nC, q2 = – 24 nC, q3 = + 35 nC

and q4 = + 16 nC, then find the electric potential at the centre P of the

square. Assume the potential to be zero at infinity.

Solution:

The distance of the point P from each charge is

r = a

2

= 1.4m

1.4

= 1m

V = V1 + V2 + V3 + V4

= 31 2 4

0 1 2 3 4

qq q q1

4 r r r r

= 1 2 3 4

0

q q q q1.

4 r

= 9

0

18 24 35 16 10 C1.

4 1m

= (9 × 109) × (45 × 10–9) V

= 405 V

Example:

Two point charges – 5 µC and + 3 µC are placed 64 cm apart. At what

points on the line joining the two charges is the electric potential zero.

Assume the potential at infinity to be zero.

Solution:

For the potential to be zero, at the point P, it will lie either between the

two given point charges or on the extended position towards the charge

of smaller magnitude.

Suppose the point P is at a distance x cm from the charge of larger

magnitude. Then from figure (a)

6 6

2 2

0

1 5 10 3 10. 0

4 x 10 64 x 10

or, 3 5

064 x x

or,

8x 3200

x 64 x

or, x = 40 cm

From the figure (b), we get

3 5

0x 64 x

or,

320 2x0

x x 64

or, x = 160 cm

Potential due to a Continuous Charge Distribution

When a charge distribution is continuous, instead of consisting of

separate point charges, we divide it into small elements, each carrying a

charge dq. Considering this differential element of charge dq as a point

charge, we determine the potential dV at the given point P due to dq,

and then integrate over the continuous charge distribution.

Assuming the zero of potential to be at infinity,

dV = 0

1 dq

4 r

Where r is the distance of P from dq.

The potential V at P due to the entire distribution of charge is given by

0

1 dqV dV

4 r

Case I : Line charge distribution

For the charge uniformly distributed along a line dq = dl, where is the

linear charge density and dl is a line element.

Case II : Surface charge distribution

For the charge uniformly distributed on a surface with surface charge

density , the charge on a surface element dA is dq = dA

Case III: Volume charge distribution

For the charge uniformly distributed in a volume with volume charge

density , the charge in a volume element dV is dq = dV.

Example :

Electric charge Q is uniformly distributed around a thin ring of radius a.

Find the potential at a point P on the axis of the ring at a distance x

from the centre of the ring.

Solution:

Since the entire charge is at distance r = 2 2x a from the point P,

therefore, the potential at P due to the total charge Q is

V = 2 2

0

1 Q

4 x a

NOTE:

At the centre, V = 0

1 Q

4 a

Various Cases:

A thin uniformly charged spherical shell

(a) Potential at an outside point:

We already know that for all points outside the sphere, the field is

calculated as if the sphere were removed and the total charge q is

positioned as a point charge at the centre of the sphere.

Hence, the potential at an outside point at a distance r from its

centre is same as the potential due to a point charge q at the

centre.

i.e., 0

1 qV

4 r

…*for an outside point+

Remember:

This is valid only when the charge distribution is uniform.

(b) Potential at an inside point:

We know that inside the spherical shell of uniform distribution of

charge, the electric field E is zero everywhere. Hence, if a test

charge q0 is moved inside the sphere no work is done on that

charge against the electric force. Hence for any infinitesimal

displacement anywhere inside the sphere, W = 0 or q0 V = 0 or,

V = 0. Hence V (potential) remains constant inside the sphere and

is equal to its value at the surface

i.e., 0

1 qV

4 R

…[for an inside point, r < R]

A Charged Conducting Sphere

Since the charge on a conducting sphere is uniformly distributed on its

surface, therefore, it acts just like a spherical shell of charge uniformly

distributed in it. So

V = 0

1 q

4 r

...[for an outside point, r > R]

and V = 0

1 q

4 R

...[for an inside point, r < R]

Important Points about Conducting Sphere:

For a charged spherical shell

(1) Vinside = Vsurface = REsurface = constant

(2) Voutside = rEoutside

It increases or decreases with increasing r depending on whether q < 0 or

q > 0.

(3) Figure shows the variation of V with r for a positively charged

spherical shell. V is maximum inside and on the surface of the

sphere.

(4) Figure shows the variation of V with r for a negatively charged

spherical shell.

V is minimum inside and on the surface of the sphere. It is

maximum (= 0) at infinity.

(5) The maximum potential (Vm) upto which a spherical shell can be

raised is

Vm = REm

where Em is the electric field magnitude at which air becomes

conductive (known as dielectric strength of air). Its value is about

3 × 106 N/C.

Thus, a conducting sphere of 1 m radius can be charged upto a

maximum potential of 3 × 106 volt in air.

Potential Due to Uniformly Distributed Charge in a Spherical volume

Case I:

At a point P outside the sphere:

As we know for all external points, the total charge q is supposed to be

concentrated at the centre of the sphere. So the potential at an outside

point P, distant r from the centre O is

VP = 0

1 q

4 r

…potential at an outside point i.e., for r > R

Case II:

At a point P inside the sphere:

Let us divide the entire sphere into two parts, one is a sphere of radius

OP = r and the other is the spherical shell of internal radius r and external

radius R. As in case I, the potential at P due to the charge of sphere of

radius r is

V1 = 0

1 q'

4 r

=

3

0

4r

1 34 r

= 2

0

r

3

where 3

q

4R

3

For finding the potential due to the other part, let us consider a thin

spherical element of radius x and thickness dx. Its volume is 4πx2dx, and

hence it contains the charge dq = (4πx2dx)p = 4πpx2dx

The potential at P due to this element is

dV = 0

1 dq

4 x

= 2

0

1 4 x dx

4 x

= 0

1xdx

The potential at P due to the spherical shell of internal radius rand

external radius R is

V2 = dV

= 0

xdx

=

R2

0 r

x

2

= 2 2

0

R r2

So, the net potential at P is

V = V1 + V2

= 2 22

0 0

R rr

3 2

= 2 2

0

3R r

6

= 2 2

3

0

q3R r

4R 6

3

or, V = 2 2

3

0

q3R r

8 R

….[potential at an internal point i.e., r < R]

(1) Putting r = R, we get Vsurface = 0

q

4 R

(2) Putting r = 0,

we get Vcentre = 0

3q

8 R

= surface

3V

2

(3) Figure shows the Variation of V with r. For

0 < r < R, the graph is parabolic.

For r > R, the graph is hyperbolic.

EQUIPOTENTIAL SURFACES

"An equipotential surface is a surface on which the electric potential V

is same at every point".

The electric field in a region can be represented by a family of

equipotential surfaces, each of which corresponds to different potential.

Figure (a) and (b) show different arrangement of charges and

corresponding field lines, and cross-sections of equipotential surfaces for

(a) the uniform electric field due to a large plane sheet of charge (b) the

electric field associated with a positive point charge.

Figure (c) represents the equipotential surfaces for an electric dipole.

Figure (d) represents the equipotential surfaces for two identical positive

charges.

Example :

Some equipotential surfaces are shown in the figure below. What can

you say about the magnitude and the direction of the electric field?

Solution:

First we will find the direction of the E noting that the E lines are

always perpendicular to the local equipotential surfaces and points in

that direction in which the potential is decreasing. So it will be as shown

below

Note that we have drawn the E lines perpendicular to all equipotential

surfaces and it is pointing in that direction in which the potential is

decreasing. So, the total angle E is making with the positive x-axis is

120° as shown.

Now, how to calculate the magnitude of the electric field! For this we

need to jump from one equipotential surface to another through

shortest (perpendicular) distance.

Then calculate,

E = Change in potential

Distance travelled

Let us suppose we are jumping from 20 V line to 10 V line from B to A.

Note that the BA line is the shortest length between the two

equipotential surfaces. The potential difference between the two lines is

(20 V – 10 V) = 10 volt. Now, calculate the length AB = 10 sin 30° [cm] = 5

cm = 5 × 10–2 m

Therefore,

E = 2

10 volt

5 10 m

= 200 [volt/m]

Hence, the complete answer is that the E has the value 200 V/m and

makes 120° with positive x-axis.

Example :

The electric potential existing in space is V(x, y, z) = B(xy + yz + zx). Find

the expression for the electric field at point P(1, 1, 1) and its magnitude

if B = 10 S.I. unit.

Solution:

The formula E = dV

dr

in extended form is written as follows:

V V V

E i j kx y z

where, V V V

, andx y z

are partial derivatives of potential V with respect to x, y and z

respectively. For partial derivative you should note that if say you are

calculating V

,x

then other variables y and z will be treated as constant.

Now, If we write

x y zE iE jE kE ,

then

Ex = V

,x

y

VE

y

z

Vand E .

z

In the question

V = B(xy + yz + zx)

So,

Ex = V

x

= – B(y + z)

because variables y and z are treated constant.

Now, put the values of B, y and z. Given B = 10 and (x, y, z) are (1, 1, 1)

respectively.

So,

Ex = V

x

= – B(y + z)

= – 10 (1 + 1)

= – 20 N/C

Similarly,

Ey = V

y

= – B(x + z)

[now treating x and z constant]

therefore

Ey = – 10 (1 + 1)

= – 20 N/C

and Ez = V

z

= – B (x + y)

= – 20 N/C

therefore,

x y zE E i E j E k

or, E 20i 20j 20k N/C

Now, if we want the magnitude of E , then it's equal to

2 2 2

x y zE E E

= 2 2 2

20 20 20

= 20 3

Example:

An electric field E i20 j30 netwon/coulomb exists in the

space. If the potential at the origin is taken to be zero, find the

potential at (2m, 2m).

Solution:

We have

E = dV

dr

so, it can be written in vector form as

dV = E.dr

Note, you can write E as x y zE i E j E k

and dr dxi dy j dzk

therefore,

x y zE.dr E .dx E .dy E .dz

In the given question the z component of E or the point is not given. So

you can write

x yE.dr E .dx E .dy

Now,

dV = E.dr

= 20i 30 j . dxi dy j

or, dV = – 20 dx – 30 dy

Now we will have to integrate it within limits. Given V = 0 when x = 0 and

y = 0 (lower limit) and we have to calculate V when x = 2 and y = 2 (upper

limit)

therefore,

y 2V x 2

0 x 0 y 0

dV 20 dx 30 dy

or, V 2 2

0 0 0V 20 x 30 y

or, V – 0 = – 20 (2 – 0) – 30 (2 – 0)

or, V = – 40 – 60

= – 100 volt.

POTENTIAL ENERGY OF A SYSTEM OF CHARGES

"The electric potential energy of a system of fixed point charges is equal

to the work that must be done by an external agent to assemble the

system, bringing each charge in from an infinite distance".

Let us first consider a simple case of two point charges q1 and q2

separated by a distance r. To find the electric potential energy of this

two-charge system, we consider the charges q1 and q2 initially at infinity.

When one of the two charges (say q1) is brought from infinity to its place

(say A), no work is done, because no electrostatic force acts on it. But

when we next bring the other charge q2 from infinity to its location (B),

we must do work because no, exerts an electrostatic force on q2 during

its motion.

Assuming the potential at infinity to be zero, the potential at B due to

the charge q1 at A is

V = 1

0

q1

4 r

The work done by the external agent in bringing the charge q2 from

infinity to B is

W = q2V

= 1 2

0

q q1

4 r

So, the potential energy is

U = W

= 1 2

0

q q1

4 r

This formula U = 1 2

0

q q1

4 r is true for any sign of q1. If q1 and q2 are of

same sign, q1 . q2 > 0, so the potential energy U is positive. If q1 and q2

are of opposite signs, q1q2 < 0, and hence U is negative.

Now, we consider a three-charges system shown in the figure.

First, we suppose that three charges are at infinity. Then we bring them

one by one to their locations.

As before, the work done by the external agent in bringing the two

charges q1 and q2 (one by one is)

W1 = 1 2

0 12

q q1

4 r …(ii)

Now, the potential at C, due to q1 (at A) and q2 (at B) is

V = 1 2

0 13 23

q q1

4 r r

So, when q3 is brought from infinity to C, the work done on it by the

external agent is

W2 = q3V

= 1 3 2 3

0 13 23

q q q q1

4 r r

The total work done is W = W1 + W2

= 1 3 2 31 2

0 12 13 23

q q q qq q1

4 r r r

Hence, the potential energy of the system is

U = W

= 1 3 2 31 2

0 12 13 23

q q q qq q1

4 r r r

or, U = U12 + U13 + U23

This can be generalized for any system containing n point charges q1, q2,

q3..........qn. The potential energy of the system will be written as

U = (U12 + U13 + ………. + U1n) + (U23 + ……… + U2n)

+ (U34 + ……… + U3n) + …………. + U(n – 1)n

or, U = ij

i j

U

= i j

i j 0 ij

q .q1

4 r

Thus, the potential energy of a system of charges is the sum of potential

energies of air possible pairs of charges that can be formed by taking

charges from the system.

POTENTIAL ENERGY IN AN EXTERNAL FIELD

(i) Potential Energy of a single charge:

The work done by an external agent in bringing a charge g from

infinity to the present location in the electric field is stored in the

form of potential energy of g.

If V be the electric potential at the present location of the point

charge g, then the work done in bringing it from infinity to the

present location is qV.

The potential energy of q at r in an external field is U q. r

….*potential energy+

where V r is the potential at r.

(ii) Potential energy of a system of two charges:

Suppose two charges q1 and q2 are located at 1 2r and r

respectively in an external field.

To find the potential energy of the two charge system, first bring

q1 from infinity to 1r . The work done in this process is

W1 = q1 V( 1r ),

where V(1r ) is the potential at

1r due to the external field E .

Next, we bring q2 from infinity to 2r .

The work done on q2 against the external field E is

W2 = q2 V( 2r ),

where V(2r ) is the potential at

2r due to the external field E only.

The work done on q2 against the field of q1 is

W3 = 1 2

0 12

q q,

4 r

where r12 is the separation between q1 and q2.

The potential energy of the two-charge system is

U = W1 + W2 + W3

= 1 21 2 2

0 12

q qqV r q V r

4 r

Example :

Two protons are separated by a distance Ft. What will be the speed of

each proton when they reach infinity under their mutual repulsion?

Solution:

Let the charge on each proton be Q. Initially they are at rest, so their

kinetic energy (KE) = 0 and they have only potential energy equal to 2

0

Q.

4 R

Finally at infinity, they have infinite separation so their potential energy

will become zero. (because R = now.) And total kinetic energy is, (1

2

mv2 for each proton) 2x 1

2 mv2 = mv2 (where m is the mass of each

proton and v is the speed with which they are moving at infinity). Now,

total initial energy = total final energy

or, 2

2

0

Qmv

4 R

Therefore,

v = 0

Q.

4 mR

Example:

A bullet of mass 2 gm is moving with a speed of 10 m/s. If the bullet

has a charge of 2 micro-coulomb, through what potential it be

accelerated starting from rest, to acquire the same speed?

Solution:

Use the relation

qV = 1

2mv2;

Here, q = 2 × 10–6 coulomb;

m = 2 × 10–3 kg;

v = 10 m/s

Therefore,

V = 2mv

2q

=

23

6

2 10 10

2 2 10

= 1

6

10

2 10

= 5 × 104 volt

= 50 kV

Example :

Three equal charges Q are at the vertices of an equilateral triangle of

side A. How much work is done in bringing them closer to an

equilateral triangle of side A

2?

Solution:

In this problem first we have to calculate the initial and final potential

energy and then the work done is final potential energy minus initial

potential energy.

Initial potential energy 2

0

1 3Q

4 A (for each combination, the energy is

2

0

Q

4 A and we have three such combinations)

Final potential energy

= 2

0

3Q

A4

2

Now, work done

W = 2 2

00

3Q 3Q

A 4 A4

2

Therefore,

W = 2

0

3Q

4 A

Example :

Two particles have equal masses of 5 g each and opposite charges of 4

× 1 0–5 C and – 4 × 10–5 C. They are released from rest with a separation

of 1 m between them. Find the speed of particles when the separation

is reduced to 50 cm.

Solution:

Again in this question we have to do the conservation of mechanical

energy. Initially, there is only potential energy but finally there is both

kinetic and potential energy.

Initial state energy:

Bodies are at rest so kinetic energy = 0

Potential energy = 1 2

0

Q Q

4 R

= 9 5 59 10 4 10 4 10

1

= – 14.4 joule

Final state energy:

Let us suppose both the charges approach each other with individual

speed v. (both will have same speed because masses are same and

momentum will be conserved. Since initial momentum is zero, they will

have final momentum also equal to zero. Since momentum is vector,

they will have equal and opposite momentum)

Final kinetic energy of both

= 21

2 mv2

= mv2

Final potential energy

= 1 2

0

Q Q

4 R

= 9 5 59 10 4 10 4 10

0.5

= – 28.8 joule

Now,

initial total energy = final total energy

– 14.4 = mv2 – 28.8

therefore,

mv2 = 14.4;

or, v = 3

14.4

5 10

= 54 m/s

Work done in rotating an Electric Dipole in an Electric Field:

If a dipole, placed in a uniform electric field, is rotated from its

equilibrium position, work has to be done against the electric field

torque.

If a small work dW is done (against the field torque) in rotating the

dipole through a small angle dӨ, then dW = Torque × angular

displacement or dW = Ө × dӨ = pE sin Ө dӨ

Thus, the total work done against the field torque for a deflection ' '

from equilibrium is

W = 0

pEsin d

= 0pE cos

W = pE(1 – cos Ө)

This is the formula for the work done in rotating an electric dipole

against an electric field through an angleӨ from the direction of the field

(equilibrium position). The work done in rotating from the position 1 to

Ө2 is given

by W = pE(cos Ө1 – cos Ө2).

Case (i)

If the dipole be rotated through 90° from the direction of the field, then

the work done:

W = pE(1 – cos 90°)

= pE (1 – 0) = pE.

Case (ii)

The work done in rotating the dipole through 180° from the direction of

the field

W= pE (1 – cos 180°)

= pE [1 – (–1)]

= 2pE.

NOTE:

The work done by the field torque on the dipole is

W = pE (cos Ө2 – cos Ө1)

= – pE(1 – cos Ө),

when Ө1 = 0°, Ө2 = Ө.

Potential Energy of an Electric Dipole in an Electric Field.

The potential energy of an electric dipole in an electric field is defined as

the work done in bringing the dipole from infinity to inside the field.

An electric dipole (+q, –q) is brought from infinity to a uniform electric

field E in such a way that the dipole moment p is always in the direction

of the field. Due to the field E , a force F qE acts on the charge +q

in the direction of the field, and force F qE on the charge –q in

the opposite direction. Hence, in bringing the dipole in the field, work

will be done on the charge +q by an external agent, while work will be

done by the field itself on the charge –q. But, as the dipole is brought

from infinity into the field, the charge –q covers 2l distance more than

the charge +q. Therefore, the work done on –q will be greater. Hence the

'net' work done in bringing the dipole from infinity into the field

= force on charge (–q) × additional distance moved

= – qE × 2l

= – pE.

This work is the potential energy U0 of the electric dipole placed in the

electric field parallel to it:

U0 = – pE

In this position the electric dipole is in stable equilibrium inside the field.

On rotating the dipole through an angle Ө work will have to be done on

the dipole. This work is given by

W = pE(1 – cos Ө).

This will result in an increase in the potential energy of the dipole.

Hence, the potential energy of the dipole in the position Ө will be given

by

UӨ = U0 + W

= – pE + pE (1 – cos Ө).

or UӨ = – pE cos Ө

or, U p.E

This is the general equation of the potential energy of the electric dipole.

DIPOLE IN A UNIFORM EXTERNAL FIELD

Consider a small dipole (+q, –q) of length 2l, placed in uniform electric

field of intensity E . Forces of magnitude qE each act on the charge +q,

and –q as shown in figure. These forces are equal, parallel and opposite,

and hence net force on the dipole is zero but these forces constitute a

couple. The moment of this couple i.e., torque is

Ʈ = force × perpendicular distance

= F × 2l sin Ө

= qE 2l sin Ө

= 2ql E sin Ө

or = pE sin Ө,

where 2ql = p (electric dipole moment).

Ʈ= pE sin Ө

or, p E

If the dipole is placed perpendicular to the electric field (Ө= 90° or sin Ө

= 1), then the couple acting on it will be maximum. If this be Өmax, then

Өmax = PE.

If E = 1 NC–1, then p = Өmax. Hence, electric dipole moment is the torque

acting on the dipole placed perpendicular to the direction of a uniform

electric field of intensity 1 NC–1.

Content Builder

Period of Dipole in Uniform Electric Field

1. If dipole is parallel or antiparallel to the field (i.e., = 0° or 180°),

torque is zero. Hence dipole is in equilibrium position.

Consider an electric dipole of dipole moment p making an angle

'Ө' with the field direction. It experiences a torque pE sin

tending to bring it in the direction of field. Therefore, on being

released, the dipole oscillates about an axis through its centre of

mass and perpendicular to the field. If l is the moment of inertia

of the dipole about the axis of rotation, then the

equation of motion is:

2

2

dldt

= – pE sin Ө

For small amplitudes sin Ө - Ө.

Thus, 2

2

d pE

dt l

= – Ѡ2Ө,

where pE

l

This is an SHM whose period of oscillation is

T = 2

= l

2 .pE

2. Potential energy of the electric dipole at an angle Ө with electric

field is U P.E = PE cos Ө.

3. Work done is rotating the dipole from angle Ө1 to Ө2 is W = PE

(cos Ө1 – cosӨ2).

Example :

An electric dipole consists of two opposite charges of magnitude 0.2 µC

each, separated by a distance of 2 cm. The dipole is placed in an

external field of 2 × 10s N/C. What maximum torque does the field

exert on the dipole?

Solution:

Torque = pE sin Ө

For maximum value of torque, sin = 1

maximum torque = pE

= (0.2 × 10–6 × 2 × 10–2) (2 × 105)

= 8 × 10–4 Nm

Motion of a Charged Particle In a Uniform Electric Field

Along x-axis,

ux = v0,

ax = 0,

Sx = x,

vx = v0

S = 21

ut at2

x = v0t

Along y-axis,

uy = 0,

ay = qE

,m

y = 21 qEt

2 m

y =

2

0

1 qE x

2 m v

2

2

0

1 qExy

2 mv

Ө y Ө x2

i.e., path is parabolic.

Deflection on the screen Y = (y + D tan Ө)

tan Ө = y

2

x 0 0

v qEt qEx

v mv mv

So tan Ө= 2

2

0

qEx 2

2mv x

y

tanx

2

which means that the tangent at point P intersect the line AB at its mid

point i.e. at distance x

2 from B.

x

Y D tan2

(i) Electric Field at a point on the Axis of a Dipole (end-on position).

Suppose an electric dipole (q, –q) of length 2l is situated in a

medium of dielectric constant K. Let P be a point on the axis at a

distance r metre from the mid-point O of the dipole. We have to

determine the intensity of the electric field at P. (In figure)

Let E and E2 be the intensities of the electric fields at P due to the

charges +q and –q of the dipole, respectively. The distance of the

point P from the charge +q is (r – l) and the distance from charge –

q is (r + l). Therefore,

E1 =

2

0

1 q

4 K r l

(away from charge +q)

and E2 =

2

0

1 q

4 K r l

(towards the charge –q)

The fields E1 and E, are along the same line but in opposite

directions. Therefore, the resultant field E at the point P will be

E = E1 – E2

=

2 2

0 0

1 q 1 q

4 K 4 Kr l r l

=

2 2

0

1 1 q

4 K r l r l

directed from –q to +q

=

2

2 2

0

4q r

4 K r l

or E =

2

2 2

0

2pr

4 K r l

directed from –q to +q.

If l is very small compared to r(l << r), the term l2 may be

neglected in comparison to r2. Then, the electric field at the point

P due to the dipole is given by

E = 4

0

1 2pr

4 K r

= 3

0

1 2p

4 K r

directed from –q to +q.

For vacuum (or air) K = 1.

E = 3

0

1 2p

4 r

or, 3

0

1 2pE

4 r

The direction of the electric field E is along the axis of the dipole

from the negative charge towards the positive charge, i.e., is in

the direction of dipole moment.

(ii) Electric field at a point on the Equatorial Line of a Dipole

(Broadside-on position).

In this case, the point P is situated on the right-bisector of the

dipole AB at a distance r metre from its mid-point. E1 and E2 are

the intensities of the electric fields at P due to the charges +q and

–q of the dipole, respectively.

The distance of P from each charge is

2 2 2 2r l AP BP r l .

Therefore,

E1 = 2 2

0

1 q

4 K r l

(in the direction AP )

and E2 = 2 2

0

1 q

4 K r l

(in the direction PB ).

The magnitudes of E1 and E2 are equal (but directions are

different). On resolving E1 and E2 into two components, that is,

parallel and perpendicular to AB, the components perpendicular

to AB (E1 sin Ө and E2 sin Ө) cancel each other (because they are

equal and opposite) while the components parallel to AB (E1 cos Ө

and E2 cos Ө), being in the same direction, add up (In Figure).

Hence, the resultant electric field at the point P is

E = E1 cos Ө + E2 cos Ө

= 2 2 2 2

0 0

1 q 1 qcos cos

4 K 4 Kr l r l

= 2 2

0

2qcos

4 K r l

= 2 2 2 2

0

2q

4 K r l r l

or E =

3/2

2 2

0

2ql

4 K r l

=

3/2

2 2

0

p

4 K r l

For vacuum (or air) K = 1.

E =

3/2

2 2

0

p

4 K r l

For a small dipole r >> l.

Therefore,

E = 3

0

1 p

4 r

or 3

0

1 pE

4 r

The direction of the electric field is parallel to the axis of the

dipole.

Content Builder

Electric Field due to a Dipole at any Point for a Small Dipole

Suppose the point P on the line making an angle with the axis of a

small dipole (l << r). The components of the dipole moment p along and

perpendicular to OP are p cos and p sin respectively.

For the component p cos , the point P is on its axis, so the field at P due

to this component is

Er = 3

0

1 psin.

4 r

For the component p sin , the point P lies on its equatorial line, so the

field at P due to this component is

3

0

1 psinE .

4 r

The resultant field at P is

E = 2 2

rE E

or, E = 2

3

0

1 p. 3cos 1

4 r

If E makes an angle Ө with OP, then

tan Ө = r

E 1tan

E 2

or, 1 1tan tan

2

ELECTRIC FLUX

(i) It is denoted by 'Ө'.

(ii) It is a scalar quantity.

(iii) It is defined as the total number of lines of force passing normally

through a curved surface placed in the filed.

(iv) It is given by the dot product of E and normal infinitesimal area

ds integrated over a closed surface–

d E.ds

E.ds

= Eds cos

where Ө = angle between electric field and normal to the area

(v) (a) if Ө = 0, Ө = Eds (maximum)

(b) if Ө = 90°, Ө = zero

(vi) Unit:

(a) Newton – metre2 / coulomb.

(b) Volt – meter

(vii) Dimension: [M L3 T–3 A–1]

(viii) Flux due to a positive change goes out of the surface while that

due to negative change comes into the surface.

(ix) Flux entering is taken as negative while flux leaving is taken as

positive.

(x) Value of electric flux is independent of shape and size of the

surface.

(xi) Flux is associated with all vectors.

(xii) If only a dipole is present in the surface then net flux is zero.

(xiii) Net flux of a surface kept in a uniform electric field is zero.

(xiv) Net flux from a surface is zero does not imply that intensity of

electric field is also zero.

GAUSS'S LAW

This law states that electric flux E through any closed surface is equal to

0

1

times the net charge 'q' enclosed by the surface i.e.,

E E.ds

= 0

q

NOTE:

The closed surface can be hypothetical and then it is called a Gaussian

surface.

If the closed surface enclosed a number of charges q1, q2 ........... qn etc.

then

= E.ds

= 0

q

= 1 2 n

0

q q ........q

Flux is –

(i) Independent of distances between charges inside the surface and

their distribution.

(ii) Independent of shape, size and nature of surface.

(iii) Dependent on charges enclosed by surface, their nature and on

the medium.

(iv) Net flux due to a charge outside the surface will be zero.

(v) If Q = 0, then = 0 but it is not necessary that E = 0.

(vi) Gauss law is valid only for the vector fields which obey inverse

square law.

(vii) Gauss's and coulomb's law are comparable.

Note –

(i) A charge q is placed at the centre of a cube,

then……………………

(a) Total flux through cube = 0

q

(b) Flux through each surface = 0

q

6

(ii) A charge q is placed at the centre of a face of a cube, then total

flux through cube = 0

q

2

How A second cube can be assumed adjacent to the first cube total flux through

both cubes = 0

q,

So flux through each cube =

0

q

2

(iii) Now, q is placed at a corner then the flux will be 0

q.

8

Gauss's Law

Example:

A hemispherical surface of radius R is kept in a uniform electric field E

such that E is parallel to the axis of hemi-sphere, Net flux from the

surface will be–

Solution:

E.ds

= E . R2

= (E) (Area of surface perpendicular to E)

= E . R2.

Example:

A rectangular surface of length 4m and breadth 2m is kept in an electric

field of 20 N/c. Angle between the surface and electric field is 30°.

What is flux thought this surface?

NOTE:

Angle between surface and E is given to be 30°. This is not the 'Ө' used

in our formula 'Ө' is the angle between normal to surface and E.

So here Ө = 90 – 30 = 60°.

Solution:

Ө = EA cos Ө

= 20 × 8 cos 60

= 80 V-m

Example:

In the following, find out the emerging electric flux through S1 and S2

where

[q1 = 1µc, q2 = 2µc, q3 = –3µc]

Solution:

11

0

q

= 6

12

10

8.85 10

= 1.13 × 105 V.m

Example:

A charge 'q' is placed at the centre of a cube of side 'a'. If the total flux

passing through cube and its each surface be ф1 and ф2 respectively

then ф1 : ф2 will be –

(A) 1 : 6

(B) 6 : 1

(C) 1 : 6a2

(D) 6a2 : 1

Solution:

(B) When q is placed at the centre of cube then total flux passing

through cube is

1

0

q

and flux through each surface is

2

0

q

6

1 2:

= 6 : 1

Example:

If charges q

2 and 2q are placed at the centre of face and at the corner,

of a cube. Then total flux through cube will be –

(A) 0

q

2 (B)

0

q

(C) 0

q

6 (D)

0

q

8

Solution:

(A) Flux through cube, when q

2 is placed at the centre face, is

2

0

2q

8

= 0

q,

4

Total flux = 1 2

= 0 0

q q

4 4

= 0

1 q

2

Example:

Flux entering a closed surface is 2000 V-m. Flux leaving that surface is

8000 V-m. Find the charge inside surface.

Solution:

Net flux = фout – фin

ф = (8000 – 2000)

= 6000 V-m

0

q

q = (6000) (8.85 × 10–12)

= 0.53 µc

APPLICATION OF GAUSS'S LAW

Electric field due to a charged conducting sphere/ Hollow conducting or

insulating sphere.

(i) In all the three type of spheres, charge resides only on the outer

surface of the sphere in order to remain in minimum potential

energy state.

Case 1:

OP = r > R

2

0

q qE r

4 r

= 2

2

0

1 Rr

r

( = surface charge density)

Case 2:

r = R

0

E r

Case 3:

r < R

E 0

i.e. At point interior to a conducting or a hollow sphere, electric

field intensity is zero.

(iii) For points outside the sphere, it behaves like all the charge is

present at the centre.

(iv) Intensity of electric field is maximum at the surface.

Imp.

(v) Electric field at the surface is always perpendicular to the surface.

(vi) For points, near the surface of the conductor, E = 0

perpendicular to the surface.

(vii) Graphically,

Electric potential

Case 1:

r < R

Vin = 0

1 Q

4 R

Case 2:

r = R

Vsurface = 0

1 Q

4 R

Case 3:

r > R

Vout = 0

1 Q

4 r

(i) For points interior to a conducting or a hollow sphere, potential is

same everywhere and equal to the potential at the surface.

(iii) at r = , V = 0

NOTE:

Here, we see that E inside the sphere is zero but V ≠ 0. So E = 0 does

not imply V = 0. This presents a good example for it. Similarly V = 0 does

not imply E = 0.

Electric field due to solid insulating sphere

A charge given to a solid insulating sphere is distributed equally

throughout its volume

Electric Field

Case 1:

r > R (point is outside the sphere)

2

0

1 QE r

4 r

Case 2:

r = R (point is at the surface)

2

0

1 QE r

4 R

= Emax = Esurface

Case 3:

r < R (point is inside the sphere)

3

0

1 QE r r

4 R

= 0

r

3

inE r

at r = 0, E = 0

(i) Graphically

(ii) Again, for points outside the sphere, it behaves as all the charge is

present at the centre.

(iii) For points outside, it obeys inverse square law.

(iv) Intensity of electric field at infinity is zero.

(v) Intensity at the surface is maximum and is equal to 2

0

1 Q.

4 R

(vi) Again, it is perpendicular to the surface at the surface.

(vii) Intensity is zero at the centre and for points inside the sphere, it is

directly proportional to distance of the point from the centre.

Electric Potential

Case 1:

r > R

Vout = 0

1 Q

4 r

Case 2:

r = R

Vsurface = 0

1 Q

4 R

Case 3:

r < R

Vin = 2 2

3

0

Q 3R r1

4 2R

Vcentre = 0

3 1 Q

2 4 R (Imp)

Vcentre = surface

3V

2

(i) Graphically

(ii) Again, Ecentre = 0, but Vcentre

(iii) Electric potential at infinity is zero.

(iv) Electric potential is maximum at the centre.

Example:

A solid insulating sphere of radius R is given a charge. If inside the

sphere at a point the potential is 1.5 times that of the potential at the

surface, this point will be –

(A) At the centre

(B) At distance 3

2R from the centre

(C) Potential will be same inside and on the surface of sphere, so

given information is inadequate.

(D) Insulating bodies cannot be given charge.

Solution:

(A) Potential at the centre of insulating sphere is given by

Vin = 2 2

3

0

Q 3R r1

4 2R

…(1)

and on the surface,

Vsurface = 0

1 Q

4 R …(2)

given that,

Vin = surface

3V

2

2 2

3

Q 3R r 3 Q

2R 2 R

r 0

Hence the point will be at the centre.

Example:

Two concentric spheres of radii r & R (r < R) are given the charges q and

Q respectively. Find the potential difference between two spheres.

Solution:

Potential at the inner sphere = potential due to inner + potential due to

outer sphere

1

0 0

1 q 1 QV

4 r 4 R

(potential at points inside is same everywhere and is equal to potential

at the surface).

Potential at outer sphere V2 = potential due to inner + potential due to

outer sphere

= 0 0

1 q 1 Q

4 R 4 R

potential difference = V1 – V2

= 0

1 q q

4 r R

0

q 1 1V

4 r R

NOTE:

Here, we see that 'V' depends only on the charge of inner sphere.

Example:

In the following fig, of charged spheres A, B & C whose charge densities

are σ, – σ & σ and radii a, b & c respectively what will be the value of VA

& VB.

Solution:

VA = 2 2 24 a 4 b 4 c

k k ka b c

= 0

a b c

q

A

VB = 2 2 24 a 4 b 4 c

kb b c

= 2

0

ab c

b

Ex. Figure shows a uniformly charged sphere of radius R and total charge

Q. A point charge q is situated outside the sphere at a distance r from

centre of sphere. Find out the following:

(i) Force acting on the point charge q due to the sphere.

(ii) Force acting on the sphere due to the point charge.

Sol. (i) Electric field at the position of point charge

2

KQˆE r

r

so, 2

KqQr

r

2

KqQF

r

(ii) Since we know that every action has equal opposite reaction so

sphere 2

KqQˆF r

r

sphere 2

KqQF

r

Ex. Figure shows a uniformly charged thin sphere of total charge Q and

radius R. A point charge q is also situated at the centre of the sphere.

Find out the following:

(i) Force on charge q

(ii) Electric field intensity at A.

(iii) Electric field intensity at B.

Sol. (i) Electric field at the centre of the uniformly charged hollow sphere

= 0

So force on charge q = 0

(ii) Electric field at A

A Sphere qE E E

= 2

Kq0

r;

r = CA

E due to sphere = 0, because point lies inside the charged hollow

sphere.

(iii) Electric field BE at point B = Sphere qE E

= 2 2

KQ Kqˆr r

r r

=

2

K Q qr

r;

r = CB

• Here we can also assume that the total charge of sphere is concentrated

at the centre, for calculation of electric field at B.

Ex. Two concentric uniformly charged spherical shells of radius R1 and R2

(R2 > R1) have total charges Q1 and Q2 respectively Derive an expression

of electric field as a function of r for following positions.

(i) r < R1

(ii) R1 < r < R2

(iii) r > R2

Sol. (i) for r < R1,

therefore point lies inside both the spheres

net inner outerE E E 0

(ii) for R1 < r < R2,

therefore point lies outside inner sphere but inside outer sphere :

Enet = Einner + Eouter

= 12

KQr 0

r

= 12

KQr

r

(iii) for r > R2

point lies outside inner as well as outer sphere therefore.

Enet = Einner + Eouter

= 1 22 2

KQ KQˆ ˆr r

r r

= 1 2

2

K Q Qr

r

Ex. A solid non conducting sphere of radius R and uniform volume charge

density has its centre at origin. Find out electric field intensity in

vector form at following positions :

(i)

R, 0, 0

2

(ii)

R R, , 0

2 2

(iii) (R, R, 0)

Sol. (i) at

R, 0, 0

2:

Distance of point from centre

=

2

2 2R0 0

2

= R

R,2

so point lies inside the sphere so

0

rE

3

=

0

Ri

3 2

(ii) At

R R, , 0

2 2;

distance of point from centre

=

2 2

2R R0

2 2

= R = R

so point lies at the surface of sphere, therefore

3

KQE r

R

=

3

3

4K R

R R3 ˆ ˆi jR 2 2

=

0

R Rˆ ˆi j3 2 2

(iii) The point is outside the sphere

so, 3

KQE r

r

=

3

3

4K R

3 ˆ ˆRi Rj2R

=

0

ˆ ˆRi Rj6 2

(b) Electric Field inside a Cavity of Non-conducting Charged Body:

Consider the sphere shown in figure charged uniformly with charge

density coul/m3. Inside the sphere a spherical cavity is created with

centre at C.

Now we wish to find electric field strength inside the cavity. For this we

consider a point P in the cavity at a position vector x from the centre of

sphere and at a position vector y from the centre of cavity as shown.

If 1E , be the electric field strength at P due to the complete charge of

the sphere (inside cavity also) then we known electric field strength

inside a uniformly charged sphere is given as

1

0

xE ;

3

2

0

yE

3

Now the electric field due to the charged sphere in the cavity at point P

can be given as

net 1 2E E E

[As now charge of cavity is removed]

=

0

a

3

[As x y a]

This shows that the net electric field inside the cavity is uniform and in

the direction of a i.e. along the line joining the centre of spheres and

cavity.

Ex. Two conducting hollow spherical shells of radii R and 2R carry charges –Q

and 3Q respectively. How much charge will flow into the earth if Inner

shell is grounded?

Sol. When inner shell is grounded to the Earth then the potential of inner

shell will become zero because potential of the Earth is taken to be zero.

Kx K3Q

0R 2R

x = 3Q

,2

the charge that has increased

=

3Q

Q2

= Q

2

hence charge flows into the Earth

= Q

2

Ex. An isolated conducting sphere of charge Q and radius R is connected to a

similar uncharged sphere (kept at a large distance) by using a high

resistance wire. After a long time what is the amount of heat loss?

Sol. When two conducting spheres of equal radius are connected charge is

equally distributed on them. So we can say that heat loss of system

H = U1 – U1

=

2 2

2

0 0 0

Q QQ 4 40

8 R 8 R 8 R

=

2

0

Q

16 R

Ex. The two conducting spherical shells are joined by a conducting wire

and cut after some time when charge stops flowing. Find out the

charge on each sphere after that.

Sol. After cutting the wire, the potential of both the shells is equal

Thus, potential of inner shell

Vin =

K 2Q xKx

R 2R

= k x 2Q

2R

and potential of outer shell

Vout =

K 2Q xKx

2R 2R

= KQ

R

As Vout = Vin

K x 2QKQ

R 2R

2Q x 2Q

x 0

So charge on inner spherical shell = 0

and outer spherical shell = – 2Q.

Ex. Find charge no each spherical shell after joining the inner most shell and

outer most shell by a conducting wire. Also find charges on each

surface.

Sol. Let the charge on the innermost sphere be x.

Finally potential of shell 1 = Potential of shell 3

K 2Q K 6Q xKx

R 2R 3R

=

K 2Q K 6Q xKx

3R 3R 3R

3x – 3Q + 6Q – x = 4Q;

2x = Q;

x = Q

2

Ex. There are 4 concentric shells A, B, C and D of radius of a, 2a, 3a, 4a

respectively. Shells B and D are given charges +q and –q respectively.

Shell C in now earthed. Find the potential difference VA – VC.

Sol. Let shell C acquires charge 'q' which will be such that final potential of C

is zero.

C

kq kq' kqV 0

3a 3a 4a

kq kq' kq

3a 3a 4a

1 1q' 3q

4 3

q

q'4

As VC = 0

VA – VC = VA

Now calculating VA we get

VA =

qk

kq kq4

2a 3a 4a

VA = kq

6a

or A C

kqV V

6a

Ex. Figure shows three large metallic plates with charges – Q, 3Q and Q

respectively. Determine the final charges on all the surfaces.

Sol. We assume that charge on surface 2 is x. Following conservation of

charge, we see that surfaces 1 has charge (–Q – x). The electric field

inside the metal plate is zero so fields at P is zero.

Resultant field at P –

EP = 0

0 0

Q x x 3Q Q

2A 2A

Q x x 4Q

5Q

x2

• We see that charges on the facing surfaces of the plates are of equal

magnitude and opposite sign. This can be in general proved by gauss

theorem also. Remember this it is important result. Thus the final charge

distribution on all the surface is as shown in figure:

(A) TOTAL ELECTROSTATIC ENERGY OF A SYSTEM OF CHARGES:

Total electrostatic potential energy of system of charges can be given as

U = self energy of all charged bodies + Interaction energy of all pairs

of charged bodies.

Let us consider some cases to understand this concept. Figure shows two

uniformly charged non conducting spheres of radius R1 and R2 and

charged with charges Q1 and Q2 respectively separated by a distance r.

To find the total electrostatic energy of this system, we can write as

U = Uself + Uinteraction

U = 2 2

1 2 1 2

1 2

3KQ 3KQ KQ Q

5R 5R r

(B) ELECTRIC ENERGY OF A SYSTEM OF CONCENTRIC SHELLS:

Figure shown two concentric shells of radii a and b charged uniformly

with charges q1 and q2. Here the total energy of this system can be given

as

Utotal = self energy of inner shell + self energy of outer shell + interaction

energy of the two shell

= 2 2

1 2 1 2Kq Kq Kq q

2a 2b b

Alternative Method:

We know that the total electrostatic energy of a system in stored in the

form of field energy of the system hence here we can calculate the total

electrostatic energy of the system by integrating the field energy density

in the space surrounding the shells where electric field exist.

Total field energy in the electric field associated with the system shown

in figure can be given as

U =

22b

1 22 210 02 2

0 0

K q qKq1 14 x dx 4 x dx

2 x 2 r

=

22

1 1 2

1 1 1 1 1Kq K q q

2 a b 2 b

= 2 2 2 2 2

1 1 1 2 2Kq Kq Kq Kq Kq1

2 a 2b b 2b b

= 2 2

1 2 1 2Kq Kq Kq q

2b 2b b

Ex. Figure shows a shell of radius R having charge q, uniformly distributed over

it. A point charge q is placed at the centre of shell. Find work required

to Increase radius of shell from R to R1 as shown In figure (b).

Sol. Work = Ui – Uf

U1 = SEq + SEq1 + lE

= 2

1 1q

Kq Kq qSE

2R R

Uf = 2

1 1q

1 1

Kq Kq qSE

2R R

work done = Ui – Uf

= 2 2

1 1 1 1

1 1

Kq Kq q Kq Kq q

2R R 2R R

(Try this problem yourself using the energy density formula)

Ex. Find the electrostatic energy stored in a cylindircal shell of length ,

inner radius a and outer radius b, coaxial with a uniformly charged wire

with linear charge density C

.m

Sol. For this we consider an elemental shell of radius x and width dx. The

volume of this shell dV can be given as

dV 2 x .dx

The electric field due to the wire at the shell is

2K

Ex

The electrostatic field energy stored in the volume of this shell is

dU = 2

0

1E .dV

2

or dU =

2

0

1 2K.2 x .dx

2 x

The total electrostatic energy stored in the above mentioned volume can

be obtained by integrating the above expression within limits from a to b

as

U = dU

=

2a

0

b

1 2K2 x .dx

2 x

or U =

b2

0 a

1dx

4 x

or U =

b2

0 a

1dx

4 x

or U =

2

0

bn

4 a