12 TH PHYSICS

Example 8.6: A wire is bent in a circle of –q
radius 10 cm. It is given a charge of 250µC V2 = 4SH0 r2
which spreads on it uniformly. What is the
The electrostatic potential is the work

electric potential at the centre ? done by the electric field per unit charge,
Solution : Given : W
q = 250 µC = 250 × 10-6 C ©¨§V Q · .
R = 10 cm = 10-1 m ¹¸
As V V == 4?π1 0 qr = 9u109 1u02 510u10 6 The potential at C due to the dipole is,
= 2.25 × 107 volt
VC V1 V2 q ª1 - 1º
b) Electric potential due to an electric dipole: 4SH 0 « »
¬ r1 r2 ¼

By geometry, 2 r 2 2 2 r cos T
r r 2 2 2 r cos T
1
2

r 2

We have studied electric and magnetic r 2 r2 § 1 2 2  cos T ·
1 ¨ r2 r ¸
dipoles in XIth Std. Figure 8.8 shows an © ¹

electric dipole AB consisting of two charges 2 
2 § r2 r ·
+q and -q separated pb,y a finite distance 2ℓ. r2 ¨ 1 2 cos T ¸
Its dipole moment is of magnitude p = q× r 2 ¹
©

2l, directed from -q to +q. The line joining the For a short dipole, 2  << r and

centres of the two charges is called dipole axis. If r >>   r is small ∴ 2 can be neglected
A straight line drawn perpendicular to the axis r2 2
and passing through centre O of the electric
dipole is called equator of dipole. 2 r2 § 1 r cos T ·
¨© ¹¸
In order to determine the electric potential ?r 1
due to a dipole, let the origin be at the centre 2
(O) of the dipole. r
2 r 2 § 1 cos T ·
©¨ ¹¸
r2
· 12
¸¹
? r1 r ¨©§1 2 cos T
r
1
r2 r ©¨§1 2 cos T
r ·2
¸¹

1 1 ¨©§1 2 cos T · 12 and
? r r ¹¸

r1

Fig. 8.8: Electric potential due to an electric 1 1 ©¨§1 2 cos T · 12
dipole. r2 r r ¹¸

Let C be any point near the electric dipole ?VC q ª 1 § 2  cos T 1 2
4SH 0 « r ©¨ r
at a distance r from the centre O inclined at an V1 V2 ¬« 1 ·
¸¹
angle θ with axis of the dipole. r1 and r2 are
the distances of point C from charges +q and 1 ©¨§1 2  cos T 1 2 º
r r »
-q, respectively. · ¼»
¹¸
Potential at C due to charge +q at A is,
+q
V1 = 4SH 0 r1 Using binomial expansion, ( 1 + x)n = 1 +
nx, x << l and retaining terms up to the first
order of  only, we get
Potential at C due to charge -q at B is,
r

192

VC 4SqH0 1 «¬ª¨§©1  · §©¨1  · º Solution: Given
r r ¹¸ r ¹¸ »¼
cos T cos T p = 1×10-9 Cm

? VC 4S144HSS0qqHHpoorrcro 2¬«ªª¬«s12rT  cro ∵csoTspº¼»T  º r = 0.3 m
r »¼
1 cos T a) Potential at a point on the axial line

1 p 9 u 109 u 1u10 9
V = 4π0 r 2 = =100 volt
0.3 2

b) Potential at a point on the equatorial line

qu2 =0

c) Potential at a point on a line making an

Electric potential at C, can also be expressed angle of 60° with the dipole axis is

as, VC 1 p .r 1 p cosT 9 u 109 u 1u10 9 cos 60q
4SH0 r3 V = 4S0 r 2 =
0.3 2

1 p .r § r· = 50 volt
4SH 0 r2 ¨©¨ r ¸¹¸
VC , r c) Electrostatics potential due to a system of
where r is charges:

a unit vector along the position We now extend the analysis to a system of

vector, OC = r charges.

i) Potential at an axial point,θ = 00 (towards

+q) or 1800 (towards – q)

Vaxial r1 p
4SHo r 2

i.e. This is the maximum value of the potential.

ii) Potential at an equatorial point, θ = 90° and

V=0

Hence, the potential at any point on the Fig. 8.9: System of charges.
equatorial line of a dipole is zero. This is
the minimum value of the magnitude of the Consider a system of charges q1, q2 .........
potential of a dipole.
qn at distances r1, r2 ...... rn respectively from
Thus the plane perpendicular to the line
between the charges at the midpoint is an point P. The potential V1 at P due to the charge
equipotential plane with potential zero. The
work done to move a charge anywhere in this q1 is V1 = 1 q1
plane (potential difference being zero) will be 4SH 0 r1
zero.
Similarly the potentials V2, V3 ........Vn at
Example 8.7: A short electric dipole has
P due to the individual charges q2, q3 ...........qn

are given by q2 q3 , 1 qn
1 r2 1 r3 4SH0 rn
V2 4SH 0 , V3 4SH 0 Vn

dipole moment of 1 × 10-9 C m. Determine By the superposition principle, the
the electric potential due to the dipole at a
point distance 0.3 m from the centre of the potential V at P due to the system of charges is
dipole situated
a) on the axial line b) on the equatorial line the algebraic sum of the potentials due to the
c) on a line making an angle of 60° with the
dipole axis. individual charges.
∴ V = V1 + V2 + ... + Vn

= 1 § q1 + q2 + ----- + qn ·
4SH 0 ¨ r1 r2 rn ¸
© ¹

193

1 n qi As V1 + V2 = 0
4SH 0 i 1 ri
¦ Or, V 1 ª q1 q2 º
4π0 «¬ x 0.16 »¼
For a continuous charge distribution, x 0

summation should be replaced by integration. ª 5 u 10 8 3u10 8 º
x 0.16 »
9 × 109 « x ¼ 0
¬
Use your brain power

∴x = 0.40 m, x = 40 cm

Is electrostatic potential necessarily zero at 8.5 Equipotential Surfaces:
a point where electric field strength is zero? An equipotential surface is that surface,
Justify.
at every point of which the electric potential is
Example 8.8: Two charges 5 × 10-8 C and the same. We know that,

-3 × 10-8 C are located 16 cm apart. At what The potential (V) for a single charge q is

point (s) on the line joining the two charges given by V = 1 q
4SH0 r
is the electric potential zero ? Take the
If r is constant then V will be constant.
potential at infinity to be zero. Hence, equipotential surfaces of single point
charge are concentric spherical surfaces
Solution : As shown below, suppose the two centered at the charge. For a line charge, the
shape of equipotential surface is cylindrical.
point charges are placed on x- axis with the

positive charge located at the origin O.

q1= 5 × 10-8 C q = -3 × 10-8 C
2

Let the potential be zero at the point P and

OP = x. For x < 0 ( i.e. to the left of O), the

potentials of the two charges cannot add up

to zero. Clearly, x must be positive. If x lies

between O and A, then

V1 + V2 = 0, where V1 and V2 are the Fig. 8.10 : Equipotential surfaces.
potentials at points O and A, respectively. Equipotential surfaces can be drawn

q1= 5 × 10-8 C q2= -3 × 10-8 C through any region in which there is an electric

field.

By definition the potential difference

1 ª q1 q2 º 0 between two points P and Q is the work done
4SH 0 ¬« x 0.16 x »¼ per unit positive charge displaced from Q to P.
∴ VP – VQ = WQP
If points P and Q lie on an equipotential
ª 5 u 10 8 3u10 8 º 0 surface, Vp = VQ.
9 × 109 ¬« x » º ∴  WQP = 0
5 0.16 x ¼ ¼»
ª x Thus, no work is required to move a test
⇒9 ×109 × 10-8 ¬« 3 x 0 charge along an equipotential surface.
0.16 a) If dx is the small distance over the
equipotential surface through which unit
5 − 3 0 positive charge is carried then
⇒x 0.16 −x =
dW E . d x E dx cos T 0
∴x = 0.10 m, x = 10 cm

The other possibility is that x may also

lie on extended OA.

194

? cos T = 0 or T = 900
i.e. E ⊥ d x as shown in Fig. 8.11

Hence electric field intensity E is always
normal to the equipotential surface i.e., for any
charge distribution, the equipotential surface
through a point is normal to the electric field
at that point.

Fig. 8.13: Equipotential surfaces for a dipole.

 Fig. 8.14: Equipotential surfaces for two
Fig. 8.11: Equipotential surface ⊥ to E identical positive charges.

b) If the field is not normal, it would have a Fig. 8.15: (a) Between
nonzero component along the surface. So to 2 plane metallic
move a test charge against this component sheets.
work would have to be done. But by the
definition of equipotential surfaces, there
is no potential difference between any two
points on an equipotential surface and hence
no work is required to displace the charge on
the surface. Therefore, we can conclude that
the electrostatic field must be normal to the
equipotential surface at every point, and vice
versa.

Do you know?

Equipotential surfaces do not intersect each (b) When one of the sheet is replaced by a
other as it gives two directions of electric charged metallic sphere.
fields at intersecting point which is not pos-
sible. Like the lines of force, the equipotential
surface give a visual picture of both the
direction and the magnitude of electric field in
a region of space.

Fig. 8.12: Equipotential surfaces for a Example 8.9: A small particle carrying a
uniform electric field. negative charge of 1.6 × 10-19 C is suspended
in equilibrium between two horizontal
metal plates 10 cm apart having a potential

195

difference of 4000 V across them. Find the To calculate the electric potential energy

mass of the particle. of the two charge system, we assume that the

Solution: Given : two charges q1 and q2 are initially at infinity.
We then determine the work done in bringing
q = 1.6 × 10-19 C
the charges to the given location by an external
dx = 10 cm = 10 × 10-2 m = 10-1 m
agency.
dV = 4000 V
In bringing the first charge q1 to position
E = −dV = − 4000
dx 10−1 A r1 , no work is done because there is no
external field against which work needs to be
= - 4 × 104 Vm-1
done as charge q2 is still at infinity i.e., W1 =
As the charged particle remain suspended 0. This charge produces a potential in space

in equilibrium, g iven by V1 = 4S1H0 qr 11
--- (8.14)
F = mg = qE

∴m
= qE = 1.6u10 19 4u104 Where r1 is the distance of point A from
g 9.8

= 0.653 × 10-15 kg the origin.

m = 6.53 × 10-16 kg When we bring charge q2 from infinity to

8.6 Electrical Energy of Two Point Charges B r2 at a distance r12, from q1, work done is
and of a Dipole in an Electrostatic Field: W= 24=S(qHp10ort1e2n×tiaql2a,t(Bwhdeuree to charge q1) × q2
AB = r12) --- (8.15)
When two like charges lie infinite distance
apart, their potential energy is zero because This work done in bringing the two
no work has to done in moving one charge at
infinite distance from the other. But when they charges to their respective locations is stored
are brought closer to one another, work has
to be done against the force of repulsion. As as the potential energy of the configuration of
electrostatic force is conservative, this work
gets stored as the potential energy of the two two charges. 1 q1q2 --- (8.16)
charges. Electrostatic potential energy of 4SH 0 r12
a system of point charges is defined as the ?U =
total amount of work done to assemble the
Equation (8.16) can be generalised for a
system of charges by bringing them from
system of any number of point charges.
infinity to their present locations.
Example 8.10: Two charges of magnitude
a) Potential energy of a system of 2 point
5 nC and −2 nC are placed at points
charges:
(2 cm, 0, 0) and (20 cm, 0, 0) in a region of

space, where there is no other external field.

Find the electrostatic potential energy of

the system.

Solution : Given

q1= 5 nC = 5 × 10-9 C

q2 = -2 nC = -2 × 10-9 C
r = (20 – 2) cm = 18 cm = 18 × 10-2 m
O 1 q1q2
U = 4π0 r
Fig. 8.16: System of two point charges.
Let us consider 2 charges q1 and q2 with = 9u109 u5u10 9 u 2u10 9
position vectors r1 and r2 relative to some 18u 10 2
origin (O). = -5 × 10-7 J = -0.5 × 10-6 J = -0.5 µJ

196

b) Potential energy for a system of N point (c) Potential energy of a single charge in an

charges: external field:

Equation (8.16) gives an expression for Above, we have obtained an expression

potential energy for a system of two charges. for potential energy of a system of charges

We now analyse the situation for a system of when the source of the electric field, i.e.,

N point charges. charges and their locations, were specified.

In bringing a charge q3 from ∞ to C In this section, we determine the potential
( r3 ) work has to be done against electrostatic
energy of a charge (or charges) in an external
forces of both q1and q2
∴ W3 = (potential at C due to q1 and q2 )× q3 field E which is not produced by the given
+ × q3
1 ª q1 q2 º charge (or charges) whose potential energy we
4SH 0 « r13 r23 »
= ¬ ¼ wish to calculate. The external sources could

be known, unknown or unspecified, but what is

= 1 ª q1q3 + q2q3 º known is the electric field E or the `electrostatic
4SH 0 « r13 r23 »
¬ ¼ potential V due to the external sources.

Similarly in bringing a charge q4 from Here we assume that the external field

∞ to D r4 work has to be done against is not affected by the charge q, if q is very

electrostatic forces of q1, q2, and q3 small. The external electric field E and the

1 ª q1 q4 q2 q4 q3 q4 º corresponding external potential V may vary
4SH 0 « r14 r24 r34 »
W4 = ¬ + + ¼ pfrooimntIpfPoVinh(tartvo)inpigsoitnphtoe. seitxiotenrnvaelcptoortenrti,althaetnanbyy

Proceeding in the same way, we can

write the electrostatic potential energy of a definition, work done in bringing a unit positive

system of N point charges at r1, r2 ....rN as charge from ∞ to the point P is equal to V.
¦U 1 q j qk
∴ Work done in bringing a charge q, from ∞

4SH 0 all pairs rjk to the given point in the external field
is qV ( r ).
Example 8.11: Calculate the

electrostatic potential energy This work is stored in the form of potential

of the system of charges energy of a system of charge q. 

shown in the figure. ∴PE of a system of a single charge q at r in

Solution : Taking zero of potential energy at an external field i srg iven by
PE qV
∞, we get potential energy (PE) of the system --- (8.17)

of charges 1 q j qk (d) Potential energ y o f a system of two
PE = 4π0 rjk
∑ charges in an external field:

1 ªqq q q q q In order to find the potential energy of
4SH 0
« r r r a system of two charges q1 and q2 located at
¬ r1 and r2 respectively in an external field, we
calculate the work done in bringing the charge
q q q q q q º
» q1 from ∞ to r1.
r r2 r2 ¼
From (8.17),in the said process work done
1 ªq2 q2 q2 q2 q2 q2 º = q1V ( r 1 ) --- (8.18)
« »
4SH 0 ¬ r r r r r 2r 2 ¼ To bring the charge q2 to r2, the work is

1 ª 2q2 º ª 2q2 º done not only against the external field E but

4SH 0 « r 2 » « 4SH 0 r » also against the field due to q1.
¬ ¼ «¬ ¼»

197

∴Work done on q2 against the external field r = 16 cm = 0.16 m

= q2 V ( r2 ) and W4SqoH1rk0qd2r1o2n,e on q2 against the a) Electrostatic potential energy of the
field due to
q1 = system of two charges is
1 q1q2
where r12 = distance between q1and q2. V= 4π0 r

By the Principle of superposition for = 9u109 u 2 u10 6 u 4 u10 6

fields, we add up the work done on q2 against 0.16

the two fields. = 0.45 J

∴ Work done in bringing q2 to r2 ³ ³(b EP)=EIn)−tdd=hrVe 4eqπ∴l1eq0cV2trr=i c+fi eqlE1dV,dtro( tr=1a)lpotential energy
--- (8.19)
q2 V § o · q1 q2 + q2 V ( r2 )
¨© ¸¹ 4SH 0 r12
r2

Thus from (8.18) and (8.19) potential A dr ,V = A
r2 r
energy of the system ∴ Tota l PE = 4qπ1q02r +
Aq Aq
= Total work done in assembling the 1 + 2
r
configuration + q1 q2 1 r2
4SH 0r12
= q1 V r1 + q2 V r2 8u105 u 2u10 6
= -0.45+ 0.08 +
Example 8.12: Two charged particles 8u105 u 4u10 6

having equal charge of 3 ×10-5 C each are   0.08

brought from infinity to a separation of

30 cm. Find the increase in electrostatic = -0.45 -20 + 40

potential energy during the process. = 19.55 J

Solution : Taking the potential energy (PE) (e) Potential energy of a dipole in an external
field:
at ∞ to be zero,

Increase in PE = present PE

q1 q2 9u109 u(3u10 5 )2
V = 4π0 r =
0.3

9u9u109 u10 10 81
= 3u10 1 = 3 = 27 J

Example 8.13: Fig. 8.17 : Couple acting on a dipole.

a) Determine the electrostatic potential Consider a dipole with charges -q and
energy of a system consisting of two +q separated by a finite distance 2 , placed

charges -2 µC and +4 µC (with no external in a uniform electric field E . It experiences a
field) placed at (-8 cm, 0, 0) and (+8 cm, 0, torqueτ which tends to rotate it.
0) respectively. τ = p × E or W pE sinT
b) Suppose the same system of charges is
now placed in an external electric field In order to neutralize this torque, let us
E = A (1/r2), where A = 8 × 105 cm-2, what assume an external torqueτ ext is applied,

which rotates it in the plane of the paper

would be the electrostatic potential energy from angle θ0 to angle θ , without angular
of the configuration acceleration and at an infinitesimal angular

Solution: Given : speed. Work done by the external torque

q1 = -2 µC = -2 × 10-6 C, r1= 0.08 cm TT
q2= +4 µC = +4 × 10-6 C, r2 = 0.08 cm
W ³ Wext T dT ³ pE sinT dT
T0 T0

198

T 8.7 Conductors and Insulators, Free Charges
pE >- cos T @ TT and Bound Charges Inside a Conductor:
pE ª¬- cos T - - cosT0 ¼º a) Conductors and Insulators:

pE >- cosT cosT0 @ When you come in contact with wires in
wet condition or while opening the window of
pE >cosT0 - cosT @ your car, you might have experienced a feeling
of electric shock. Why don’t you get similar
This work done is stored as the potential experiences with wooden materials?

energy of the system in the position when the The reason you get a shock is that
dipole makes an angle θ with the electric there occurs a flow of electrons from one body
to another when they come in contact via
field. The zero potential energy can be chosen rubbing or moving against each other. Shock
as per convenience. We can choose U (θ0 ) is basically a wild feeling of current passing
= 0, giving through your body.

?U T U T0 pE cosT0 - cosT Conductors are materials or substances
which allow electricity to flow through them.
a) If initially the dipole is perpendicular to the This is because they contain a large number
S of free charge carriers (free electrons). In a
field E i.e., T0 2 then metal the outer (valence) electrons are loosely
bound to the nucleus and are thus free for
conductivity, when an external electric field is
applied.

U T pE cos S 2 - cosT Metals, humans, earth and animal bodies
are all conductors. The main reason we get
- pE cos T electric shocks is that being a good conductor
our human body allows a resistance free path
U T - p. E for the current to flow from the wire to our
body.
b) If initially the dipole is parallel to the field Under electrostatic conditions the conductors
have following properties.
E then T0 0 U T pE cos 0 - cos T 1. In the interior of a conductor, net

U T pE 1 - cos T electrostatic field is zero.
2. Potential is constant within and on the
Example 8.14: An electric dipole consists
surface of a conductor.
of two opposite charges each of magnitude 3. In static situation, the interior of a

1µC separated by 2 cm. The dipole is placed conductor can have no charge.
4. Electric field just outside a charged
in an external electric field of 105 N C-1.
conductor is perpendicular to the surface
Find: of the conductor at every point.
5. Surface charge density of a conductor
(i) The maximum torque exerted by the could be different at different points.

field on the dipole

(ii) The work the external agent will have

to do in turning the dipole through 180°

starting from the position θ = 0°

Solution: Given :

p = q × 2ℓ = 10-6 × 2 × 10-2 = 2 × 10-8 cm

E = 105 NC-1

(i) τ = p E sin 90°= 2 × 10-8 × 105 × 1
max

= 2 × 10-3 Nm

(ii) W = pE ( cosθ1 − cosθ2 )

= 2 × 10-8 × 105 × (cos 0- cos 180°)

= 2 × 10-3 ( 1 + 1 ) = 4 × 10-3 J

199

Electrostatic shielding : In insulators, the electrons are tightly
• To protect a delicate instrument from bound to the nucleus and are thus not available
for conductivity and hence are poor conductors
the disturbing effects of other charged of electricity. There are no free charges since
bodies near it, place the instrument all the charges are bound to the nucleus. An
inside a hollow conductor where E = 0. insulating material can be considered as a
This is called electrostatic shielding. collection of molecules that are not easily
• Thin metal foils are used in making the ionized. An insulator can carry any distribution
shields. of external electric charges on its surface or in
• During lightning and thunder storm it is its interior and the electric field in the interior
always advisable to stay inside the car can have non zero values unlike conductors.
than near a tree in open ground, since 8.8 Dielectrics and Electric Polarisation:
the car acts as a shield.
Faraday Cages: Dielectrics are insulates which can be
• It is an enclosure which is used to block used to store electrical energy. This is because
the external electric fields in conductive when such substances are placed in an external
materials. field, their positive and negative charges
• Electro-magnetic shielding: MRI get displaced in opposite directions and the
scanning rooms are built in such a molecules develop a net dipole moment. This
manner that they prevent the mixing is called polarization of the material and such
of the external radio frequency signals materials are called dielectrics.
with the MRI machine.
In every atom there is a positively
b) Free charges and Bound charges inside charged nucleus and there are negatively
materials: charged electrons surrounding it. The negative
charges form an electron cloud around the
The electrical behaviour of conductors positive charge. These two oppositely charged
and insulators can be understood on the basis regions have their own centres of charge
of free and bound charges. (where the effective charge is located). The
centre of negative charge is the centre of
In metallic conductors, the electrons in mass of negatively charged electrons and that
the outermost shells of the atoms are loosely of positive charge is the centre of mass of
bound to the nucleus and hence can easily get positively charged protons in the nucleus.
detached and move freely inside the metal.
When an external electric field is applied, they Thus, dielectrics are insulating materials
drift in a direction opposite to the direction of or non- conducting substances which can be
the applied electric field. These charges are polarised through small localised displacement
called free charges. of charges. e.g. glass, wax, water, wood , mica,
rubber, stone, plastic etc.
The nucleus, which consist of the positive
ions and the electrons of the inner shells, Dielectrics can be classified as polar
remain held in their fixed positions. These dielectrics and non polar dielectrics as
immobile charges are called bound charges. described below.
Polar dielectrics:
In electrolytic conductors, positive and
negative ions act as charge carriers but their A molecule in which the centre of mass
movements are restricted by the electrostatic of positive charges (protons) does not coincide
force between them and the external electric with the centre of mass of negative charges
field. (electrons), because of the asymmetric shape
of the molecules is called polar molecule as
shown in Fig. 8.18 (a). They have permanent

200

dipole moments of the order of 10-30 Cm. They Polarization of a non-polar dielectric in an
act as tiny electric dipoles, as the charges are external electric field:
separated by a small distance. The dielectrics
like HCℓ, water, alcohol, NH3 etc are made of In the presence of an external electric
polar molecules and are called polar dielectrics. field Eo, the centres of the positive charge
Water molecule has a bent shape with its two in each molecule of a non-polar dielectric is
O - H bonds which are inclined at an angle of pulled in the direction of E , while the centres
about 105°. It has a very high dipole moment
of 6.1 × 10-30 Cm. Fig. 8.18 (b) and (c) show o
the structure of HCl and H2O, respectively.
of the negative charges are displaced in the
(a) opposite direction. Therefore, the two centres
are separated and the molecule gets distorted.
Fig. 8.18. (a) A polar molecule. The displacement of the charges stops when
the force exerted on them by the external field
is balanced by the restoring force between the
charges in the molecule.

Each molecule becomes a tiny dipole
having a dipole moment. The induced dipole

moments of different molecules add up giving

a net dipole moment to the dielectric in the

presence of the external field.

(b) (c) Fig. 8.20 (a) Shows the non polar dielectric in
absence of electric field while.
Fig. 8.18. Examples of Polar molecules
(b) HCI (c) H2O.

Non Polar dielectrics:
A molecule in which the centre of mass of

the positive charges coincides with the centre
of mass of the negative charges is called a non
polar molecule as shown in Fig. 8.19 (a). These
have symmetrical shapes and have zero dipole
moment in the normal state. The dielectrics
like hydrogen, nitrogen, oxygen, CO2, benzene,
methane are made up of nonpolar molecules
and are called non polar dielectrics. Structures

of H2 and CO2 are shown in Fig. 8.19 (b) and
(c), respectively.

Fig. 8.20 (b) shows it in presence of an
(b) external field.

Polarization of a polar dielectric in an

(a) external electric field:

The molecules of a polar dielectric have

tiny permanent dipole moments. Due to thermal

(c) agitation in the material in the absence of any

Fig. 8.19. (a) Nonpolar molecule. Examples of external electric field, these dipole moments
Nonpolar molecules (b) H (c) CO . are randomly oriented as shown in Fig. 8.21

22

201

(a). Hence the total dipole moment is zero. Reduction of electric field due to polarization

When an external electric field is applied the of a dielectric:

dipole moments of different molecules tend to When a dielectric is placed in an external
align with the field. As a result the dielectric electric field, the value of the field inside the
develops a net dipole moment in the direction dielectric is less than the external field as a
of the external field. Hence the dielectric is result of polarization. Consider a rectangular
polarized. The extent of polarization depends dielectric slab placed in a uniform electric
on the relative values of the two opposing field E acting parallel to two of its faces.
energies. Since the electric charges are not free to move
about in a dielectric, no current results when it

is placed in an electric field. Instead of moving

the charges, the electric field produces a slight

rearrangement of charges within the atoms,

resulting in aligning them with the field. This

is shown in Fig. 8.20 and Fig. 8.21. During the

process of alignment charges move only over

distances that are less than an atomic diameter.

Fig. 8.21 (a) Shows the polar dielectric in As a result of the alignment of the dipole
absence of electric field while. moments there is an apparent sheet of positive
charges on the right side and negative charges

on the left side of the dielectric. These two

sheets of induced surface charges produce an

electric field E0 called the polarization field
in the insulator which opposes the applied
electric field E . The net field E ' , inside the

Fig. 8.21 (b) shows it in presence of an dielectric is the vector sum of the applied field
E and the polarization field E0
∴ E' = E - E0 (in magnitude)
This is shown in Fig. 8.22 (a), (b) and (c).

external field.

1. The applied external electric field which

tends to align the dipole with the field.

2. Thermal energy tending to randomise the

alignment of the dipole.

The polarization in presence of a strong

external electric field is shown in Fig. 8.21 (b)

Thus, both polar and nonpolar dielectric Fig. 8.22 (a) When a dielectric is placed in an
develop net dipole moment in the presence of external electric field, the dipoles become aligned.
an electric field.

The dipole moment per unit volume is

called polarization and is denoted by P . For

linear isotropic dielectrics P = χe E .
constant called
χe  is a electric

susceptibility of the dielectric medium.

It describes the electrical behaviour of a

dielectric. It has different values for different

dielectrics. Fig. 8.22 (b) Induced surface charges on the
dielectric establish a polarization field E0 in the
For vacuum χ = 0. interior.
e

202

electrical component which allows current to
pass through it and dissipates heat but can’t

store electrical energy. So there was a need

to develop a device that can store electrical
energy. The most common arrangement for

this consists of a set of conductors (conducting

plates) having charges on them and separated

Fig. 8.22 (c) The nEet field E′ is a vector sum of by a dielectric material.
and E0 . The conductors 1 and 2 shown in the Fig.

8.23 have charges +Q and -Q with potential

Do you know? difference, V = V1 - V2 between them. The
electric field in the region between them is
If we apply a large enough electric field, we
can ionize the atoms and create a condition proportional to the charge Q.

for electric charge to flow like a conductor.

The fields required for the breakdown of
dielectric is called dielectric strength.

The greater the applied field, greater is the

degree of alignment of the dipoles and hence

greater is the polarization field.
The induced dipole moment disappears

when the field is removed. The induced dipole Fig. 8.23: A capacitor formed by two conductors.
moment is often responsible for the attraction
of a charged object towards an uncharged The potential difference V is the work
insulator such as charged comb and bits of
paper. done to carry a unit positive test charge from
Table 1:Dielectric constants of various materials:
the conductor 2 to conductor 1 against the field.

As this work done will be proportional to Q,
then V ∝ Q and the ratio Q is a constant.
V
Material Min Max ∴ C = Q
V
Air 1 1

Ebonite 2.7 2.7 The constant C is called the capacitance

Glass 3.8 14.5 of the capacitor, which depends on the size,

Mica 49 shape and separation of the system of two

Paper 1.5 3 conductors.

Paraffin 23 The SI unit of capacitance is farad (F).

Porcelain 5 6.5 Dimensional formula is [M-1 L-2T4A2].

Quartz 55 1 farad = 1 coulomb/1volt

Rubber 24 A capacitor has a capacitance of one

Wood dry 1.4 2.9 farad, if the potential difference across it rises

Metals ∞∞ by 1volt when 1 coulomb of charge is given

8.9 Capacitors and Capacitance, to it. In practice farad is a big unit, the most

Combination of Capacitors in Series and commonly used units are its submultiples.

Parallel: 1µF = 10-6F

In XIthStd. you have studied about resistors, 1nF=10-9F

resistance and conductance. A resistor is an 1pF = 10-12F

203

Uses of Capacitors based on the shape of the conductors.

Principle of a capacitor: Combination of Capacitors:
When there is a combination of capacitors
To understand the principle of a capacitor
to be used in a circuit we can sometimes
let us consider a metal plate P1 having area A. replace it with an equivalent capacitor or a
Let some positive charge +Q be given to this single capacitor that has the same capacitance
as the actual combination of capacitors. The
plate. Let its potential be V. Its capacity is effective capacitance depends on the way the
individual capacitors are combined. Here we
given by C1 = Q discuss two basic combinations of capacitors
V which can be replaced by a single equivalent
capacitor.
Now consider another insulated metal (a) Capacitors in series:

plate P2 held near the plate P1. By induction a When a potential difference (V ) is
negative charge is produced on the nearer face applied across several capacitors
connected end to end in such a way that
and an equal positive charge develops on the sum of the potential difference across all the
capacitors is equal to the applied potential
farther face of P2 (Fig. 8.24 (a)). The induced
negative charge lowers the potential of plate difference V, then the capacitors are said to be

P1, while the induced positive charge raises its connected in series.
potential.

(a) (b)

Fig. 8.24: (a) and (b) Parallel plate capacitor.
As the induced negative charge is closer

to P1 it is more effective, and thus there is a Fig. 8.25: Capacitors in series.
net reduction in potential of plate P1. If the
outer surface of P2 is connected to earth, the In series arrangement as shown in Fig.
induced positive charges on P2 being free,
flows to earth. The induced negative charge on 8.25, the second plate of first conductor is

connected to the first plate of the second

P2 stays on it, as it is bound to positive charge conductor and so on. The last plate is connected
of P1. This greatly reduces the potential of P2,
(Fig 8.24 (b)). If V1 is the potential on plate P2 to earth. In a series combination, charges on
due to charge (- Q) then the net potential of the
the plates (± Q)are the same on each capacitor.

Potential difference across the series

system will now be +V-V1. combination of capacitor is V volt,

Hence the capacity C2 = Q ∴C2 > C1 where V= +V1Q+ V2 +QV3
V - V1 ∴V = Q C2 + C3
C1
Thus capacity of metal plate P1, is

increased by placing an identical earth

connected metal plate P2 near it.
Such an arrangement is called capacitor.

It is symbolically shown as .
⊥⊥

If the conductors are plane then it is

called parallel plate capacitor. We also have Fig. 8.26: Effective capacitance of three
spherical capacitor, cylindrical capacitor etc. capacitors in series.

204

Let Cs represent the equivalent capacitance In this combination all the capacitors
shown in Fig. 8.26, thenV = Q have the same potential difference but the
Cs plate charges (± Q1) on capacitor1, (± Q2)
Q =Q +Q +Q on the capacitor 2 and (± Q3) on capacitor 3
∴ Cs C1 C2 C3 are not necessarily the same. If charge Q is
applied at point A then it will be distributed to
1 = 1 + 1 + 1 the capacitors depending on the capacitances.
∴ Cs C1 C2 C3 ∴Total charge Q can be written as Q = Q1 +
Q2 + Q3 = C1 V + C2 V + C3V
( for 3 capacitors in series)
Let Cp be the equivalent capacitance of
This argument can be extended to yield the combination then Q = CpV
∴C pV = C1V + C2 V + C3 V
an equivalent capacitance for n capacitors ∴Cp = C1 + C2 + C3

connected in series which is equal to the sum The general formula for effective
capacitance Cp for parallel combination of n
of the reciprocals of individual capacitances capacitors follows similarly
Cp = C1 + C2+ .............. + Cn
of the capacitors. If all capacitors are equal then Ceq = nC

1 1 1 ................. 1 Remember this
? Ceq C1 C2 Cn

If all capacitors are equal then
= C1eq Cn=or Ceq C
n

Remember this

Series combination is used when a Capacitors are combined in parallel when
high voltage is to be divided on several we require a large capacitance at small
capacitors. Capacitor with minimum potentials.
capacitance has the maximum potential
difference between the plates. Example 8.15 When 108 electrons are
transferred from one conductor to another, a
b) Capacitors in Parallel: potential difference of 10 V appears between
The parallel arrangement of capacitors the conductors. Find the capacitance of the
two conductors.
is as shown in Fig. 8.27 below, where the Solution : Given :
insulated plates are connected to a common Number of electrons n = 108
terminal A which is joined to the source of V = 10 volt
potential, while the other plates are connected ∴charge transferred
to another common terminal B which is Q = ne = 108 × 1.6 × 10-19
earthed. (∵ e = 1.6 × 10-19 C)

= 1.6 × 10-11 C

∴ Capacitance between two conductors

C = Q = 1.6 u 10 11 = 1.6 u 10 10 F
V 10

Example 8.16: From the figure given below

find the value of the capacitance C if the

equivalent capacitance between A and B is

to be 1 µF. All other capacitors are in micro

Fig. 8.27: Parallel combination of capacitors. farad.

205

A parallel plate capacitor consists of two
thin conducting plates each of area A, held
parallel to each other, at a suitable distance d
apart. The plates are separated by an insulating
medium like paper, air, mica, glass etc. One of
the plates is insulated and the other is earthed
as shown in Fig. 8.28.

Solution : Given :

C1 = 8 µF , C2 = 4 µF , C3 = 1µF ,
C4 = 4 µF , C5 = 4 µF
The effective capacitance of C4 and C5 in
parallel

= C4 + C5 = 4 + 4 = 8 µF Fig. 8.28: Capacitor with dielectric.

The effective capacitance of C3 and 8 µF in When a charge +Q is given to the insulated

series plate, then a charge -Q is induced on the inner
1u8 8

= 1 8 = 9 µF
The capacitance 8 µF is in parallel with

the series combination of C and C . Their face of earthed plate and +Q is induced on
12

effective combination is its farther face. But as this face is earthed the

C1C2 + 8 8×4 8 32 µF charge +Q being free, flows to earth.
C1 + C2 9 ⇒ 12 + 9⇒ 9
In the outer regions the electric fields due
32
This capacitance of 9 µF is in series with to the two charged plates cancel out. The net

C and their effective capacitance is given to field is zero. V
E = 2VH0 2H 0
be 1µF - = 0

32 u C
9
32 1 In the inner regions between the two
9
C capacitor plates the electric fields due to the

32 32 two charged plates add up. The net field is thus
9 9
? u C C E = 2VH0 + 2VH0 = HV0 = AQH0 --- (8.20 )


= 1.39 µF

8.10 Capacitance of a Parallel Plate The direction of E is from positive to
Capacitor Without and With Dielectric negative plate.
Medium Between the Plates:
Let V be the potential difference between
In section 8.8 we have studied the
behaviour of dielectrics in an external field. Let the 2 plates. Then electric field between the
us now see how the capacitance of a parallel
plate capacitor is modified when a dielectric is plates is given by
introduced between its plates. V
a) Capacitance of a parallel plate capacitor E = d or V = Ed --- (8.21)
without a dielectric:
Substituting Eq. (8.20) in Eq. (8.21) we
get V = Q d

Aε 0
Capacitance of the parallel plate capacitor

is given by

206

Remember this Let E0 be the electric field intensity
between the plates before the introduction of

(1) If there are n parallel plates then there the dielectric slab. Then the potential difference

will be (n-1) capacitors, hence between the plates is given by V0 = E0d,
where Eo V Q and
C = (n - 1) Aε 0 AH o ,
d Ho

(2) For a spherical capacitor, consisting σ is the surface charge density on the plates.

of two concentric spherical conducting Let a dielectric slab of thickness t (t < d) be

shells with inner and outer radii as a and b introduced between the plates of the capacitor.

respectively, the capacitance C is given by The field E0 polarizes the dielectric, inducing
charge - Qp on the left side and +Qp on the right
C = 4SH 0 § ab · side of the dielectric as shown in Fig. 8.29.
¨ b-a ¸
© ¹ These induced charges set up a field Ep
inside the dielectric in the opposite direction of
(3) For a cylindrical capacitor, consisting

of two coaxial cylindrical shells with radii E0. The induced field is given by

of the inner and outer cylinders as a and b, Vp Qp ª«V p Qp º
Ho AH o ¬
and length ℓ, the capacitance C is given by Ep »
2SH0  A ¼
C

loge b The net field (E) inside the dielectric
a
reduces to E0- Ep.
C = Q = Q = AH0 Hence,
b) Capa citaVnce o©§¨fAQaHdp0 a·¹¸ralledl plate c-a-p- a(8c.i2to2r) E = Eo - Ep = Eo «ª Eo Eo = k º ,
k «¬ -E »
¼»
p

where k is a constant called the dielectric

with a dielectric slab between the plates: constant. Q or Q AKH0 E --- (8.23)
?E AH0 K
Let us now see how Eq. (8.22) gets

modified with a dielectric slab in between the Remember this

plates of the capacitor. Consider a parallel

plate capacitor with the two plates each of area The dielectric constant of a conductor is

A separated by a distance d. The capacitance infinite.

of the capacitor is given by The field Ep exists over a distance t and E0
Aε 0 over the remaining distance (d - t) between the
C0 = d
capacitor plates. Hence the potential difference

between the capacitor plates is

V = Eo d - t + E t

= Eo d - t + Eo t §©¨ E E0 ·
k k ¹¸

= Eo ¬ª« d - t + t º
k »¼


= cAaQHpoac«ª¬idta-nct e+oktf º
The »¼
Fig. 8.29: Dielectric slab in the capacitor. the capacitor on the

introduction of dielectric slab becomes

207

C =Q = Q = AH 0 8.11 Displacement Current:
V d- -t +
Q § t + d · § d t ·
AH 0 ¨© k ¹¸ ¨© k ¸¹

Special cases:

1. If the dielectric fills up the entire space then
AH 0 k
t = d ?C = d = k C0

∴ capacitance of a parallel plate capacitor
=C
increases k times i.e. k C0

2. If the capacitor is filled with n dielectric slabs Fig. 8.31: Displacement current in the space
between the plates of the capacitor.
of thickness t1, t2....... tn then this arrangement is
equivalent to n capacitors connected in series We know that electric current in a DC
circuit constitutes a flow of free electrons. In
as shown in Fig. 8.30. AH 0 a circuit as shown in Fig 8.31, a parallel plate
capacitor with a dielectric is connected across a
?C = § t1 t2 tn · DC source. In the conducting part of the circuit
¨ k1 k2 kn ¸
© + + ............. + ¹



free electrons are responsible for the flow of

current. But in the region between the plates

of the capacitor, there are no free electrons

available for conduction in the dielectric.

As the circuit is closed, the current flows

through the circuit and grows to its maximum

value (ic) in a finite time (time constant of the
circuit). The conduction current, i is found

c

to be same everywhere in the circuit except

inside the capacitor. As the current passes

through the leads of the capacitor, the electric

field between the plates increases and this in

Fig. 8.30 : Capacitor filled with n dielectric slabs. turn causes polarisation of the dielectric. Thus,

3. If the arrangement consists of n capacitors there is a current in the dielectric due to the

in parallel with plate areas A1, A2, .............. An movement of the bound charges. The current

and plate separation d due to bound charges is called displacement
C = H0
k An kn
d 1 A then
if n
A1 + A2 k2 + .........+ current (id) or charge- separation current.
A2 .............. An = We can now derive an expression between
A1 =
ic and id.
C = AH 0 k1 + k2 + .........+ kn From Eq (8.23) we can infer that the
dn
charge produced on the plates of a capacitor is

4. If the capacitor is filled with a conducting due to the electric field E.

slab (k = ∞) then q = Akε E
0
§ d ·
C = ¨ d- t ¸ C ∴ C > Co DifferenddtiqtatinAgktHhe0 daEbove equation, we get
¹ o dt --- (8.24)
©
The capacitance thus increases by a factor

§d· dq/dt is the conduction current (ic)in the
¨ ¸
© d - t ¹ conducting part of the circuit.

208

ic dq AkH 0 dE (ii) Capacitance C′ Aε0 k
dt dt d
=
dE ic dE
dt AkH 0 ? dt v ic (for fixed value of A) 8.85 u 10 12 u 4 u10 4 u 6.7

= 2u10 3

The rate of change of electric field (dE/dt) = 7.90 × 10-12 F

across the capacitor is directly proportional to Example 8.18: In a capacitor of capacitance

the current (ic) flowing in the conducting part 20 µF, the distance between the plates is 2
of the circuit.
mm. If a dielectric slab of width 1 mm and
The quantity on the RHS of Eq (8.24) is
dielectric constant 2 is inserted between the
having the dimension of electric current and is
plates, what is the new capacitance ?
caused by the displacement of bound charges
Solution: Given
in the dielectric of the capacitor under the
C = 20 µF = 20 × 10-6 F
influence of the electric field. This current,
d = 2 mm = 2 × 10-3 m
called displacement current (i ), is equivalent
d t = 1 × 10-3 m

to the rate of flow of charge (dq/dt=ic) in Aε k =2 AH 0
the conducting part of the circuit. In the C= d and
0
absence of any dielectric between the plates C′ = t
d –t k
of the capacitor, k =1 (for air or vacuum), the t
k
displacement current id = Aε0 (dE/dt). C d –t +
As a broad generalization of displacement ⇒ ⇒ CC20'' == §¨© 2ud10 3

current in a circuit containing a capacitor, it   1  u10 3  1u10 3  ·¸
2 ¹
can be stated that the displacement currents do

not remain confined to the space between the 2u10 3

plates of a capacitor. A displacement current

(id) exists at any point in space where, time- ⇒ C′ = 26.6 µF
varying electric field (E) exists (i.e. dE/dt ≠0).
8.12 Energy Stored in a Capacitor:
Example 8.17 A parallel plate capacitor Acapacitor is a device used to store energy.

has an area of 4 cm2 and a plate separation Charging a capacitor means transferring
electron from one plate of the capacitor to the
of 2 mm other. Hence work will have to be done by the
battery in order to remove the electrons against
(i) Calculate its capacitance the opposing forces. These opposing forces
arise since the electrons are being pushed to
(ii) What is its capacitance if the space the negative plate which repels them and
electrons are removed from the positive plate
between the plates is filled completely with which tends to attract them. In both the cases,
the forces oppose the transfer from one plate to
a dielectric having dielectric constant of another. As the charges on the plate increases,
opposition also increases.
constant 6.7.
This work done is stored in the form of
Solution : Given electrostatic energy in the electric field between
the plates, which can later be recovered by
A = 4 cm2 = 4 × 10-4 m2 discharging the capacitor.

d = 2 mm = 2 × 10-3 m

ε0 = 8.85 × 10-12 C2 / Nm2
(i) Capacitance C
= Aε0
d

= 8.85 u 10 12 u 4 u10 4 = 1.77 × 10-12 F
2u10 3

209

Consider a capacitor of capacitance C The potential difference between the plates
being charged by a DC source of V volts as
shown in Fig. 8.32. is maintained constant at 400 volt. What is

the change in the energy of capacitor if the

slab is removed ?

Solution : Energy stored in the capacitor

with air 1 1
2 2
Ea= CV2 = ×3×10 –9 × (400)2

= 24 × 10–5 J

Fig. 8.32: Capacitor charged by a DC source. when the slab of dielectric constant 3

During the process of charging, let q' is introduced between the plates of the

be the charge on the capacitor and V be the capacitor, the capacitance of the capacitor

p otentiaCl d=ifVfqe'r ence between the plates. Hence increases to
A small amount of work is done if a small
C′ = kC

C′ = 3 × 3 × 10–9 = 9 × 10–9 F

Energy stored in the capacitor with the

charge dq is further transferred between the dielectric (E )
d
plates. q' 1
C
? dW V dq dq Ed = 2 C ' V2
1

Total work done in transferring the charge Ed = 2 × 9 × 10-9 × (400)2
Q q ' dq Q = 72 × 10-5 J
W dw 1 dq
³ ³ ³ q'
Change in energy = Ed– Ea = (72 - 24) × 10-5
= 48 ×10–5 J
OC C O

1 ª q ' 2 ºQ 1 Q2 There is, therefore, an increase in the
« »
C ¬« 2 ¼» 0 2C energy on introducing the slab of dielectric

material.

This work done is stored as electrical 8.13 Van de Graaff Generator:
Van de Graaff generator is a device used
potential energy U of the capacitor. This work
to develop very high potentials of the order of
done can be expressed in different forms as 107 volts. The resulting large electric fields are
used to accelerate charged particles (electrons,
follows. protons, ions) to high energies needed for
experiments to probe the small scale structure
?U = 1 Q2 = 1 CV 2 = 1 QV Q =CV of matter and for various experiments in
2 C 2 2 Nuclear Physics.

Observe and discuss It was designed by Van de Graaff (1901-
1967) in the year 1931.
The energy supplied to the battery is QV Principle: This generator is based on
(i) the phenomenon of Corona Discharge
but energy stored in the electric field is
11 (action of sharp points),
2 QV. The rest half 2 QV of energy is (ii) the property that charge given to a hollow
wasted as heat in the connecting wires and
battery itself. conductor is transferred to its outer surface
and is distributed uniformly over it,
Example 8.19: A parallel plate air capacitor (iii) if a charge is continuously supplied to an
has a capacitance of 3 × 10–9 Farad. A slab insulated metallic conductor, the potential
of dielectric constant 3 and thickness 3 cm of the conductor goes on increasing.
completely fills the space between the plates.

210

Construction: filled with nitrogen at high pressure. A small
Fig. 8.33 shows the schematic diagram of quantity of Freon gas is mixed with nitrogen to
ensure better insulation between the vessel S
Van de Graaff generator. and its contents. A metal plate M held opposite
to the brush A on the other side of the belt is
Fig. 8.33: Schematic diagram of van de Graff connected to the vessel S, which is earthed.
generator. Working: The electric motor connected to the
pulley P1 is switched on, which begins to rotate
P1 P2 = Pulleys setting the conveyor belt into motion. The DC
BB = Conveyer belt supply is then switched on. From the pointed
A = Spray brush ends of the spray brush A, positive charge is
C = Collector brush continuously sprayed on the belt B. The belt
D = Dome shaped hollow conductor carries this charge in the upward direction,
E = Evacuated accelerating tube which is collected by the collector brush C and
I = Ion source sent to the dome shaped conductor.
P = DC power supply
S = Steel vessel filled with nitrogen As the dome is hollow, the charge is
M = Earthed metal plate distributed over the outer surface of the dome.
An endless conveyor belt BB made of an Its potential rises to a very high value due to
insulating material such as reinforced rubber the continuous accumulation of charges on it.
or silk, can move over two pulleys P1 and The potential of the electrode I also rises to
P2. The belt is kept continuously moving by this high value.
a motor (not shown in the figure) driving the
lower pulley (P1). The positive ions such as protons or
The spray brush A, consisting of a large deuterons from a small vessel (not shown in
number of pointed wires, is connected to the the figure) containing ionised hydrogen or
positive terminal of a high voltage DC power deuterium are then introduced in the upper part
supply. From this brush positive charge can of the evacuated accelerator tube. These ions,
be sprayed on the belt which can be collected repelled by the electrode I, are accelerated in
by another similar brush C. This brush is the downward direction due to the very high
connected to a large, dome-shaped, hollow fall of potential along the tube, these ions
metallic conductor D, which is mounted on acquire very high energy. These high energy
insulating pillars (not shown in the figure). E charged particles are then directed so as to
is an evacuated accelerating tube having an strike a desired target.
electrode I at its upper end, connected to the Uses: The main use of Van de Graff generator
dome-shaped conductor. is to produce very high energy charged particles
To prevent the leakage of charge from having energies of the order of 10 MeV. Such
the dome, the pulley and belt arrangement, high energy particles are used
the dome and a part of the evacuated tube 1. to carry out the disintegration of nuclei of
are enclosed inside a large steel vessel S,
different elements,
2. to produce radioactive isotopes,
3. to study the nuclear structure,
4. to study different types of nuclear reactions,
5. accelerating electrons to sterilize food and

to process materials.

Internet my friend

1. https://en.m.wikipedia.org
2. hyperphyrics.phy-astr.gsu.edu
3. https://www.britannica.com/science
4. https://www.khanacademy.org>in-i

211

Exercises

Q1. Choose the correct option

i) A parallel plate capacitor is charged and qQ (D) qQ
(C) 6SH0 L 4SH0 L
then isolated. The effect of increasing

the plate separation on charge, potential, v) A parallel plate capacitor has circular

capacitance respectively are plates of radius 8 cm and plate separation

(A) Constant, decreases, decreases 1mm. What will be the charge on the

(B) Increases, decreases, decreases plates if a potential difference of 100 V

(C) Constant, decreases, increases is applied?

(D) Constant, increases, decreases (A) 1.78 × 10-8 C (B) 1.78 × 10-5 C

ii) A slab of material of dielectric constant (C) 4.3 × 104 C (D) 2 × 10-9 C

k has the same area A as the plates of a Q2. Answer in brief.

parallel plate capacitor and has thickness i) A charge q is moved from a point A

(3/4d), where d is the separation of the above a dipole of dipole moment p to

plates. The change in capacitance when a point B below the dipole in equitorial

the slab is inserted between the plates is plane without acceleration. Find the

(A) C AH 0 § k 3 · work done in this process.
d ¨© 4k ¸¹

(B) C AH 0 § 2k ·
d ¨© k 3 ¹¸
AH 0 k 3
(C) C d § 2k ·
¨© ¸¹

(D) C AH 0 § 4k ·
d ¨© k 3 ¸¹

iii) Energy stored in a capacitor and ii) If the difference between the radii of the

dissipated during charging a capacitor two spheres of a spherical capacitor is

bear a ratio. increased, state whether the capacitance

(A) 1:1 (B) 1:2 will increase or decrease.

(C) 2:1 (D) 1:3 iii) A metal plate is introduced between

iv) Charge +q and -q are placed at points the plates of a charged parallel plate

A and B respectively which are distance capacitor. What is its effect on the

2L apart. C is the mid point of A and B. capacitance of the capacitor?

The work done in moving a charge +Q iv) The safest way to protect yourself from

along the semicircle CRD as shown in lightening is to be inside a car. Justify.

the figure below is v) A spherical shell of radius b with charge
Q is expanded to a radius a. Find the

work done by the electrical forces in the

process.

3. A dipole with its charges, -q and +q

located at the points (0, -b, 0) and (0 +b,

0) is present in a uniform electric field E.

(A) qQ (B) qQ The equipotential surfaces of this field
6SH0 L 2SH0 L are planes parallel to the YZ planes.

212

(a) What is the direction of the electric in which all the dipoles are perpendicular
to the field, θ2 = 90°.[Ans: 1.575 × 10-3 J]
field E? (b) How much torque would the 11. A charge 6 µC is placed at the origin
and another charge –5 µC is placed on
dipole experience in this field? the y axis at a position A (0, 6.0) m.

4. Three charges -q, +Q and -q are placed

at equal distance on straight line. If the

potential energy of the system of the

three charges is zero, then what is the

ratio of Q:q?

5. A capacitor has some dielectric between

its plates and the capacitor is connected

to a DC source. The battery is now

disconnected and then the dielectric is

removed. State whether the capacitance, a) Calculate the total electric potential

the energy stored in it, the electric field, at the point P whose coordinates are

charge stored and voltage will increase,

decrease or remain constant. (8.0, 0) m

6. Find the ratio of the potential differences b) Calculate the work done to bring

that must be applied across the parallel a proton from infinity to the point

and series combination of two capacitors P ? What is the significance of the

C1 and C2 with their capacitances in the negative sign ?
ratio 1:2, so that the energy stored in
[Ans: (a) Vp = 2.25 × 103 V
these two cases becomes the same.
(b) W = -5.4 × 10-16 J]
7. Two charges of magnitudes -4Q and
12. In a parallel plate capacitor with air
+2Q are located at points (2a, 0) and (5a,
between the plates, each plate has an
0) respectively. What is the electric flux
area of 6 × 10–3 m2 and the separation
due to these charges through a sphere of
between the plates is 2 mm. a) Calculate
radius 4a with its centre at the origin?
the capacitance of the capacitor, b) If this
8. A 6 µF capacitor is charged by a 300
capacitor is connected to 100 V supply,
V supply. It is then disconnected
what would be the charge on each plate?
from the supply and is connected to
c) How would charge on the plates be
another uncharged 3µF capacitor. How
affected if a 2 mm thick mica sheet of
much electrostatic energy of the first
k = 6 is inserted between the plates while
capacitor is lost in the form of heat and
the voltage supply remains connected ?
electromagnetic radiation ?
[Ans: (a) 2.655 × 10-11 F,
[Ans: 9 × 10-2 J]
(b) 2.655 × 10-9 C, (c) 15.93 × 10-9 C]
9. One hundred twenty five small liquid
13. Find the equivalent capacitance between
drops, each carrying a charge of
P and Q. Given, area of each plate = A
0.5 µC and each of diameter 0.1 m form
and separation between plates = d.
a bigger drop. Calculate the potential at 2 Aε0 4 Aε
d d
the surface of the bigger drop. [Ans: (a) (b) 0 ]

[Ans: 2.25 × 106 V]

10. The dipole moment of a water molecule

is 6.3 × 10–30 Cm. A sample of water

contains 1021 molecules, whose dipole

moments are all oriented in an electric

field of strength 2.5 × 105 N /C. Calculate

the work to be done to rotate the dipoles

from their initial orientation θ1 = 0 to one

213

9. Current Electricity

Can you recall?

• There can be three types of electrical Fig 9.1: Electric network.
conductors: good conductors (metals), For a steady current flowing through an
semiconductors and bad conductors electrical network of resistors, the following
(insulators). Kirchhoff 's laws are applicable.
9.2.1 Kirchhoff’s First Law: (Current law/
• Does a semiconductor diode and resistor Junction law)
have similar electrical properties? The algebraic sum of the currents at a
junction is zero in an electrical network, i.e.,
• Can you explain why two or more
resistors connected in series and parallel n
have different effective resistances?
¦ Ii 0 , where Ii is the current in the ith
9.1 Introduction:
In XIth Std. we have studied the origin of i1

electrical conductivity, in particular for metals. conductor at a junction having n conductors.
We have also studied how to calculate the
effective resistance of two or more resistances P
in series and in parallel. However, a circuit
containing several complex connections Fig. 9.2: Kirchhoff first law.
of electrical components cannot be easily Sign convention:
reduced into a single loop by using the rules
of series and parallel combination of resistors. The currents arriving at the junction are
More complex circuits can be analyzed considered positive and the currents leaving
by using Kirchhoff’s laws. Gustav Robert the junction are considered negative.
Kirchhoff (1824-1887) formulated two rules
for analyzing a complicated circuit. In this Consider a junction P in a circuit where
chapter we will discuss these laws and their six conductors meet (Fig.9.2). Applying the
applications. sign convention, we can write
9.2 Kirchhoff’s Laws of Electrical Network: I1 - I2 + I3 +I4 -I5 -I6 = 0 --- (9.1)

Before describing these laws we will Arriving currents I1, I3 and I4 are considered
define some terms used for electrical circuits. positive and leaving currents I2, I5 and I6 are
Junction: Any point in an electric circuit where considered negative.
two or more conductors are joined together is Equation (9.1) can also be written as
a junction. I1 + I3 + I4 = I2 +I5 + I6
Loop: Any closed conducting path in an
electric network is called a loop or mesh. Thus the total current flowing towards the
Branch: A branch is any part of the network junction is equal to the total current flowing
that lies between two junctions. away from the junction.

In Fig. 9.1, there are two junctions,
labeled a and b. There are three branches:
these are the three possible paths 1, 2 and 3
from a to b.

214

Example 9.1: Figure shows currents in a sense. Applying the sign conventions to Eq.

part of electrical circuit. Find the current X ? (9.2), we get,

Solutions: At junction B, -I1R1-I3R5-I1R3+ε1= 0 BFDCB in
∴ε1= I1R1+ I3R5+ I1R3
current I1 is split into I2 and
I therefore I = I + I Now consider the loop

3 123 anticlockwise direction. Applying the sign

Substituting values we get conventions, we get,
I2R2 I3R5 I2R4 H 2 0
I3 = 14 A
At C, I5 = I3 + I4 therefore ∴ H 2 I2R2 I3R5 I2R4
I = 16 A
Remember this
5
Kirchhoff’s first law is consistent with
At D, I5 = I6 + I7 therefore the conservation of electrical charge while
I6 = 7 A the voltage law is consistent with the law of
conservation of energy.
9.2.2 Kirchhoff’s Voltage Law:
Some charge is received per unit time
The algebraic sum of the potential due to the currents arriving at a junction. For
conservation of charge, same amount of charge
differences (products of current and resistance) must leave the junction per unit time which
leads to the law of currents.
and the electromotive forces (emfs) in a closed
Algebraic sum of emfs (energy per unit
loop is zero. --- (9.2) charge) corresponds to the electrical energy
supplied by the source. According to the law of
¦ IR ¦H 0 conservation of energy, this energy must appear
in the form of electrical potential difference
Sign convention: across the electrical elements/devices in the
loop. This leads to the law of voltages.
1. While tracing a loop through a resistor,

if we are travelling along the direction

of conventional current, the potential

difference across that resistance is

considered negative. If the loop is traced

against the direction of the conventional

current, the potential difference across that

resistor is considered positive.

2. The emf of an electrical source is positive

while tracing the loop within the source Steps usually followed while solving a
problem using Kirchhoff’s laws:
from the negative terminal of the source to i) Choose some direction of the currents.
ii) Reduce the number of variables using
its positive terminal. It is taken as negative
Kirchhoff’s first law.
while tracing within the source from iii) Determine the number of independent

positive terminal to the negative terminal. loops.
iv) Apply voltage law to all the independent
Fig. 9.3: Electrical network.
Consider an electrical network shown in loops.
Fig. 9.3. v) Solve the equations obtained
Consider the loop ABFGA in clockwise
simultaneously.
vi) In case, the answer of a current variable

is negative, the conventional current is
flowing in the direction opposite to that
chosen by us.

215

Example 9.2: Two batteries of 7 volt and Applying Kirchhoff second law,
13 volt and internal resistances 1 ohm and 2
ohm respectively are connected in parallel (i) loop EFCDE,
with a resistance of 12 ohm. Find the
current through each branch of the circuit 3I2 4 I1 10 0 --- (2)
and the potential difference across 12-ohm 4 I1 3I2 10
resistance. (ii) loop FABCF
Solutions: Let the currents passing through 4 I3 3I2 5 0
the two batteries be I1 and I2.

Applying Kirchhoff second law to the
loop AEFBA,

4 I3 3I2 5 --- (3)
From Eq. (1) and Eq. (2)

4 I3 I2 3I2 = 10 --- (4)

12 I1 I2 1I1 7 0 --- (1) 3I2 4 I3 4 I2 10
12 I1 I2 1I1 7 --- (2)
4 I3 7 I2 10
For the loop CEFDC From Eq. (3) and Eq. (4)
10I2 5
12 I1 I2 2 I2 13 0 I2 0.5A

12 I1 I2 2 I2 13 Negative sign indicates that I2 current
flows from F to C

From Eq. (2) 4 I1 3 0.5 10

From (1) and (2) 2 I2 I1 13 7 6 I1 = 2.12A
∴ I3 I1 I2 2.12 0.5 1.62A
I1 2 I2 6

Substituting I1 value in (2)
I=2 83=58 2.237 A
I1 2 I2 6 9.3 Wheatstone Bridge:
85 Resistance of a material changes due to
I1 2u 38 6 1.526 A
several factors such as temperature, strain,
I I1 I2 1.526 A 2.237 A 0.711 A humidity, displacement, liquid level, etc.
Therefore, measurement of these properties
Potential difference across 12 Ω resistance is possible by measuring the resistance.
V IR 0.711u12 8.53V Measurable values of resistance vary from
a few milliohms to hundreds of mega ohms.
Example 9.3: For the given network, find
the current through 4 ohm and 3 ohm. Depending upon the resistance range (milliohm
Assume that the cells have negligible to tens of ohm, tens of ohm to hundreds of ohms,
internal resistance. hundreds of ohm to mega ohm, etc.), various
Solution: Applying Kirchhoff first law methods are used for resistance measurement.
At junction F, Wheatstone’s bridge is generally used to
I1 = I3 – I2 I1 I2 I3 --- (1) measure resistances in the range from tens of
ohm to hundreds of ohms.

216

The Wheatstone Bridge was originally A special case occurs when the current

developed by Charles Wheatstone (1802- 1875) passing through the galvanometer is zero. In

to measure the values of unknown resistances. this case, the bridge is said to be balanced.

It is also used for calibrating measuring Condition for the balance is Ig = 0. This
condition can be obtained by adjusting the
instruments, voltmeters, ammeters, etc.

Four resistances P, Q, R and S are values of P, Q, R and S. Substituting Ig = 0 in
Eq. (9.4) and Eq. (9.5) we get,
connected to form a quadrilateral ABCD as

shown in the Fig. 9.4. A battery of emf ε along – I1P + I2S = 0 ∴ I1P = I2S --- (9.6)

with a key is connected between the points A – I1Q + I2R = 0 ∴ I1Q = I2R --- (9.7)

and C such that point A is at higher potential Dividing Eq. (9.6) by Eq. (9.7), we get
P S
with respect to the point C. A galvanometer Q = R --- (9.8)

of internal resistance G is connected between

points B and D. This is the condition for balancing the

When the key is closed, current I flows Wheatstone bridge.

through the circuit. It divides into I1 and I2 at If any three resistances in the bridge are
point A. I1 is the current through P and I2 is the
current through S. The current I gets divided at known, the fourth resistance can be determined

1 by using Eq. (9.8).

point B. Let Ig be the current flowing through Example 9.4: At what value should the
the galvanometer. The currents flowing variable resistor be set such that the bridge
is balanced? If the source voltage is 30 V
through Q and R are respectively (I1 – Ig) and find the value of the output voltage across
(I + I ), XY, when the bridge is balanced.

1g

From the Fig. 9.4,

I = I1 + I2 --- (9.3)

Consider the loop ABDA. Applying

Kirchhoff’s voltage law in the clockwise sense

shown in the loop we get,

– I1P – IgG + I2S = 0 --- (9.4) XY

Now consider loop BCDB, applying

Kirchhoff’s voltage law in the clockwise sense

shown in the loop we get,

– (I1 – Ig) Q + (I2 + Ig) R + Ig G = 0 --- (9.5) When the bridge is balanced

P/Q=R/S
Q = PS / R
1.36 u103 u 4.4 u103
300 19.94 u103:


Total resistance of the arm

ADC = 19940 + 4400 = 24340 Ω

To find output voltage across XY:

Potential difference across

Fig. 9.4 : Wheatstone bridge. AC = I1 u 24340 30
From these three equations (Eq. (9.3),
(9.4), (9.5) we can find the current flowing I1 = 30 A
through any branch of the circuit. 24340

Potential difference across

217

AD = I1 ×19940 Temporary contact with the wire AB can be

30 u19940 / 24340 24.58V established with the help of the jockey. A cell

of emf ε along with a key and a rheostat are

I2 30 30 A connected between the points A and B.
1360 300 1660
A suitable resistance R is selected from

So, Potential difference across resistance box. The jockey is brought in contact
30
AB= I2 u1360 1660 u1360 24.58 V with AB at various points on the wire AB and

Vout VB VD the balance point (null point), D, is obtained.

VA VB VA VD The galvanometer shows no deflection when

the jockey is at the balance point.

VAB VAD Let the respective lengths of the wire

= 24.58-24.58 = 0V between A and D, and that between D and C
be  x and  R . Then using the conditions for
Application of Wheatstone bridge:
Figure 9.4 is a basic circuit diagram of the balance, we get
Wheatstone bridge, however, in practice
the circuit is used in different manner. In all X = RAD
cases it is used to determine some unknown R RDB
resistance. Few applications of Wheatstone
bridge circuits are discussed in the following where R and R are resistance of the parts
article. AD DB
9.3.1 Metre Bridge:
AD and DB of the wire resistance of the wire. If

l is length of the wire, ρ its specific resistance,

and A its area of cross section then
U AD U DB
RAD A RDB A

X = RAD Ux / A
R RDC UR / A

∴ X = x
Therefore, R = R
 R --- (9.9)
X  x

R
Knowing R,  x and  R , the value of the

unknown resistance can be determined.

Fig. 9.5: Metre bridge. Example 9.5: Two resistances 2 ohm and 3
Metre bridge (Fig. 9.5) consists of a ohm are connected across the two gaps of the
wire of uniform cross section and one metre metre bridge as shown in figure. Calculate
in length, stretched on a metre scale which is the current through the cell when the bridge
fixed on a wooden table. The ends of the wire is balanced and the specific resistance of the
are fixed below two L shaped metallic strips. material of the metre bridge wire. Given the
A single metallic strip separates the two L resistance of the bridge wire is 1.49 ohm and
shaped strips leaving two gaps, left gap and its diameter is 0.12 cm.
right gap. Usually, an unknown resistance X is Solution: When the bridge is balanced, the
connected in the left gap and a resistance box resistances 2 and 3 ohm are in series and the
is connected in the other gap. One terminal of total resistance is 5 ohm.
a galvanometer is connected to the terminal C Let R1 be the resistance of the wire =1.49
on the central strip, while the other terminal Ω, and R2 be the total resistance (2+3)=5 Ω
of the galvanometer carries the jockey (J).

218

Rp R1R2 1.49 u 5 1.15: detect whether there is a current
R1 R2 1.49 5 through the central branch. This is
possible only by tapping the jokey.
The current through the cell
Applications:
= H 2 1.74 A • The Wheatstone bridge is used for
Rp 1.15
RS r 2 measuring the values of very low resistance
Specific resistance of the wire U l precisely.
0.12 • We can also measure the quantities such
l 1m, r 2 0.06cm, R 1.49 : as galvanometer resistance, capacitance,
inductance and impedance using a
1.49 u 3.14 u 0.06 u10 2 2 Wheatstone bridge.
1
U Do you know?
RS r2
l Wheatstone bridge along with operational
amplifier is used to measure the physical
1.68 u10 6 : m parameters like temperature, strain, etc.

Remember this Observe and discuss

1. Kelvin’s method to determine the
resistance of galvanometer (G) by using
meter bridge.

Source of errors. The galvanometer whose resistance (G) is
1. The cross section of the wire may not to be determined is connected in one gap
and a known resistance (R) in the other gap.
be uniform. Working :
2. The ends of the wire are soldered to the 1.  A suitable resistance is taken in the

metallic strip where contact resistance resistance box. The current is sent
is developed, which is not taken into round the circuit by closing the key.
account. Without touching the jockey at any
3. The measurements of  x and  R may point of the wire, the deflection in the
not be accurate. galvanometer is observed.
To minimize the errors 2. The rheostat is adjusted to get a suitable
(i) The value of R is so adjusted that the deflection Around (2/3)rd of range.
null point is obtained to middle one 3.  Now, the jockey is tapped at different
third of the wire (between 34 cm and points of the wire and a point of contact
66 cm) so that percentage error in D for which, the galvanometer shows
the measurement of  x and  R are no change in the deflection, is found.
minimum and nearly the same. 4.  As the galvanometer shows the same
(ii) The experiment is repeated by deflection with or without contact
interchanging the positions of unknown
resistance X and known resistance box
R.
(iii) The jockey should be tapped on the
wire and not slided. We use jockey to

219

between the point B and D, these two The resistances in the arms P and Q

points must be equipotential points. are fixed to desired ratio. The resistance
in the arm R is adjusted so that the
5.  The length of the bridge wire between

the point D and the left end of the galvanometer shows no deflection. Now the

wire is measured. Let lg be the length bridge is balanced. The unknown resistance
of the segment of wire opposite to the X = RQ / P , where P and Q are the fixed

galvanometer and lr be the length of resistances in the ratio arms and R is an
the segment opposite to the resistance
adjustable known resistance.

box. If L is the length of the wire and r is
its radius then the specific resistance of the
Calculation :

Let RAD and RDC be the resistance of material of the wire is given by
the two parts of the wire AD and DC
XSr2
respectively. Since bridge is balanced U L

G = R AD
R R DC
Do you know?
? R AD lg ? G lg
? R DC lr R lr Wheatstone Bridge for Strain
G lg Measurement:
R                      {lg + lr =
100 - lg 100 cm} Strain gauges are commonly used
for measuring the strain. Their electrical
G § lg lg · R resistance is proportional to the strain in
©¨¨ 100 - ¸¸¹ the device. In practice, the range of strain
gauge resistance is from 30 ohms to 3000
Using this formula, the unknown resistance ohms. For a given strain, the resistance
of the galvanometer can be calculated. change may be only a fraction of full range.
2. Post Office Box Therefore, to measure small resistance
changes with high accuracy, Wheatstone
A post office box (PO Box) is a bridge configuration is used. The figure
practical form of Wheatstone bridge as below shows the Wheatstone bridge where
shown in the figure. the unknown resistor is replaced with a
strain gauge as shown in the figure.

It consists of three arms P, Q and R. In these circuit, two resistors R1 and
The resistances in these three arms are R2 are equal to each other and R3 is the
adjustable. The two ratio arms P and Q variable resistor. With no force applied
contain resistances 10 ohm, 100 ohm and
1000 ohm each. The third arm R contains to the strain gauge, rheostat is varied and
resistances from 1 ohm to 5000 ohm. The
unknown resistance X forms the fourth
resistance. There are two tap keys K1 and
K.

2

220

finally positioned such that the voltmeter Therefore, the potential difference per unit
will indicate zero deflection, i.e., the bridge
is balanced. The strain at this condition length of the wire is,
represents the zero of the gauge. VAB HR
L = L(R r)
If the strain gauge is either stretched As
or compressed, then the resistance changes. long as ε remains constant, VAB will
This causes unbalancing of the bridge. This
produces a voltage indication on voltmeter remain constant. VAB L
which corresponds to the strain change. If L is known as potential
the strain applied on a strain gauge is more,
then the voltage difference across the meter gradient along AB and is denoted by K.
terminals is more.  If the strain is zero, then
the bridge balances and meter shows zero Potential gradient can be defined as potential
reading.
difference per unit length of wire.
This is the application of precise
resistance measurement using a Wheatstone Fig. 9.6: Potentiometer.
bridge. Consider a point C on the wire at distance
 from the point A, as shown in the figure.
9.4 Potentiometer: The potential difference between A and C is
VAC. Therefore,
A voltmeter is a device which is used for VAC = K  i.e. VAC ∝ 
Thus, the potential difference between two
measuring potential difference between two points on the wire is directly proportional to
the length of the wire between them provided
points in a circuit. An ideal voltmeter which the wire is of uniform cross section, the current
through the wire is the same and temperature
does not change the potential difference to be of the wire remains constant. Uses of
potentiometer are discussed below.
measured, should have infinite resistance so 9.4.2 Use of Potentiometer:
A) To Compare emf. of Cells
that it does not draw any current. Practically,

a voltmeter cannot be designed to have an

infinite resistance. Potentiometer is one such

device which does not draw any current from

the circuit. It acts as an ideal voltmeter. It is

used for accurate measurement of potential

difference.

9.4.1 Potentiometer Principle:

A potentiometer consists of a long wire AB

of length L and resistance R having uniform

cross sectional area A. (Fig. 9.6) A cell of emf

ε having internal resistance r is connected

across AB as shown in the Fig. 9.6. When the

circuit is switched on, current I passes through

the wire. Fig. 9.7: Emf comparison by
H individual method.

Current through AB, I = R r Method I : A potentiometer circuit is set up
by connecting a battery of emf ε , with a key
Potential difference across AB is K and a rheostat such that point A is at higher

VAB = I R R
H
VAB = (R r)

221

potential than point B. The cells whose emfs When two cells are connected so that

are to be compared are connected with their their negative terminals are together or their

positive terminals at point A and negative positive terminals are connected together as

terminals to the extreme terminals of a two- shown in Fig. 9.8 (b).
In this case their emf oppose each other
way key K1K2. The central terminal of the two
ways key is connected to a galvanometer. The and effective emf of the combination of two
cells is ε1 – ε2 ( ε1 > ε2 assumed). This method
other end of the galvanometer is connected to of connecting two cells is called the difference
method. Remember that this combination of
a jockey (J). (Fig. 9.7) Key K is closed and cells is not a parallel combination of cells.

then, key K is closed and key K2 is kept open.
1

Therefore, the cell of emf ε comes into circuit.
1

The null point is obtained by touching the

jockey at various points on the potentiometer Fig. 9.8 (a):Sum method.
wire AB. Let 1 be the length of the wire
between the null point and the point A. 1
corresponds to emf ε1 of the cell. Therefore,
ε1 = K 1

where K is the potential gradient along the

potentiometer wire.

Now key K1 is kept open and key K2 is Fig. 9.8 (b): Difference method.
closed. The cell of emf ε2 now comes in the
Circuit is connected as shown in Fig.9.9.
circuit. Again, the null point is obtained with When keys K1 and K3 are closed the cells ε1
the help of the Jockey. Let 2 be the length of and ε2 are in the sum mode. The null point
is obtained using the jockey. Let 1 be the
the wire between the null point and the point length of the wire between the null point

A. This length corresponds to the emf ε2 of and the point A. This corresponds to the emf
( ε1 + ε 2 ).
the cell.
∴ ε1 + ε2 = k 1
∴ ε2 = K2 Now the key K1 and K3 are kept open and
keys K2 and K4 are closed. In this case the two
From the above two equations we get cells are in the difference mode. Again the null
point is obtained. Let 2 be the length of the
H1 1 --- (9.10)
H2 2 wire between the null point and the point A.
Thus, we can compare the emfs of the two This corresponds to ε1 - ε2
∴ ε1 - ε2 = 2
cells. If any one of the emfs is known, the

other can be determined.

Method II: The emfs of cells can be compared

also by another method called sum and

difference method.

When two cells are connected so that the

positive terminal of the first cell is connected

to the negative terminal of the second cell
as shown in Fig 9.8 (a). The emf of the two

cells are added up and the effective emf of

the combination of two cells is ε1 + ε2 . This Fig. 9.9: Emf comparison, sum and difference
method of connecting two cells is called the method.

sum method.

222

From the above two equations, The length of the wire 2 between the
null point and point A is measured. This
H1 H2 1
By comHp1o neHn2do an2d dividendo method, we corresponds to the voltage between the null

get, point and point A. ∴ H1 k1 1
V k2 2
H1 1 2 --- (9.11) ∴ V = k2
H2 1 2
Consider the loop PQSTP.

Thus, emf of two cells can be compared. ε1 = IR + Ir and

B) To Find Internal Resistance (r) of a Cell: V = IR
∴ H1 IR Ir R r 1
The experimental set up for this method R 2
V IR
consists of a potentiometer wire AB connected
§ 1 1·¸
in series with a cell of emf ε , the key K1, and ∴r R ¨ 2 ¹ --- (9.12)
rheostat as shown in Fig. 9.10. The terminal A
©

is at higher potential than terminal B. A cell This equation gives the internal resistance of
the cell.
of emf ε1 whose internal resistance r1 is to be C) Application of potentiometer:
determined is connected to the potentiometer
The applications of potentiometer
wire through a galvanometer G and the jockey discussed above are used in laboratory. Some
practical applications of potentiometer are
J. A resistance box R is connected across the given below.
cell ε1 through the key K2. 1) Voltage Divider: The potentiometer can
be used as a voltage divider to continuously
Fig. 9.10 : Internal resistance of a cell. change the output voltage of a voltage supply
(Fig. 9.11). As shown in the Fig. 9.11,
The key K1 is closed and K2 is open. The potential V is set up between points A and B
circuit now consists of the cell ε , cell ε1 , and of a potentiometer wire. One end of a device is
the potentiometer wire. The null point is then connected to positive point A and the other end
obtained. Let 1be length of the potentiometer is connected to a slider that can move along
wire between the null point and the point A. wire AB. The voltage V divides in proportion
of lengths l1 and l2 as shown in the figure 9.11.
This length corresponds to emf ε1 .
∴ ε1 = k 1 where k is potential gradient of the Fig. 9.11 :
potentiometer wire which is constant.
Potentiometer as
Now both the keys K1 and K2 are closed so a voltage divider.
that the circuit consists of the cell ε , the cell
2) Audio Control: Sliding potentiometers, are
ε1 , the resistance box, the galvanometer and commonly used in modern low-power audio
the jockey. Some resistance R is selected from systems as audio control devices. Both sliding

the resistance box and null point is obtained.

223

(faders) and rotary potentiometers (knobs) difference of the order 10–6 volt can
are regularly used for frequency attenuation, be measured with it. Least count of a
loudness control and for controlling different potentiometer is much better compared to
characteristics of audio signals. that of a voltmeter.
3) Potentiometer as a senor: If the slider of Demerits:
a potentiometer is connected to the moving Potentiometer is not portable and direct
part of a machine, it can work as a motion measurement of potential difference or emf is
sensor. A small displacement of the moving not possible.
part causes changes in potential which is 9.5 Galvanometer:
further amplified using an amplifier circuit. A galvanometer is a device used to detect
The potential difference is calibrated in terms weak electric currents in a circuit.  It has a
of the displacement of the moving part. coil pivoted (or suspended) between concave
pole faces of a strong laminated horse shoe
Example 9.7: In an experiment to magnet.  When an electric current passes
through the coil, it deflects.   The deflection is
determine the internal resistance of a cell proportional to the current passing through the
coil. The deflection of the coil can be read with
of emf 1.5 V, the balance point in the open the help of a pointer attached to it. Position
of the pointer on the scale provided indicates
cell condition at is 76.3 cm. When a resistor the current passing through the galvanometer
or the potential difference across it. Thus, a
of 9.5 ohm is used in the external circuit of galvanometer can be used as an ammeter or
voltmeter with suitable modification. The
the cell the balance point shifts to 64.8 cm galvanometer coil has a moderate resistance
(about 100 ohms) and the galvanometer itself
of the potentiometer wire. Determine the has a small current carrying capacity (about
1 mA).
internal resistance of the cell.

Solution: Open cell balancing length

l = 76.3 cm

1

Closed circuit balancing length

l2 = 64.8 cm External resistance R = 9.5 Ω
Internal resistance r
§ l1 l2 · R
¨ l2 ¸
© ¹

§ 76.3 64.8 · u 9.5
¨© 64.8 ¸¹

1.686 :

9.4.3 Advantages of a Potentiometer Over Fig. 9.12 Internal structure of galvanometer.
a Voltmeter:
Merits: 9.5.1 Galvanometer as an Ammeter:
i) Potentiometer is more sensitive than a
Let the full scale deflection current and
voltmeter.
ii) A potentiometer can be used to measure the resistance of the coil G of moving coil

a potential difference as well as an emf galvanometer (MCG ) be I and G. It can be
of a cell. A voltmeter always measures s
terminal potential difference, and as it
draws some current, it cannot be used to converted into an ammeter, which is a current
measure an emf of a cell.
iii) Measurement of potential difference or measuring instrument. It is always connected
emf is very accurate in the case of a
potentiometer. A very small potential in series with a resistance R through which the

current is to be measured.

224

To convert a moving coil galvanometer ∴ GIg = S (I – Ig)
(MCG ) into an ammeter
? S ©¨¨§ I Ig ¸¸¹· G --- (9.13)
To convert an MCG into an ammeter, the Ig
modifications necessary are
1. Its effective current capacity must be Equation 9.13 is useful to calculate the

increased to the desired higher value. range of current that the galvanometer can
2. Its effective resistance must be decreased.
measure.
The finite resistance G of the galvanometer
when connected in series, decreases the (i) If the current I is n times current Ig, then
current through the resistance R which is
actually to be measured. In ideal case, an I = n Ig. Using this in the above expression we
ammeter should have zero resistance.
3. It must be protected from the possible get S GI g OR S G
damages, which are likely due to the nIg Ig n 1
passage of an excess electric current to be
passed. This is the required shunt to increase the range
In practice this is achieved by connecting
a low resistance in parallel with the n times.
galvanometer, which effectively reduces
the resistance of the galvanometer. This low (ii) Also if Is is the current through the shunt
resistance connected in parallel is called shunt resistance, then the remaining current (I – Is)
(S). This arrangement is shown in Fig. 9.13. will flow through galvanometer. Hence
Uses of the shunt:
G (I – Is) = S Is
S Is
i.e. G I – G Is = GI

i .e. ?S IIsIs+ G Is =
§ S G ·
¨© ¸¹
G

This equation gives the fraction of the

total current through the shunt resistance.

a. It is used to divert a large part of total Example 9.8: A galvanometer has a
current by providing an alternate path
and thus it protects the instrument from resistance of 100 Ω and its full scale
damage. deflection current is 100 µ A. What shunt

b. It increases the range of an ammeter. resistance should be added so that the
c. It decreases the resistance between the
ammeter can have a range of 0 to 10 mA ?
points to which it is connected. Solution: Given IG = 100 µ A = 0.1 mA
The shunt resistance is calculated as The upper limit gives the maximum current
follows. In the arrangement shown in the figure, to be measured, which is I = 10 mA .
Ig is the current through the galvanometer.
Therefore, the current through S is The galvanometer resistance is G = 100 Ω.
(I – Ig) = Is
Now

n 10 100? s G 100 10 0 :
0.1 n 1 100 1 99

Example 9.9: What is the value of the shunt

resistance that allows 20% of the main

current through a galvanometer of 99 Ω?

Solution: Given

G = 99 Ω and IG =(20/100)I = 0.2 I

Fig. 9.13 Ammeter. Now IG G 0.2 I u 99 0.2u 99
Since S and G are parallel, I IG 0.8
∴ GIg = S Is S I 0.2I 24.75 :

225

9.5.2 Galvanometer as a Voltmeter: where Ig is the current flowing through the
A voltmeter is an instrument used to galvanometer.

measure potential difference between two Eq. (9.14) gives the value of resistance X.
points in an electrical circuit. It is always
connected in parallel with the component If nV V V
across which voltage drop is to be measured. Vg ( Ig ˜G) is the factor by which
A galvanometer can be used for this purpose.
To Convert a Moving Coil Galvanometer the voltage range is increased, it can be shown

that X = G (nv-1)

into a Voltmeter. Example 9.10: A Galvanometer has a
To convert an MCG into a Voltmeter the
resistance of 25 Ω and its full scale deflection
modifications necessary are:
1. Its voltage measuring capacity must be current is 25 µA. What resistance should be

increased to the desired higher value. added to it to have a range of 0 -10 V?
2. Its effective resistance must be increased,
Solution: Given G = 25 µA.
and
3. It must be protected from the possible Maximum voltage to be measured is

damages, which are likely due to excess V =10 V.
applied potential difference.
All these requirements can be fulfilled, if The Galvanometer resistance G = 25 Ω.
we connect a resistance of suitable high value
(X) in series with the given MCG. The resistance to be added in series,
A voltmeter is connected across the points V 10
where potential difference is to be measured. If X IG G 25 u10 6 25
a galvanometer is used to measure voltage, it
draws some current (due to its low resistance), 399.975u103:
therefore, actual potential difference to be
measured decreases. To avoid this, a voltmeter Example 9.11: A Galvanometer has a
resistance of 40 Ω and a current of 4 mA is
should have very high resistance. Ideally, it
needed for a full scale deflection . What is
should have infinite resistance.
the resistance and how is it to be connected

to convert the galvanometer (a) into an

ammeter of 0.4 A range and (b) into a

voltmeter of 0.5 V range?
Solution: Given G = 40 Ω and IG = 4 mA
(a) To convert the galvanometer into an

ammeter of range 0.4 A,

I IG S IGG

0.4 0.004 S 0.004 u 40

Fig. 9.14 : Voltmeter. S 0.004 u 40 0.16 0.4040 :
0.396 0.396

A very high resistance X is connected in

series with the galvanometer for this purpose as (b)To convert the galvanometer into a

shown in Fig. 9.14. The value of the resistance voltmeter of range of 0.5 V

X can be calculated as follows. V IG G X
0.5 0.004 40 X
If V is the voltage to be measured, then
0.5
V = Ig X + Ig G. X 0.004 40 85 :

∴ Ig X = V – Ig G
V
?X Ig G , --- (9.14)

226

Comparison of an ammeter and a voltmeter:

AMMETER VOLTMETER antimony-bismuth thermo-couple is shown
in a diagram.
1. It measures 1. It measures
For this thermo couple the current
current. potential difference flows from antimony to bismuth at the cold
junction. (ABC rule). For a copper-iron
2. It is connected in 2. It is connected in

series. parallel.

3. It is an MCG with 3. It is an MCG with

low resistance. high resistance.

(Ideally zero) (Ideally infinite)

4. Smaller the shunt, 4. Larger its

greater will r e s i s t a n c e ,

be the current greater will be the

measured. potential difference

5. Resistance of measured.

ammeter is 5. Resistance of couple (see diagram) the current flows

RA S ˜G G voltmeter is from copper to iron at the hot junction,
S G n RV G X G ˜ nV
This effect is reversible. The direction

of the current will be reversed if the hot

and cold junctions are interchanged.

THERMOELECTRICITY The thermo emf developed in a
When electric current is passed through
thermocouple when the cold junction is
a resistor, electric energy is converted into
thermal energy. The reverse process, viz., at 00 C and the hot junction is at T°C is
conversion of thermal energy directly into 1
electric energy was discovered by Seebeck given by H D T 2 ET 2
and the effect is called thermoelectric effect.
Seebeck Effect Here α and b are called the
thermoelectric constants. This equation
If two different metals are joined to form tells that a graph showing the variation of
a closed circuit (loop) and these junctions ε with temperature is a parabola.
are kept at different temperatures, a small
emf is produced and a current flows through Do you know?
the metals. This emf is called thermo emf
this effect is called the Seebeck effect Accelerator in India:
and the pair of dissimilar metals forming Cyclotron for medical applications.
the junction is called a thermocouple. An

Picture credit: Director, VECC, Kolkata,
Department of Atomic Energy, Govt. of India

227

Exercises

1. Choose the correct option. (A) 2 Ω (B) 4 Ω
i) Kirchhoff’s first law, i.e., ΣI = 0 at a
(C) 8 Ω (D) 16 Ω
junction, deals with the conservation of
(A) charge (B) energy 2. Answer in brief.
(C) momentum (D) mass
ii) When the balance point is obtained in the i) Define or describe a Potentiometer.

potentiometer, a current is drawn from ii) Define Potential Gradient.
(A) both the cells and auxiliary battery
(B) cell only iii) Why should not the jockey be slided
(C) auxiliary battery only
(D) neither cell nor auxiliary battery along the potentiometer wire?
iii) In the following circuit diagram, an
iv) Are Kirchhoff’s laws applicable for both
infinite series of resistances is shown.
Equivalent resistance between points A AC and DC currents?
and B is 
v) In a Wheatstone’s meter-bridge

experiment, the null point is obtained in

middle one third portion of wire. Why is

it recommended?

vi) State any two sources of errors in meter-

bridge experiment. Explain how they can

be minimized.

vii) What is potential gradient? How is it

measured? Explain.

viii) On what factors does the potential

(A) infinite (B) zero gradient of the wire depend?

(C) 2 Ω (D) 1.5 Ω ix) Why is potentiometer preferred over a

iv) Four resistances 10 Ω, 10 Ω, 10 Ω  and voltmeter for measuring emf?

15 Ω form a Wheatstone’s network. What x) State the uses of a potentiometer.

shunt is required across 15 Ω resistor to xi) What are the disadvantages of a

balance the bridge potentiometer?

(A) 10 Ω (B) 15 Ω xii) Distinguish between a potentiometer and

(C) 20 Ω (D) 30 Ω a voltmeter.

v) A circular loop has a resistance of 40 Ω. xiii) What will be the effect on the position

Two points P and Q of the loop, which are of zero deflection if only the current

one quarter of the circumference apart flowing through the potentiometer wire is

are connected to a 24 V battery, having (i) increased (ii) decreased.

an internal resistance of 0.5 Ω. What is 3. Obtain the balancing condition in case of

the current flowing through the battery. a Wheatstone’s network.

(A) 0.5 A (B) 1A 4. Explain with neat circuit diagram,

(C) 2A (D) 3A how you will determine the unknown

vi) To find the resistance of a gold bangle, resistance by using a meter-bridge.

two diametrically opposite points of the 5. Describe Kelvin’s method to determine

bangle are connected to the two terminals the resistance of a galvanometer by using

of the left gap of a metre bridge. A a meter bridge.

resistance of 4 Ω is introduced in the right 6. Describe how a potentiometer is used

gap. What is the resistance of the bangle to compare the emfs of two cells by

if the null point is at 20 cm from the left connecting the cells individually.

end?

228

7. Describe how a potentiometer is used 15. When two cells of emfs. ε1 and ε2
are connected in series so as to assist
to compare the emfs of two cells by

combination method. each other, their balancing length on a

8. Describe with the help of a neat circuit potentiometer is found to be 2.7 m. When

diagram how you will determine the the cells are connected in series so as to

internal resistance of a cell by using oppose each other, the balancing length

a potentiometer. Derive the necessary is found to be 0.3 m. Compare the emfs

formula. of the two cells.

9. On what factors does the internal [Ans: 1.25]

resistance of a cell depend? 16. The emf of a cell is balanced by a length

10. A battery of emf 4 volt and internal of 120 cm of potentiometer wire. When

resistance 1 Ω is connected in parallel with the cell is shunted by a resistance of 10

another battery of emf 1 V and internal Ω, the balancing length is reduced by 20

resistance 1 Ω (with their like poles cm. Find the internal resistance of the

connected together). The combination is cell.

used to send current through an external [Ans: r = 2 Ohm]

resistance of 2 Ω. Calculate the current 17. A potential drop per unit length along a

through the external resistance. wire is 5 x 10-3 V/m. If the emf of a cell

[Ans: 1 A] balances against length 216 cm of this

11. Two cells of emf 1.5 Volt and 2 Volt potentiometer wire, find the emf of the

having respective internal resistances of 1 cell.

Ω and 2 Ω are connected in parallel so as [Ans: 0.01080 V]

to send current in same direction through 18. The resistance of a potentiometer wire is

an external resistance of 5 Ω. Find the 8 Ω and its length is 8 m. A resistance box

current through the external resistance. and a 2 V battery are connected in series

[Ans: 5/17 A] with it. What should be the resistance in

12. A voltmeter has a resistance 30 Ω. What the box, if it is desired to have a potential

will be its reading, when it is connected drop of 1μV/mm?

across a cell of emf 2 V having internal [Ans: 15992 ohm]

resistance 10 Ω? 19. Find the equivalent resistance between

[Ans: 1.5 V] the terminals of A and B in the network

13. A set of three coils having resistances shown in the figure below given that the

10 Ω, 12 Ω and 15 Ω are connected in resistance of each resistor is 10 ohm.

parallel. This combination is connected I2

in series with series combination of three I1-I2 I-I
2

coils of the same resistances. Calculate

the total resistance and current through I
1 I-I
the circuit, if a battery of emf 4.1 Volt is
2

used for drawing current.

[Ans: 0.1 A] E

14. A potentiometer wire has a length of 1.5 [Ans: 14 Ohm]

m and resistance of 10 Ω. It is connected 20. A voltmeter has a resistance of

in series with the cell of emf 4 Volt and 100 Ω. What will be its reading when it

internal resistance 5 Ω. Calculate the is connected across a cell of emf 2 V and

potential drop per centimeter of the wire. internal resistance 20 Ω?

[Ans: 0.0178 V/cm] [Ans: 1.66 V]

229

10. Magnetic Fields due to Electric Current

Can you recall? Try this

• Do you know that a magnetic field is Fig. 10.1 (a) Fig. 10.1 (b)
produced around a current carrying wire?
You can show that wires having
• What is right hand rule?
• Can you suggest an experiment to draw currents passing through them, (a) in

magnetic field lines of the magnetic field opposite directions repel and (b) in the
around the current carrying wire?
• Do you know solenoid? Can you compare same direction attract.
the magnetic field due to a current
carrying solenoid with that due to a bar Hang two conducting wires from
magnet?
an insulating support. Connect them to
Do you know?
a cell first as shown in Fig. 10.1 (a) and
You must have noticed high tension power
transmission lines, the power lines on the later as shown Fig. 10.1 (b), with the help
big tall steel towers. Strong magnetic fields
are created by these lines. Care has to be of binding posts. You will notice that the
taken to reduce the exposure levels to less
than 0.5 milligauss (mG). wires in (a) repel each other and those in (b)

10.1 Introduction: come closer, i.e., they attract each other as
In this Chapter you will be studying how
soon as the current starts. The force in this
magnetic fields are produced by an electric
current. Important foundation for further experiment is certainly not of electrostatic
developments will also be laid down.
origin, even through the current is due to the
Hans Christian Oersted first discovered
that magnetic field is produced by an electric electrons flowing in the wires. The overall
current passing through a wire. Later, Gauss,
Henry, Faraday and others showed that charge neutrality is maintained throughout
magnetic field is an important partner of
electric field. Maxwell’s theoretical work the wire, hence the electrostatic forces are
highlighted the close relationship of electric
and magnetic fields. This resulted into several ruled out.
practical applications in day today life,
for example electrical motors, generators, You have learnt in Xth Std. that if a
communication systems and computers. magnetic needle is held in close proximity of
a current carrying wire, it shows the direction
In electrostatics, we have considered of magnetic field circling around the wire.
static charges and the force exerted by them on Imagine that a current carrying wire is grabbed
other charge or test charge. We now consider with your right hand with the thumb pointing
forces between charges in motion. in the direction of the current, then your fingers
curl around in the direction of the magnetic
field (Fig. 10.2).

230

I 
(ii) If the charge is stationary, v =0, the

force = 0, even if B ≠ 0.

From Eq. (10.4) it may be observed that

the force on the charge due to electric field

depends on the strength of the electric field

and the magnitude of the charge. However, the

magnetic force depends on the velocity of the

Fig. 10.2: Right Fig. 10.3: Force on wire charge and the cross product of the velocity
hand thumb rule. 2 due to current in wire 1. vector v the magnetic field vector B , and the
charge q. 
How can one account for the force on
Consider the vectors v and B with certain
the neighbouring current carrying wire? The angle between them. Then v × B will be a

magnetic field due to current in the wire 1 at vector perpendicular to the plane containing
the vectors v and B (Fig. 10.4).
any point on wire 2 is directed into the plane

of the paper. The electrons flow in a direction

opposite to the conventional current. Then the

wire 2 experiences a force F towards wire 1.

10.2 Magnetic Force:

From the above discussion and Fig. 10.3,

you must have realized that the directions of
v , B and F follow a vector cross product

relationship. Actually the magnetic force Fm Fig. 10.4: The cross product is in the direction of
on an electron with a charge -e, moving with the unit vector perpendicular to both v and B .
velocity v in a magnetic field B is
Fm = -e( v × B ) --- (10.1) Thus the vectors v and F are always
perpendicular to each other. Hence. F . v = 0,
In general for a charge q, the magnetic
for any magnetic field B . Magnetic force Fm
force will be  is thus perpendicular to the displacement and
Fm = q( v × B )
--- (10.2) hence the magnetic force never does any work

If both electric field E and the magnetic on moving charges.

field B are present, the net force on charge q The magnetic forces may change the

moving with the velocity v in direction of motion of a charged particle but

F = q[ E +( v × B )] --- (10.3) they can never affect the speed.

= q E +q( v × B ) = F e + Fm --- (10.4) Interestingly, Eq. (10.2) leads to the

Justification for this law can be found in definition ofunits of B . From Eq. (10.2),
F = q |v×B | = qvB sin θ
, --- (10.5)
experiments such as the one described in Fig. where θ is the angle between v and B

10.1 (a) and (b). The force described in Fig. and is unit vector in the direction of force .

(10.4) is known as Lorentz force. Here F e is If the force F is 1 N acting on the

the force due to electric field and Fm is the charge of 1 C moving with a speed of 1m s-1
force due to magnetic field.
perpendicular to B , then we can define the

There are interesting consequences of the unit o∴D..f.iuBBmn.e=itnoqsFifvoBnailslyCN, ..ms .

Lorentz force law. 
(i) If the velocity v of a charged particle is

parallel to the magnetic field B , the magnetic

force is zero. [B] = [F/qv]

231

This SI unit is called tesla (T) 10.3 Cyclotron Motion:
1 T = 104 gauss. Gauss is not an SI unit, In a magnetic field, a charged particle
but is used as a convenient unit.
typically undergoes circular motion. Figure
Can you recall? 10.5 shows a uniform magnetic field directed
perpendicularly into the plane of the paper
Electromagnetic crane: How does it work? (parallel to the -ve z axis).

Do you know? y

Magnetic Resonance Imaging (MRI) R
technique used for medical imaging
requires a magnetic field with a strength of BF x
1.5 T and even upto 7 T. Nuclear Magnetic
Resonance experiments require a magnetic Fig. 10.5: Charged
field upto 14 T. Such high magnetic fields particle moving in
can be produced using superconducting coil a magnetic field.
electromagnet. On the other hand, Earth’s
magnetic field on the surface of the Earth is Figure 10.5 shows a particle with charge
about 3.6 ×10-5 T = 0.36 gauss.
q moving with a speed v, and a uniform
Example 10.1: A charged particle travels
with a velocity v through a uniform magnetic field B is directed into the plane

magnetic field B as shown in the following of the paper. According to the Lorentz force
figure, in three different situations. What is
the direction of the magnetic force Fm due law, the magnetic force on the particle will
to the magnetic field, on the particle?
act towards the centre of a circle of radius R,
B
and this force will provide centripetal force to

sustain a uniform circular motion.

Thus

qvB = mv 2 --- (10.6)
R

∴mv = p = qBR --- (10.7)

Equation (10.7) represents what is known

as cyclotron formula. It describes the circular

motion of a charged particle in a particle

B accelerator, the cyclotron.

Fig. (a) Fig. (b)

Do you know?

B Let us look at a charged particle which
is moving in a circle with a constant speed.
Fig. (c) This is uniform circular motion that you
have studied earlier. Thus, there must be
Solution: In Fig. (a), the direction of the a net force acting on the particle, directed
vector v × B will be in the positive y towards the centre of the circle. As the speed
is constant, the force also must be constant,
direction. Hence Fm will be in the positive always perpendicular to the velocity of the
y direction. In Fig. (b) v × B will be in the particle at any given instant of time. Such a
sdairmecetidoinre. cHtieonnc.eItnheFifgo.rc(ec) Fvm force is provided by the uniform magnetic
positive x will
be in the and field B perpendicular to the plane of the
circle along which the charged particle
B are antiparallel, the angle between them moves.

is 180° and because sin 180° = 0, Fm will be

equal to zero.

232

Remember this accelerates due to the potential difference
between the two Ds and again performs
Field penetrating into the paper is semicircular motion in the other D. Thus the
represented as ⊗, while that coming out of ion is acted upon by the electric field every
the paper is shown by . time it moves from one D to the other D. As the
electric field is alternating, its sign is changed
10.3.1 Cyclotron Accelerator: in accordance with the circular motion of the
Particle accelerators have played a key ion. Hence the ion is always accelerated, its
energy increases and the radius of its circular
role in providing high energy (MeV to GeV) path also increases, making the entire path a
particle beams useful in studying particle- spiral (See Fig. 10.6).
matter interactions and some of these are also
useful in medical treatment of certain tumors/ Fig 10.6: Schematic diagram of a Cyclotron
diseases.
with the two Ds. A uniform magnetic field B is
The Cyclotron is a charged particle perpendicular to the plane of the paper, coming
accelerator, accelerating charged particles to out. The ions are injected into the D at point P.
high energies. It was invented by Lawrence An alternating voltage is supplied to the Ds. The
and Livingston in the year 1934 for the purpose entire assembly is placed in a vacuum chamber.
of studying nuclear structure.
Consider an ion source placed at P. An
Both electric as well as magnetic fields
are used in a Cyclotron, in combination. These ion moves in a semi circular path in one of the
are applied in directions perpendicular to each
other and hence they are called crossed fields. Ds and reaches the gap between the two Ds
The magnetic field puts the particle (ion) into
circular path and a high frequency electric in a time interval T/2, T being the period of
field accelerates it. Frequency of revolution
of a charged particle is independent of its a full revolution. Using the Cyclotron formula
energy, in a magnetic field. This fact is used
in this machine. Cyclotron consists of two Eq. (10.7),
semicircular disc-like metal chambers, D1 and
D2, called the dees (Ds). Figure 10.6 shows a mv = qBR,
schematic diagram of a cyclotron. A uniform
magnetic field B is applied perpendicular where q is the change on the ion.
to plane of the Ds. This magnetic field is 2S R m2S R
produced using an electromagnet producing a ?T v qBR
field upto 1.5 T. An alternating voltage upto
10000 V at high frequency, 10 MHz (fa), is 2S m --- (10.8)
applied between the two Ds. Positive ions are qB
produced by a gas ionizing source kept at the The frequency of revolution (Cyclotron
point O in between the two Ds. The electric
field provides acceleration to the charged frequency) is
particle (ion).

Once the ion in emitted, it accelerates due
to the negative voltage of a D and performs a
semi circular motion within the D. Whenever
the ion moves from one D to the other D, it

233

fc 1 qB 
T 2S m --- (10.9) Fm = v// × B =v.B sin (0°) = 0 --- (10.11)
Thus, v// will not be affected and the
The frequency of the applied voltage (fa) particle will move along the direction of B .

between the two Ds is adjusted so that polarity At the same time the perpendicular component
of the velocity ( v ⊥) leads to circular motion
of the two Ds is reversed as the ion arrives at
as stated above. As a result, the particle moves
the gap after completing one semi circle. This parallel to the field B while moving along a
circular path perpendicular to B . Thus the
condition fa = fc is the resonance condition. path becomes a helix (Fig. 10.7).
The ions do not experience any electric
Do you know?
field while they travel within the D. Their
Particle accelerators are important
kinetic energy increases by eV every time they for a variety of research purposes. Large
accelerators are used in particle research.
cross over from one D to the other. Here V is There have been several accelerators
in India since 1953. The Department of
the voltage difference across the gap. The ions Atomic Energy (DAE), Govt. of India, had
taken initiative in setting up accelerators
move in circular path with successively larger for research. Apart from ion accelerators,
the DAE has developed and commissioned
and larger radius to a maximum radius at a 2 GeV electron accelerator which is a
radiation source for research in science. This
which they are deflected by a magnetic field so accelerator, 'Synchrotron', is fully functional
at Raja Ramanna Centre for Advanced
that they can be extracted through an exit slit. Technology, Indore. An electron accelerator,
Microtron with electron energy 8-10 MeV
From Eq. (10.7), is functioning at Physics Department,
qBRexit Savitribai Phule Pune University, Pune.
v = m ,
Internet my friend
where Rexit is the radius of the path at the
(i) Existing and upcoming particle
exit. accelerators in India http://www.
researchgate.net
The kinetic energy of the ions/ protons
(ii) Search the internet for particle
will be accelerators and get more information.

1 q 2 B 2 R2 10.5 Magnetic Force on a Wire Carrying a
K.E. = 2 exit Current:
mv2 = --- (10.10)
2m We have seen earlier the Lorentz force
law (Eq. (10.4)). From this equation, we can
Thus the final energy is proportional to the obtain the force on a current carrying wire.
(i) Straight wire:
square of the radius of the outermost circular
Consider a straight wire of length L as
path (Rexit). shown in Fig. 10.8. An external magnetic field
10.4 Helical Motion: B is applied perpendicular to the wire, coming

So far it has been assumed that the charged

particle moves in a plane perpendicular to

magnetic field B . If such a particle has

some component of velocity parallel to B ,
( v// ) then itleads to helical motion. Since a
component v// is parallel to B , the magnetic
force Fm will be:

z
y

B
x

Fig. 10.7: Helical Motion of a charged particle
in a magnetic field B .

234

out of the plane of the paper. Let a current extended to a wire of arbitrary shape as shown

I flow through the wire under an applied in Fig. 10.9. I
potential difference. If v d is the drift velocity

of conduction electrons in the part of length

L of the wire, the charge q flowing across the

plane pp in time t will be

q=It --- (10.12) Fm
q = IL
Fig. 10.9: Wire with arbitrary shape.
vd Consider a segment of infinitesimal length

L dl along the wire. If I in the current flowing,

using Eq. (10.14), the magnetic force due to

I Fm perpendicular magnetic field B (coming out
e Vd
of the plane of the paper) is given by
d Fm = I d l × B --- (10.15)

The force on the total length of wire is

³ ³thus Fm
Fig. 10.8Electrons in the wire having drift d Fm I dl u B --- (10.16)
velocity vd experience a magnetic force
Fm If B is uniform over the whole wire,

upwards as the applied magnetic field lines ³Fm I ª¬ dl º¼ u B --- (10.17)

come out of the plane of the paper.

The magnetic force Fm on this charge, Example 10.2: A particle of charge q
according to Eq. (10.2), due to the applied follows a trajectory as shown in the figure.
magnetic field B is given by Obtain the type of the charge (positive or
negatively charged). Obtain the momentum
Fm q(vd u B) p of the particle in terms of B, L, s, q, s being
the distance travelled by the particle.
IL B˜ vd sin T n Particle trajectory: A uniform magnetic
vd field B is applied in the region pp,
perpendicular to the plane of the paper,
IL.B.sin 90qn , coming out of the plane of the paper.

where is a unit vector perpendicular to
both B and vd , in the direction of Fm

Fm = ILB --- (10.13)

This is, therefore, the magnetic force

acting on the portion of the straight wire

having length L.

If B is not perpendicular to the wire, then

the above Eq. (10.13) takes the form

Fm = I L × B , --- (10.14)

where L is the length vector directed

along the portion of the wire of length L. Solution: B is coming out of the paper.
Since the particle moves upwards, there
(ii) Arbitrarily shaped wire: must be a force in that direction. The
velocity is in the positive x direction.
In the previous section we considered

a straight wire. Equation (10.14) can be

235

 in upward direction. Calculate the current
∴ v × B is in -ve y direction. As the force
I in the loop for which the magnetic force
is in +y direction, i.e., opposite, the charge
would be exactly balanced by the force on
must be negative. According to Eq (10.5),
mass m due to gravity.
Force = Bqv in the y direction.
Bqv Solution: The current I in the loop with

∴ acceleration = m , its part in the magnetic field B causes an

where m is the mass of the particle. upward force Fm in the horizontal part of
the loop, given by
Using Newton’s equation of motion, the

distance travelled in the y direction is given Fm = IBa,
where a is the length of one arm of the
by 1
s = ut + 2 a t2 loop.

1 Bqv This force is balanced by the force due
= 0 + 2 m t2 as the initial
velocity in the y direction is zero. But in the to gravity.

∴ FIm==mBIagBa = mg
∴
same time t, the particle travels the distance

L along vt.he x direction, with uniform For this current, the wire loop will
velocity
hang in air.
∴ L = v.t

∴ s = 1 BqL2 1 Bq
2 mv 2 s
∴ momentum p = mv = L2
a

10.6 Force on a Closed Circuit in a Magnetic

Field B :
Equation (10.17) can be extended to a closed

wire circuit C

Fm I dlu B
³ C --- (10.18) 10.7 Torque on a Current Loop:
It will be very interesting to apply the
Here, the integral is over the closed circuit C.
results of the above sections to a current
For uniform B , carrying loop of a wire. You have learnt about
an electric motor in Xth Std. An electric motor
Fm I ª ³ d l º u B --- (10.19) works on the principle you have studied in the
« » preceding sections, i.e., the magnetic force on
¬C ¼ a current carrying wire due to a magnetic field.
Figure 10.10 shows a current carrying loop
The term in the bracket in Eq. (10.19) is the (abcd) in a uniform magnetic field. There will,
therefore, be the magnetic forces Fm acting in
sum of vectors along a closed circuit. Hence it opposite directions on the segments of the loop
ab and cd. This results into rotation of the loop
must be zero. about its central axis.

∴ Fm = 0 ( B uniform) --- (10.20) Without going into the details of contact
carbon brushes and external circuit, we can
Example 10.3: Consider a square loop of visualize the rotating action of a motor.
wire loaded with a glass bulb of mass m
hanging vertically, suspended in air with
its one part in a uniform magnetic field B
with its direction coming out of the plane of
the paper (). Due to the current I flowing
through the loop, there is a magnetic force

236

Fm Fm

Fm (90- )
l2 B

Fig. 10.10: A current loop in a magnetic field: F3
principle of a motor. Fig. 10.12 (b): Side view of the loop abcd at an

The current carrying wire loop is of angle θ.
Now we can calculate the net force and the
rectangular shape and is placed in the uniform
net torque on the loop in a situation depicted in
magnetic field in such a way that the segments
Fig. 10.12 (a) and (b). Let us obtain the forces
ab and cd of the loop are perpendicular to the
field B . We can use the right hand rule (Fig. on all sides of the loop. The force F 4 on side

10.11) to find out the direction of the magnetic 4 (bc) will be

force Fm . Let the pointing finger of the right F 4 = Il2 B sin (90-θ) --- (10.21)
hand show the direction of the current, let
The force F 2 on side 2 (da) will be equal
the middle finger show the direction of the
and opposite to F 4 and both act along the
magnetic field B , then the stretched thumb
shows the direction of the force. same line. Thus, F 2 and F 4 will cancel out
each other.
Let us now look at the action of rotation
The magnitudes of forces F1 and F 3 on
in detail. For this purpose, consider Fig. 10.12
sides 3 (ab) and 1 (cd) will be Il1 B sin 90° i.e.,
a, showing the rectangular loop abcd placed in Il1 B. These two forces do not act along the
same line and hence they produce a net torque.
a uniform magnetic field B such that the sides
ab and cd are perpendicular to the magnetic This torque results into rotation of the loop so

field B but the sides bc and da are not. that the loop is perpendicular to the direction of

B , the magnetic field. The moment arm is
1
2 (l2 sin θ ) about the central axis of the loop.

The torque τ due to forces F1 and F 3 will then

be 1 1
2 2
τ = (Il1 B l2 sin θ ) + (Il1 B l2 sin θ )

Fig. 10.11: The right hand rule.  = Il1 l2 B sin θ --- (10.22)

If the current carrying loop is made up of

multiple turns N, in the form of a flat coil, the

total torque will be

τ' = Nτ = N I l1 l2 B sin θ
τ' = (NIA)B sin θ --- (10.23)
A = l1l2
Here A is the area enclosed by the coil.

The above equation holds good for all flat

Fig. 10.12 (a): Loop abcd placed in a uniform (planar) coils irrespective of their shape, in a
magnetic field emerging out of the paper. uniform magnetic field.
Electric connections are not shown.

237

Can you recall?

How does the coil in a motor rotate by a full d
rotation? In a motor, we require continuous
rotation of the current carrying coil. As the Fig. 10.13: Moving coil galvanometer.
plane of the coil tends to become parallel
Larger the current is, larger is the deflection
to the magnetic field B , the current in the
coil is reversed externally. Referring to and larger is the torque due to the spring. If the
Fig. 10.10, the segment ab occupies the
position cd. At this position of rotation, the deflection is φ, the restoring torque due to the
current is reversed. Instead of from b to a,
it flows from a to b, force Fm continues spring is equal to K φ where K is the torsional
to act in the same direction so that the
torque continues to rotate the coil. The constant of the spring.
reversal of the current is achieved by using
a commutator which connects the wires Thus, K φ = NIAB, = § NAB · I --- (10.24)
of the power supply to the coil via carbon and the deflection φ ¨© K ¹¸
brush contacts.
Thus the deflection φ is proportional to
10.7.1 Moving Coil Galvanometer:
A current in a circuit or a voltage of a the current I. Modern instruments use digital

battery can be measured in terms of a torque ammeters and voltmeters and do not use such
exerted by a magnetic field on a current
carrying coil. Analog voltmeters and ammeters a moving coil galvanometer.
work on this principle. Figure 10.13 shows a
cross sectional diagram of a galvanometer. 10.8 Magnetic Dipole Moment:

It consists of a coil of several turns In the preceding section, we have dealt
mounted (suspended or pivoted) in such a way
that it can freely rotate about a fixed axis, in with a current carrying coil. This current
a radial uniform magnetic field. A soft iron
cylindrical core makes the field radial and carrying coil can be described with a vector
strong. The coil rotates due to a torque acting µ , its magnetic dipole moment. If n is a unit
on it as the current flows through it. This torque
is given by (Eq. 10.23) vector normal to the plane of the coil, the
direction of µ is the direction of n shown
τ = N I A.B, where A is the area of the
coil, B the strength of the magnetic field, N the in Fig. 10.12 (b). We can then define the
number of turns of the coil and I the current in
the coil. Here, sin θ = 1 as the field is radial magnitude of µ as
(plane of the coil will always be parallel to
the field). However, this torque is counter µ = NIA, --- (10.25)
balanced by a torque due to a spring fitted as
shown in the Fig. 10.13. where N is the number of turns of the coil, I

This counter torque balances the magnetic the current passing through the coil, A the area
torque, so that a fixed steady current I in the
coil produces a steady angular deflection φ . enclosed by each turn of the coil.

If held in uniform magnetic field B , the

torque responsible for the rotation of the coil,

according to Eq. (10.23) will be

τ = µB sinθ ,

238

θ being anangle between µ (i.e., ) and B . Example 10.4: A circular coil of conducting
∴τ = µ ×B --- (10.26) wire has 500 turns and an area 1.26×10-4 m2
is enclosed by the coil. A current 100 µA
You have learnt in XIth Std. about the torque on is passed through the coil. Calculate the
magnetic moment of the coil.
an electric dipole exerted by an electric field, Solution:

E.  --- (10.27) µ = NIA
∴τ = P × E = 500 × 100 × 10-6 × 1.26 × 10-4 Am2
= 630 × 10-8 = 6.3 × 10-6 Am2 or J/T.
Here P is the electric dipole moment.

The two expression Eq. (10.26) and Eq. (10.27)

are analogous to each other.

µB µB 10.10 Magnetic Field due to a Current :
Biot-Savart Law:
Case (i) Case (ii)
In sections 10.1 and 10.2, we have seen
Fig. 10.14: Minimum and maximum magnetic that magnetic field is produced by a current
carrying wire. Can we calculate this magnetic
potential energy of a magnetic dipole µ in a field?

magnetic field B .

10.9 Magnetic Potential Energy of a Dipole:  dl I

A magnetic dipole freely suspended in a I r dB

magnetic field possesses magnetic potential

energy because of its orientation in the field.

You have learnt about an electric dipole

in Chapter 8. Electrical Potential energy is Figure 10.15: A current carrying wire of
arbitrary shape, carrying a current I. The current
associated with an electric dipole on account in the differential length element dl produces

of its orientation in an electric field. It has been differential magnetic field d B at a point P at a
distance r from the element dl. The  indicates
shown that the potential energy U of an electric
that d B is directed into the plane of the paper.
dipole P in an electric field E is given by
U = - P . E
--- (10.28) Figure 10.15 shows an arbitrarily shaped

Analogously, the magnetic potential wire carrying a current I. dl is a length element
along the wire. The current in this element is
energy of a magnetic dipole µ in a magnetic in the direction of the length vector dl . Let
us calculate the differential field d B at the
field B is given by --- (10.29) point P, produced by the current I through the
U = - µ . B length element dl. Net magnetic field at the
point P can be obtained by superimposition
= - µB.cos θ , --- (10.30) of magnetic fields d B at that point due to
different length elements along the wire. This
where θ is the angle between µ and B .

Case (i) : If θ = 0, U = - µ B.cos(0°) = - µB

This is the minimum potential energy of a

magnetic dipole in a magnetic field i.e., when

 and  are parallel to each other. can be done by integrating i.e., summing up
of magnetic fields d B from these length
Case (ii) : If  = 180°, U = - µ.B.cos (180°)

= µB. elements. Experimentally, the magnetic fields

This is the maximum potential energy d B produced by current I in the length element
d l is
of a magnetic dipole in a magnetic field, i.e.,

when  and  are antiparallel to each other.

239

dB P0 Idl sinT --- (10.31) Here, the direction of d B is given by the
4S r 2 cross product d l × r (see Eq. (10.34)), hence
into the plane of the paper.
Here,θ is the angle between the directions (a) We now calculate the magnitude of
of dl and r . µ0 is called permeability constant
given by the magnetic field produced at P by all
current length elements in the upper half
µ = 4π × 10-7 T. m/A --- (10.32) of the infinitely long wire. This we do by
0 --- (10.33) integrating Eq. (10.35) from o to ∞.
(b) Let us now calculate the magnitude of the
≈ 1.26 × 10-6 T. m/A magnetic field produced at P by a current
length element in the lower half of the
The direction of d B is dictated by the wire. By symmetry, this magnitude is the
cross product dl × r . Vectorially, same as that from the upper half of the
wire. The direction of this field is also the
dB P0 Idl u r --- (10.24) same as from the upper half of the wire,
4S r3 going into the plane of the paper.
Adding both the contributions (a) and (b),
Equation (10.31) and Eq. (10.34) are the total magnetic field B at point P is

known as the Biot and Savart law. This inverse

square law is experimentally deduced. It may

be noted that this is still inverse square law
as r appears in the numerator and r3 in the

denominator. Using the Biot-Savart law, we

can calculate the magnetic field produced by

various distributions of currents as discussed

below: f 2 P0 f Idl sinT
4S 0 r2
2 dB

0
(i) Current in a straight, long wire: ³ ³B --- (10.36)

You are aware of the right hand thumb But r l 2 R2
R
rule which gives the direction of the magnetic R
and sinθ = sin (π-θ )= r --(10.37)
field produced by a current flowing in a wire.
l2 R2
Figure 10.16 shows a long wire of length l.

We want to calculate magnetic field B at a P0 I f Rdl
2S 0 l2 R2 l2 R2
point P which is at a perpendicular distance ³?B

R from the wire. Let us consider a current P0 I f dl
length element (the infinitesimal length d l of l2 R2
R
the wire, multiplied by the current I passing ³ 2S 0 3/ 2
through it) I. dl situated at a distance r from

the point P. Using Eq. (10.31), the magnetic ?B P0 I R 1 P0 I --- (10.38)
2S R2 2S R
field d B produced at P due to the current

length element I. dl becomes From Eq. (10.36), this is the magnetic

dB P0 Idl sinT --- (10.35) field at a point P at a perpendicular distance
4S r2
R from the infinitely straight wire. This is due

to both the upper semi-infinite part and the
θ lower semi-infinite part of the wire. Thus, the

r magnetic field B due to semi-infinite straight

R wire is P0 I
I 4S R
?B --- (10.39)
Fig. 10.16: The magnetic field d B at P going
into the plane of the paper, due to current I In Eq. (10.38) and Eq. (10.39), the field
through the wire.
is inversely proportional to the distance from

the wire.

240

f dl For infinitely long wires, this force will be

0 l 2 R2 ,3/2
³ To solve I infinite!

we substitute l = R tan θ ; dl = r sec2θ dθ Force per unit length of the wire will be

Now the limits of the integral also change. Fc P0 I1I2 --- (10.42)
2S d
l = 0, tan θ = 0 ∴ θ = 0
If the currents I1 and I2 are antiparallel,
l = ∞, tan θ = ∞ ∴ θ = π/2 the force will be repulsive.

³? IS /2 R sec2 T dT Let us consider a section of length L of
0 R3 (tan2 T 1)3/2
the wire 2. The force on this section due to the

³1 S /2 (cos2 T )3/2 sec2 T dT current in wire 1 is given by

R2 0 (sin2 T cos2 T )3/2 F = I2 B.L --- (10.43)
P0 I1I2 L --- (10.44)
³1 S /2 cosT dT 2S d F21
0
R2 We will denote this force by F21 i.e., the
force on a section of length L of wire 2 due to
> @1 sinT S /2 1 [1 0] 1
0 R2 R2
R2 the current in wire 1. Similarly, the force on

10.11 Force of Attraction between two Long a section of the same length L of wire 1 will

Parallel Wires: experience a force due to the current in wire 2.

As an application of the result obtained This force we denote as F , which is
12
in the last section, let us obtain the force of
equal and opposite to F21
attraction between two long, parallel wires
∴ F21 = - F12 --- (10.44 A)
separated by a distance d (Fig. 10.17). Let the
The force of attraction per unit length is
currents in the two wires be I1 and I2.
The magnetic field at the second wire due then, from Eq. (10.44),

to the current I1 in the first one, according to F P0 I1I2 --- (10.45)
L 2S d
Eq. (10.38),
P0 I1
B 2S d --- (10.40) If the currents I and I are flowing in
12

opposite directions, then there is a force of

I1 repulsion on the sector of length L of each of

the wires. The magnitude of the repulsive force

per unit length of the wire is also given by

F P0 I1I2 --- (10.46)
L 2S d
I2
We can summarize these result as: Parallel
Fig. 10.17 : Two long parallel wires, distance d
currents attract, antiparallel currents repel.
apart.
By the right hand rule, the direction of this The ampere: Definition of the unit of
electrical current ampere, was adopted a
field is into the plane of the paper. We now few decades ago. Consider two parallel
conducting wires having infinite length,
apply the Lorentz Force law. Accordingly, the have a separation of 1 m, and are placed
in vacuum. The constant current through
force on the wire 2, because of the current I these wires producing a force on each other
2 of magnitude 2×10-7 N per meter of their
length, is 1 ampere (A).
and the magnetic field B due to current in wire

1, is given by (Eq. 10.13).

³F I 2 § P0 I1 · dl --- (10.41)
©¨ 2S d ¹¸

The direction of this force is towards wire

1, i.e., it will be attractive force.

241