Zero Potential Due to a System of Two Point Charge.
If both charges are like then resultant potential is not zero at any finite point because potentials due to like
charges will have same sign and can therefore never add up to zero. Such a point can be therefore obtained
only at infinity.
If the charges are unequal and unlike then all such points where resultant potential is zero lies on a closed
curve, but we are interested only in those points where potential is zero along the line joining the two charges.
Two such points exist, one lies inside and one lies outside the charges on the line joining the charges.
Both the above points lie nearer the smaller charge, as potential created by the charge larger in magnitude will
become equal to the potential created by smaller charge at the desired point at larger distance from it.
I. For internal point : II. For External point :
P P Q1 Q2
Q1
Q2
x1 x2
x1
x x
(It is assumed that Q1 < Q2 ). Q1 = Q2 ⇒ x1 = x 1)
x1
Q1 Q2 x (x + x1) (Q2 /Q1 −
x1
= (x − x1) ⇒ x1 = (Q2 /Q1 + 1)
N Examples based on oscillation of charge and neutral
Example: 69 Two similar charges of +Q as shown in figure are placed at points A and B. – q charge is placed at
point C midway between A and B. – q charge will oscillate if
(a) It is moved towards A D B
A
(b) It is moved towards B +Q C –q +Q
(c) It is moved along CD
(d) Distance between A and B is reduced
Solution: (c) When – q charge displaced along CD, a restoring force act on it which causes oscillation.
Example: 70 Two point charges (+Q) and (– 2Q) are fixed on the X-axis at positions a and 2a from origin
respectively. At what position on the axis, the resultant electric field is zero
(a) Only x = 2 a (b) Only x = − 2 a (c) Both x = ± 2 a (d) x = 3a only
2
Solution: (b) Let the electric field is zero at a point P distance d from the charge +Q so at P. k.Q + k(−2Q) =0
d2 (a + d)2
⇒ 12 ⇒d= a P +Q – 2Q
d 2 = (a + d)2 ( 2 − 1) a
x
d 2a
Since d > a i.e. point P must lies on negative x-axis as shown at a distance x from origin hence
x = d − a = a − a = 2a . Actually P lies on negative x-axis so x = − 2 a .
2 −1
Example: 71 Two charges 9e and 3e are placed at a distance r. The distance of the point where the electric field
Solution: (b)
intensity will be zero is [MP PMT 1989]
( )(a) r from 9e charge (b) r from 9e charge
3 +1 1+ 1 3
( )(c) r from 3e charge (d) r from 3e charge
1− 3 1+ 1 3
Suppose neutral point is obtained at a distance x1 from charge 9e and x 2 from charge 3e
By using x1 x =r r 9e N 3e
1 +
= Q2 3e = 1 x1 x2
1+ Q1 9e 3 r
1+
Example: 72 Two point charges – Q and 2Q are separated by a distance R, neutral point will be obtained at
(a) A distance of R from – Q charge and lies between the charges.
( 2 − 1)
(b) A distance of R from – Q charge on the left side of it
( 2 − 1)
(c) A distance of R from 2Q charge on the right side of it
( 2 − 1)
(d) A point on the line which passes perpendicularly through the centre of the line joining – Q and 2Q
charge.
Solution: (b) As already we discussed neutral point will be obtained on the side of charge which is smaller in
magnitude i.e. it will obtained on the left side of – Q charge and at a distance.
l= R ⇒l= R
( 2 − 1)
2Q −1
Q
Example: 73 A charge of + 4µC is kept at a distance of 50 cm from a charge of – 6µC. Find the two points where the
Solution: (a) potential is zero
(a) Internal point lies at a distance of 20 cm from 4µC charge and external point lies at a distance of 100
cm from 4µC charge.
(b) Internal point lies at a distance of 30 cm from 4µC charge and external point lies at a distance of 100
cm from 4µC charge
(c) Potential is zero only at 20 cm from 4µC charge between the two charges
(d) Potential is zero only at 20 cm from – 6µC charge between the two charges
For internal point X, x1 = x = 50 = 20cm and for external point Y,
Q2 + 1 6 +1
Q1 4
x1 = x = 50 = 100cm Y X – 6µC
4µC
Q2 − 1 6 − 1
Q1 4 20cm
100cm 50cm
Tricky example: 9
Two equal negative charges – q are fixed at points (0, a) and (0, – a) on the y-axis. A positive charge Q
is released from rest at the point (2a, 0) on the x-axis. The charge Q will
[IIT-JEE 1984, Bihar MEE 1995, MP PMT 1996]
(a) Execute simple harmonic motion about the origin
(b) Move to the origin and remains at rest
(c) Move to infinity
(d) Execute oscillatory but not simple harmonic motion.
Solution: (d) By symmetry of problem the components of force on Q due to charges at A and B along y-axis will
cancel each other while along x-axis will add up and will be along CO. Under the action of this force
charge Q will move towards O. If at any time charge Q is at a distance x from O.
F ⇒ 2F cosθ = 2 1 −qQ × (a 2 + x 2 A–q
4πε (a 2 + x 2 ) x 2 )1 a
0
O
1 2qQx
4πε 0 + x2 3 a
( )i.e.,F= − . a2 2 –q θ
x
B Q C
2a
As the restoring force F is not linear, motion will be oscillatory (with
amplitude 2a) but not simple harmonic.
Electric Potential Energy.
(1) Potential energy of a charge : Work done in bringing the given charge from infinity to a point in
the electric field is known as potential energy of the charge. Potential can also be written as potential energy
per unit charge. i.e. V = W = U .
Q Q
(2) Potential energy of a system of two charges : Since work done in bringing charge Q2 from ∞ to
point B is W = Q2VB, where VB is potential of point B due to charge Q1 i.e. VB = 1 Q1
4πε 0 r
So, W = U2 = 1 . Q1Q2 Q1 Q2
4πε0 r A rB
This is the potential energy of charge Q2, similarly potential energy of charge Q1 will be U1 = 1 . Q1Q2
4πε0 r
Hence potential energy of Q1 = Potential energy of Q2 = potential energy of system U = k Q1Q2 (in C.G.S.
r
U = Q1Q2 )
r
Note : ≅Electric potential energy is a scalar quantity so in the above formula take sign of Q1 and Q2.
(3) Potential energy of a system of n charges : In a system of n charges electric potential energy is
calculated for each pair and then all energies so obtained are added algebraically. i.e.
U = 1 Q1Q2 + Q2Q3 + ......... and in case of continuous distribution of charge. As dU = dQ.V ⇒
4πε0 r12 r23
U = ∫ V dQ Q1
e.g. Electric potential energy for a system of three charges r12 r31
Q2 r23 Q3
Potential energy = 1 Q1Q2 + Q2Q3 + Q3Q1
4πε0 r12 r23 r31
While potential energy of any of the charge say Q1 is 1 Q1Q2 + Q3Q1
4πε0 r12 r31
Note : ≅For the expression of total potential energy of a system of n charges consider n(n − 1) number of
2
pair of charges.
(4) Electron volt (eV) : It is the smallest practical unit of energy used in atomic and nuclear physics. As
electron volt is defined as “the energy acquired by a particle having one quantum of charge 1e when
accelerated by 1volt” i.e. 1eV = 1.6 × 10−19 C × 1J = 1.6 × 10−19 J = 1.6 × 10–12 erg
C
Energy acquired by a charged particle in eV when it is accelerated by V volt is E = (charge in quanta) × (p.d. in
volt)
Commonly asked examples :
S.No. Charge Accelerated by Gain in K.E.
p.d.
(i) Proton 5 × 104 V K = e × 5 × 104 V = 5 × 104 eV = 8 × 10–15 J [JIPMER 1999]
(ii) Electron 100 V K = e × 100 V = 100 eV = 1.6 × 10–17 J [MP PMT 2000; AFMC 1999]
(iii) Proton 1V K = e × 1 V = 1 eV = 1.6 × 10–19 J [CBSE 1999]
(iv) 0.5 C 2000 V K = 0.5 × 2000 = 1000 J [JIPMER 2002]
(v) α-particle 106 V K = (2e) × 106 V = 2 MeV [MP PET/PMT 1998]
(5) Electric potential energy of a uniformly charged sphere : Consider a uniformly charged sphere
of radius R having a total charge Q. The electric potential energy of this sphere is equal to the work done in
bringing the charges from infinity to assemble the sphere.
U = 3Q2
20πε 0 R
(6) Electric potential energy of a uniformly charged thin spherical shell :
U = Q2
8πε 0 R
(7) Energy density : The energy stored per unit volume around a point in an electric field is given by
Ue = U = 1 ε 0 E 2 . If in place of vacuum some medium is present then Ue = 1 ε 0ε r E 2
Volume 2 2
Concepts
Electric potential energy is not localised but is distributed all over the field
If a charge moves from one position to another position in an electric field so it’s potential energy change and work done in this
changing is W = U f − Ui
If two similar charge comes closer potential energy of system increases while if two dissimilar charge comes closer potential
energy of system decreases.
Q q Examples based on electric potential energy
U=0
∞
Example: 74 If the distance of separation between two charges is increased, the electrical potential energy of the
system [AMU 1998]
(a) May increases or decrease (b) Decreases
(c) Increase (d) Remain the same
Solution: (a) Since we know potential energy U = k. Q1Q2
r
As r increases, U decreases in magnitude. However depending upon the fact whether both charges are
similar or disimilar, U may increase or decrease.
Example: 75 Three particles, each having a charge of 10µC are placed at the corners of an equilateral triangle of side
Solution: (c)
10cm. The electrostatic potential energy of the system is (Given 1 = 9 × 109 N − m2 /C 2 )
4πε 0
[AMU 1998]
(a) Zero (b) Infinite (c) 27 J (d) 100 J
Potential energy of the system, 10µC
U = 9 × 109 (10 × 10 −6 )2 × 3 = 27 J
10 cm 10 cm
0.1
10µC 10 cm 10µC
Example: 76 Three charges Q, +q and +q are placed at the vertices of a right-angled isosceles triangle as shown. The
net electrostatic energy of the configuration is zero if Q is equal to [IIT (Screening) 2000]
(a) − q Q
1+ 2
(b) − 2q +q +q
1+ 2 a
(c) – 2 q
(d) +q
Solution: (b) Potential energy of the configuration U = k. Qq + k.q 2 + k. Qq =0 ⇒ Q= − 2q
a a a2 2 +1
Example: 77 A charge 10 e.s.u. is placed at a distance of 2 cm from a charge 40 e.s.u. and 4 cm from another charge
of 20 e.s.u. The potential energy of the charge 10 e.s.u. is (in ergs)
(a) 87.5 (b) 112.5 (c) 150 (d) 250
Solution: (d) Potential energy of 10 e.s.u. charge is 10 esu
U = 10 × 40 + 10 × 20 = 250 erg. 2 cm 4 cm
2 4
40 esu
20 esu
Example: 78 In figure are shown charges q1 = + 2 × 10–8 C and q2 = – 0.4 × 10–8 C. A charge q3 = 0.2 × 10–8 C in
moved along the arc of a circle from C to D. The potential energy of q3 [CPMT 1986]
q3
C
(a) Will increase approximately by 76% 80 cm
(b) Will decreases approximately by 76% AB D
(c) Will remain same q1 60 cm q2
(d) Will increases approximately by 12%
q3
Solution: (b) Initial potential energy of q3 Ui = q1q3 + q2q3 × 9 × 109 C
0.8 1
Final potential energy of q3 Uf = q1q3 + q2q3 × 9 × 109
0.8 0.2
80 cm
Change in potential energy = Uf – Ui
Now percentage change in potential energy = U f − Ui × 100 A B
ui q1
60 cm D
80 cm
q2
q2q3 1 − 1 × 100
0.2
= On putting the values ~– − 76%
q1 q2
q3 0.8 + 1
Tricky example: 10
Three charged particles are initially in position 1. They are free to move and they come in position 2
after some time. Let U1 and U2 be the electrostatics potential energies in position 1 and 2. Then
(a) U1 > U2 (b) U2 > U1
(c) U1 = U2 (d) U2 ≥ U1
Solution: (a) Particles move in a direction where potential energy of the system is decreased.
Motion of Charged Particle in an Electric Field.
(1) When charged particle initially at rest is placed in the uniform field :
Let a charge particle of mass m and charge Q be initially at rest in an electric field of strength E
+Q F = QE E A E
F=QE –Q S B
Fig. (A) Fig. (B)
(i) Force and acceleration : The force experienced by the charged particle is F = QE . Positive charge
experiences force in the direction of electric field while negative charge experiences force in the direction
opposite to the field. [Fig. (A)]
Acceleration produced by this force is a = F = QE
m m
Since the field E in constant the acceleration is constant, thus motion of the particle is uniformly
accelerated.
(ii) Velocity : Suppose at point A particle is at rest and in time t, it reaches the point B [Fig. (B)]
V = Potential difference between A and B; S = Separation between A and B
(a) By using v = u + at , v = 0 + Q E t , ⇒ v = QEt
m m
(b) By using v2 = u2 + 2as , v2 = 0 + 2× QE × s v2 = 2QV E = V ⇒ v= 2QV
m m s m
(iii) Momentum : Momentum p = mv, p = m× QEt = QEt or p = m × 2QV 2mQV
m m=
(iv) Kinetic energy : Kinetic energy gained by the particle in time t is
K= 1 mv 2 1 m (QEt) 2 Q2 E 2t 2 or K 1 m 2QV QV
2 2 m = 2m 2 m
= = × =
(2) When a charged particle enters with an initial velocity at right angle to the uniform field :
When charged particle enters perpendicularly in an electric field, it describe a parabolic path as shown
(i) Equation of trajectory : Throughout the motion particle has uniform velocity along x-axis and
horizontal displacement (x) is given by the equation x = ut
Since the motion of the particle is accelerated along y–axis, we will use equation of motion for uniform
acceleration to determine displacement y. From S = ut + 1 at 2
2
We have u=0 (along y-axis) so y = 1 at 2 Y
2
E
i.e., displacement along y-axis will increase rapidly with time (since y ∝ t 2 ) P(x, y)
From displacement along x-axis t = x u X
u
v
1 QE x 2
2 m u
So y = ; this is the equation of parabola which shows y ∝ x2
(ii) Velocity at any instant : At any instant t, vx =u and vy = QEt
m
So v =| v|= v 2 + vy2 = u2 + Q2 E 2t 2 vy
x m2
β
If β is the angle made by v with x-axis than tan β vy QEt . vx
vx mu
= =
Concepts
An electric field is completely characterized by two physical quantities Potential and Intensity. Force characteristic of the field is
intensity and work characteristic of the field is potential.
If a charge particle (say positive) is left free in an electric field, it experiences a force (F = QE) in the direction of electric field and
moves in the direction of electric field (which is desired by electric field), so its kinetic energy increases, potential energy
decreases, then work is done by the electric field and it is negative.
E Examples based on motion of
Q
Example: 79 An electron (mass = 9.1 × 10 −31 kg and charge = 1.6 × 10 −19 coul.) is sent in an electric field of
intensity 1 × 106 V / m. How long would it take for the electron, starting from rest, to attain one–tenth the
velocity of light
(a) 1.7 × 10 −12 sec (b) 1.7 × 10 −6 sec (c) 1.7 × 10 −8 sec (d) 1.7 × 10 −10 sec
Solution: (b) By using v= QEt ⇒ 1 × 3 × 10 8 = (1.6 × 10 −19 ) × 106 × t ⇒ t = 1.7 × 10 −10 sec.
m 10 9.1 × 10 −31
Example: 80 Two protons are placed 10 −10 m apart. If they are repelled, what will be the kinetic energy of each
proton at very large distance
(a) 23 × 10 −19 J (b) 11.5 × 10 −19 J (c) 2.56 × 10 −19 J (d) 2.56 × 10 −28 J
Solution: (d) Potential energy of the system when protons are separated by a distance of 10 −10 m is
U = 9 × 109 × (1.6 × 10 −19 )2 = 23 × 10 −19 J P+ p+
10 −10
According to law of conservation of energy at very larger distance, this energy is equally distributed in
both the protons as their kinetic energy hence K.E. of each proton will be 11.5 × 10 −19 J.
Example: 81 A particle A has a charge +q and particle B has charge +4q with each of them having the same mass m.
When allowed to fall from rest through the same electrical potential difference, the ratio of their speeds
vA will becomes [BHU 1995; MNR 1991]
vB
(a) 2 : 1 (b) 1 : 2 (c) 1 : 4 (d) 4 : 1
Solution: (b) We know that kinetic energy K = 1 mv2 = QV . Since, m and V are same so, v2 ∝ Q ⇒
2
vA = QA = q = 1 .
vB QB 4q 2
Example: 82 How much kinetic energy will be gained by an α − particle in going from a point at 70 V to another point
at 50 V [RPET 1996]
(a) 40 eV (b) 40 keV (c) 40 MeV (d) 0 eV
Solution: (a) Kinetic energy K = Q∆V ⇒ K = (2e)(70 − 50) V = 40 eV
Example: 83 A particle of mass 2g and charge 1µ C is held at a distance of 1 metre from a fixed charge of 1mC. If the
particle is released it will be repelled. The speed of the particle when it is at a distance of 10 metres from
the fixed charge is [CPMT 1989]
(a) 100 m/s (b) 90 m/s (c) 60 m/s (d) 45 m/s
Solution: (b) According to conservation of energy
1 mC 1 µC
A Moving charge B
Fixed charge 1m
10m
Energy of moving charge at A = Energy of moving charge at B
9 × 109 × 10−3 × 10−6 = 9 × 109 × 10−3 × 10−6 + 1 × (2 × 10−3 ) v 2
1 10 2
⇒ v2 = 8100 ⇒ v = 90 m/sec
Tricky example: 11
A mass of 1g carrying charge q falls through a potential difference V. The kinetic energy acquired by it is
E. When a mass of 2g carrying the charge q falls through a potential difference V. What will be the
kinetic energy acquired by it
(a) 0.25 E (b) 0.50 E (c) 0.75 E (d) E
Solution: (d) In electric field kinetic energy gain by the charged particle K = qV. Which depends charge and potential
difference applied but not on the mass of the charged particle.
Force on a Charged Conductor.
To find force on a charged conductor (due to repulsion of like charges) imagine a small part XY to be cut
and just separated from the rest of the conductor MLN. The field in the cavity due to the rest of the conductor
is E2, while field due to small part is E1. Then E2 E1
X E2
+ +++ Y
Inside the conductor E = E1 − E2 = 0 or E1 = E2 ++ M ++
+ + N
+ + + ++
σ E2 E1 + + ++
ε0
Outside the conductor E = E1 + E2 = + Inside + ++ +
L ++ E=0 +
+ +++ + ++
Thus E1 = E2 = σ (A) (B)
2ε 0
To find force, imagine charged part XY (having charge σ dA placed in the cavity MN having field E2).
Thus force dF = (σ dA)E2 or dF = σ2 dA . The force per unit area or electric pressure is dF = σ2
2ε 0 dA 2ε 0
The force is always outwards as (±σ )2 is positive i.e., whether charged positively or negatively, this force
will try to expand the charged body.
A soap bubble or rubber balloon expands on given charge to it (charge of any kind + or –).
Equilibrium of Charged Soap Bubble.
For a charged soap bubble of radius R and surface tension T and charge density σ . The pressure due to
surface tension 4 T and atmospheric pressure Pout act radially inwards and the electrical pressure (Pel ) acts
R
radially outward.
The total pressure inside the soap bubble Pin = Pout + 4T − σ2
R 2ε 0
Excess pressure inside the charged soap bubble Pin − Pout = Pexcess = 4T − σ2 . If air pressure inside and
R 2ε 0
outside are assumed equal then Pin = Pout i.e., Pexcess = 0 . So, 4T = σ2
R 2ε 0
This result give us the following formulae Pout Pout
(1) Radius of bubble R = 8 ε0 T air + PT + Pelec
σ2 PT air + Pin
(2) Surface tension T = σ 2R Pin +
8ε 0
(3) Total charge on the bubble Q = 8πR 2ε0TR Uncharged Charged
(4) Electric field intensity at the surface of the bubble E = 8T 32π kT
ε0R = R
(5) Electric potential at the surface V = 3πRTk = 8RT
ε0
Electrostatics
Electric Dipole.
(1) General information : System of two equal and opposite charges separated by a small fixed
distance is called a dipole.
Equatorial axis
AB
– q +q Dipole axis
2l
(i) Dipole axis : Line joining negative charge to positive charge of a dipole is called its axis. It may also
be termed as its longitudinal axis.
(ii) Equatorial axis : Perpendicular bisector of the dipole is called its equatorial or transverse axis as it is
perpendicular to length.
(iii) Dipole length : The distance between two charges is known as dipole length (L = 2l)
(iv) Dipole moment : It is a quantity which gives information about the strength of dipole. It ispa vector
quantity and is directed from negative charge to positive charge along the axis. It is denoted as and is
defined as the product of the magnitude of either of the charge and the dipole length.
i.e. p = q(2l)
Its S.I. unit is coulomb-metre or Debye (1 Debye = 3.3 × 10–30 C × m) and its dimensions are
M0L1T1A1.
Note : ≅A region surrounding a stationary electric dipole has electric field only.
≅ When a dielectric is placed in an electric field, its atoms or molecules are considered as tiny
dipoles.
+ –+
≅ Water (H2O), Chloroform (CHCl3), Ammonia (NH3), HCl, CO molecules are some example of
permanent electric dipole.
+ H+
O2– ––
+ H+
(2) Electric field and potential due to an electric dipole : It is better to understand electric dipole
with magnetic dipole.
S.No. Electric dipole Magnetic dipole
(i)
System of two equal and opposite charges separated System of two equal and opposite magnetic poles
by a small fixed distance. (Bar magnet) separated by a small fixed distance.
AB –m +m
–q +q SN
2l
p M
(ii) Electric dipole moment : p = q , directed from Magnetic dipole moment : M = , directed
(2l ) m(2l )
− q to +q. It’s S.I. unit is coulomb × meter or from S to N. It’s S.I. unit is ampere × meter2.
Debye.
(iii) Intensity of electric field Intensity of magnetic field
E e e E B e e B
Equatorial line g Equatorial line g
θ θ+α E a θ θ+α
2l a SN
–q +q 2l a Ba
p Axial line M Axial line
If a, e and g are three points on axial, equatorial and If a, e and g are three points on axial, equatorial and
general position at a distance r from the centre of general position at a distance r from the centre of
dipole dipole
on axial point Ea = 1 . 2p (directed from – q to +q) on axial point Ba = µ0 . 2M (directed from S to N)
4πε 0 r3 4π r3
on equatorial point Ee = 1 . p (directed from +q to –q) on equatorial point Be = µ0 . M (directed from N to S)
4πε0 r3 4π r3
on general point Ea = 1 . p (3 cos 2 θ + 1) on general point Ba = µ0 . M (3 cos 2 θ + 1)
4πε 0 r3 4π r3
Angle between – and p is 0o, and p is 180o, Angle between – and M is 0o, and M is
Ea Ee Ba Be
E and p is (θ + α) (where tan α = 1 tan θ ) 180o, B and M is (θ + α) (where tan α = 1 tan θ )
2 2
Electric Potential – At a Va = 1 . p , At e V = 0
4πε 0 r2
At g V = 1 . p cosθ
4πε 0 r2
(3) Dipole (electric/magnetic) in uniform field (electric/magnetic)
(i) Torque : If a dipole is placed in an uniform field such that dipole (i.e. p or M ) makes an angle θ
with direction of field then two equal and opposite force acting on dipole constitute a couple whose tendency is
to rotate the dipole hence a torque is developed in it and dipole tries to align it self in the direction of field.
Consider an electric dipole in placed in an uniform A magnetic dipole of magnetic moment M is placed in
electric field such that dipole (i.e. p ) makes an angle θ uniform magnetic field B by making an angle θ as shown
with the direction of electric field as shown
+q A F = qE N F = mB
θ θ
–q F = mB S
F = qE
B
(a) Net force on electric dipole Fnet = 0 (a) Net force on magnetic dipole Fnet = 0
(b) Produced torque τ = pE sinθ (τ = P × E ) (b) torque τ = MB sinθ (τ = M × B )
(ii) Work : From the above discussion it is clear that in an uniform electric/magnetic field dipole tries to
align itself in the direction of electric field (i.e. equilibrium position). To change it’s angular position some work
has to be done.
Suppose an electric/magnetic dipole is kept in an uniform electric/magnetic field by making an angle θ1
with the field, if it is again turn so that it makes an angle θ2 with the field, work done in this process is given by
the formula
+q +q E B
θ1 θ2 M
θ1 θ2
–q
–q
W = pE(cosθ1 − cosθ 2 ) W = MB(cosθ1 − cosθ 2 )
If θ1 = 0o and θ2 = θ then W = MB(1 – cosθ)
If θ1 = 0o and θ2 = θ i.e. initially dipole is kept along
the field then it turn through θ so work done
W = pE(1 − cosθ )
(iii) Potential energy : In case of a dipole (in a uniform field), potential energy of dipole is defined as work
done in rotating a dipole from a direction perpendicular to the field to the given direction i.e. if θ1 = 90o and θ2 = θ
then –
P E M B
P M
θ θ
W = U = pE (cos 90 − cosθ ) ⇒ U = – pE cosθ W = U = MB(cos 90 − cosθ ) ⇒ U = – MB cosθ
(iv) Equilibrium of dipole : We know that, for any equilibrium net torque and net force on a particle
(or system) should be zero.
We already discussed when a dipole is placed in an uniform electric/magnetic field net force on dipole is
always zero. But net torque will be zero only when θ = 0o or 180o
When θ = 0o i.e. dipole is placed along the electric field it is said to be in stable equilibrium, because after
turning it through a small angle, dipole tries to align itself again in the direction of electric field.
When θ = 180o i.e. dipole is placed opposite to electric field, it is said to be in unstable equilibrium.
p E B M B
E E p M B M
θ = 0o θ = 90O θ = 180o θ = 0o θ = 90O θ = 180o
Unstable equilibrium Unstable equilibrium
Stable equilibrium τ=0 Stable equilibrium τ=0
Wmax = 2pE Wmax = 2MB
τ=0 τmax = pE Umax = pE τ=0 τmax = MB Umax = MB
W=0 W = pE W=0 W = MB
Umin = – pE U=0 Umin = – MB U=0
(v) Angular SHM : In a uniform electric/magnetic field (intensity E/B) if a dipole (electric/magnetic) is
slightly displaced from it’s stable equilibrium position it executes angular SHM having period of oscillation. If I
= moment of inertia of dipole about the axis passing through it’s centre and perpendicular to it’s length.
For electric dipole : T = 2π I and For Magnetic dipole : T = 2π I
pE MB
(vi) Dipole-point charge interaction : If a point charge/isolated magnetic pole is placed in dipole
field at a distance r from the mid point of dipole then force experienced by point charge/pole varies
according to the relation F ∝ 1
r3
(vii) Dipole-dipole interaction : When two dipoles placed closed to each other, they experiences a
force due to each other. If suppose two dipoles (1) and (2) are placed as shown in figure then
Both the dipoles are placed in the field of one another hence potential energy dipole (2) is
U2 = − p2 E1 cos 0 = − p2 E1 = −p2 × 1 . 2p1 1 E2 2 E1
4πε 0 r3 O O
then by using F = − dU , Force on dipole (2) is F2 = − dU 2 –q +q –q +q
dr dr
P1 P2
r
⇒ F2 = − d 1 . 2p1 p2 = − 1 . 6 p1 p2
dr r3 4πε 0 r4
4πε 0
Similarly force experienced by dipole (1) F1 = − 1 . 6 p1 p 2 so F1 = F2 = − 1 . 6 p1 p 2
4πε 0 r4 4πε 0 r4
Negative sign indicates that force is attractive. | F |= 1 . 6 p1 p2 and F ∝ 1
4πε 0 r4 r4
S. No. Relative position of dipole Force Potential energy
(i) –q +q – q +q 1 . 6p1 p2 (attractive) 1 . 2p1 p2
P1 4πε 0 r4 4πε 0 r3
P2
(ii) +q +q 1 . 3p1 p2 (repulsive) 1 . p1 p2
4πε 0 r4 4πε0 r3
P1 P2
–q –q
r
(iii) +q 1 . 3p1 p2 (perpendicular to r ) 0
4πε 0 r4
P1 – q +q
–q
P2
r
Note : ≅ Same result can also be obtained for magnetic dipole.
(4) Electric dipole in non-uniform electric field : When an electric dipole is placed in a non-
uniform field, the two charges of dipole experiences unequal forces, therefore the net force on the dipole is not
equal to zero. The magnitude of the force is given by the negative derivative of the potential energy w.r.t.
distance along the axis of the dipole i.e. F = − dU = −p . dE .
dr dr
Due to two unequal forces, a torque is produced which rotate the dipole so as to align it in the direction of
field. When the dipole gets aligned with the field, the torque becomes zero
and then the unbalanced force acts on the dipole and the dipole then moves
linearly along the direction of field from weaker portion of the field to the +q
stronger portion of the field. So in non-uniform electric field –
(i) Motion of the dipole is translatory and rotatory E′
(ii) Torque on it may be zero.
Concepts
For a short dipole, electric field intensity at a point on the axial line is double than at a point on the equatorial line on electric
dipole i.e. Eaxial = 2Eequatorial
It is intresting to note that dipole field E ∝ 1 decreases much rapidly as compared to the field of a point charge E ∝ 1 .
r3 r2
– q +q Examples based on electric dipole
Example: 84 If the magnitude of intensity of electric field at a distance x on axial line and at a distance y on equatorial
Solution: (d) line on a given dipole are equal, then x : y is [EAMCET 1994]
Example: 85
(a) 1 : 1 (b) 1 : 2 (c) 1 : 2 (d) 3 2 : 1
According to the question 1 . 2 p = 1 . p ⇒ x = (2)1 / 3 :1
4πε 0 x 4πε 0 y3 y
3
Three charges of (+2q), (– q) and (– q) are placed at the corners A, B and C of an equilateral triangle of
side a as shown in the adjoining figure. Then the dipole moment of this combination is [MP PMT 1994; CPMT 1994]
(a) qa +2q a
(b) Zero A
(c) q a 3 a
(d) 2 qa B C
3 –q a –q
Solution: (c) The charge +2q can be broken in +q, +q. Now as shown in figure we have two equal dipoles inclined
at an angle of 60o. Therefore resultant dipole moment will be
PP
pnet = p 2 + p 2 + 2pp cos 60
= 3p 60O
= 3 qa
Example: 86 An electric dipole is placed along the x-axis at the origin O. A point P is at a distance of 20 cm from this
Solution: (b) origin such that OP makes an angle π with the x-axis. If the electric field at P makes an angle θ with x-
3
Example: 87 axis, the value of θ would be
Solution: (c) [MP PMT 1997]
Example: 89
(a) π (b) π + tan −1 3 (c) 2π (d) tan −1 3
3 3 2 3 2
According to question we can draw following figure. y
E
As we have discussed earlier in theory θ = π +α
3 P
α
tan α = 1 tan π ⇒ α = tan −1 3
2 3 2 π/3
O p θx
So, θ = π + tan −1 3
3 2
An electric dipole in a uniform electric field experiences [RPET 2000]
(a) Force and torque both (b) Force but no torque (c) Torque but no force (d) No force and no
torque
In uniform electric field Fnet = 0, τnet ≠ 0
Two opposite and equal charges 4 × 10–8 coulomb when placed 2 × 10–2 cm away, form a dipole. If this
dipole is placed in an external electric field 4 × 108 newton/coulomb, the value of maximum torque and
the work done in rotating it through 180o will be
[MP PET 1996 Similar to MP PMT 1987]
(a) 64 × 10–4 Nm and 64 × 10–4 J (b) 32 × 10–4 Nm and 32 × 10–4 J
(c) 64 × 10–4 Nm and 32 × 10–4 J (d) 32 × 10–4 Nm and 64 × 10–4 J
Solution: (d) τmax = pE and Wmax = 2pE p = Q × 2l = 4 × 10–8 × 2 × 10–2 × 10–2 = 8 × 10–12 C-m
Example: 90
So, τmax = 8 × 10–12 × 4 × 108 = 32 × 10–4 N-m and Wmax = 2 × 32 × 10–4 = 64 × 10–4 J
Solution: (d) A point charge placed at any point on the axis of an electric dipole at some large distance experiences a
Example: 91 force F. The force acting on the point charge when it’s distance from the dipole is doubled is
[CPMT 1991; MNR 1986]
(a) F (b) F (c) F (d) F
2 4 8
Force acting on a point charge in dipole field varies as F∝ 1 where r is the distance of point charge
r3
from the centre of dipole. Hence if r makes double so new force F' = F .
8
A point particle of mass M is attached to one end of a massless rigid non-conducting rod of length L.
Another point particle of the same mass is attached to other end of the rod. The two particles carry
charges +q and – q respectively. This arrangement is held in a region of a uniform electric field E such
that the rod makes a small angle θ (say of about 5 degrees) with the field direction (see figure). Will be
minimum time, needed for the rod to become parallel to the field after it is set free
+q
θE
–q
(a) t = 2π mL (b) t = π mL (c) t = 3π mL (d) t =π 2mL
2pE 2 2qE 2 2pE qE
Solution: (b) In the given situation system oscillate in electric field with maximum angular displacement θ.
It’s time period of oscillation (similar to dipole)
T = 2π I where I = moment of inertia of the system and p = qL
pE
Hence the minimum time needed for the rod becomes parallel to the field is t = T = π I
4 2 pE
L 2 L 2 ML2 π ML2 π ML
2 2 2 2 2 × qL × E 2 2qE
Here I = M + M = ⇒ t = =
Tricky example: 12
An electric dipole is placed at the origin O and is directed along the x-axis. At a point P, far away from
the dipole, the electric field is parallel to y-axis. OP makes an angle θ with the x-axis then
(a) tanθ = 3 (b) tanθ = 2 (c) θ = 45o (d) tanθ = 1
2
Solution: (b) As we know that in this case electric field makes an angle θ +α with the direction of dipole
Where tanα = 1 tanθ Y E
2 θ P
Here θ +α = 90o ⇒ α = 90 − θ O P
1 1 X
2 2
Hence tan(90 −θ) = tanθ ⇒ cot θ = tanθ
⇒ tan 2 θ = 2 ⇒ tanθ = 2
Electric Flux.
(1) Area vector : In many cases, it is convenient to treat area of a surface as a vector. The length of the
vector represents the magnitude of the area and its direction is along the outward drawn normal to the area.
ds
Area ds
(2) Electric flux : The electric flux linked with any surface in an electric field is basically a measure of
total number of lines of forces passing normally through the surface. or
Electric flux through an elementary area ds is defined as the scalar product of area of field i.e.
dφ = E ⋅ ds = E ds cosθ
∫Hence flux from complete area (S) φ = E ds cosθ = ES cosθ ds E
If θ = 0o, i.e. surface area is perpendicular to the electric field, so flux linked θ
with it will be max. d
i.e. φmax = E ds and if θ = 90o, φmin = 0
(3) Unit and Dimensional Formula
S.I. unit – (volt × m) or N −C
m2
It’s Dimensional formula – (ML3T–3A– 1)
(4) Types : For a closed body outward flux is taken to be positive, while inward flux is to be negative
nE Body
Body
Positive – flux Negative-flux
(A) (B)
Gauss’s Law.
(1) Definition : According to this law, total electric flux through a closed surface enclosing a charge is
1 times the magnitude of the charge enclosed i.e. φ = 1 (Qenc. )
ε0 ε0
(2) Gaussian Surface : Gauss’s law is valid for symmetrical charge distribution. Gauss’s law is very
helpful in calculating electric field in those cases where electric field is symmetrical around the source producing
it. Electric field can be calculated very easily by the clever choice of a closed surface that encloses the source
charges. Such a surface is called “Gaussian surface”. This surface should pass through the point where electric
field is to be calculated and must have a shape according to the symmetry of source.
e.g. If suppose a charge Q is placed at the centre of a hemisphere, then to calculate the flux through this
body, to encloses the first charge we will have to imagine a Gaussian surface. This imaginary Gaussian surface
will be a hemisphere as shown.
Net flux through this closed body φ = Q •Q
ε0
Hence flux coming out from given hemisphere is φ = Q .
2ε 0
(3) Zero flux : The value of flux is zero in the following circumstances
(i) If a dipole is enclosed by a surface (ii) If the magnitude of positive and negative charges
are equal inside a closed surface
φ = 0;Qenc = 0 –q +q +q
+4q
Qenc = 0,
so, φ = 0 –5q
(iii) If a closed body (not enclosing any charge) is placed in an electric field (either uniform or non-uniform)
total flux linked with it will be zero
E
Sphere
φin = −πR 2 E φT = 0 φ out = +πR 2 E φT = 0
y ds
E
x
ds ds
a
a
φT = 0 φ in = φ out = E a 2
(4) Flux emergence : Flux linked with a closed body is independent of the shape and size of the body
and position of charge inside it
•Q •Q •Q
Q Q
•Q
φT = ε0 φT = ε0
Q φT Q
=
φT = ε0
ε0
(i) If a hemispherical body is placed in uniform (ii) If a hemispherical body is placed in non-uniform
electric field as shown below. then flux linked with
electric field then flux linked with the curved surface the curved surface.
E
R R
φcurved = +πR 2 E nˆ φcurved = 2πR 2 E nˆ
(v) If charge is kept at the centre of cube (iv) If charge is kept at the centre of a face
φ total = 1 .(Q)
ε0
Q Q
6ε 0
φ face =
First we should enclosed the charge by assuming
a Gaussian surface (an identical imaginary cube)
φ corner = Q φ edge = Q φ total = Q
8ε 0 12ε 0 ε0
φ cube = Q (i.e. from 5 face only)
2ε 0
Q
φ face = 1 Q = Q .
5 2ε 0 10ε 0
Concept
In C.G.S. ε0 = 1 . Hence if 1C charge is enclosed by a closed surface so flux through the surface will be φ = 4π .
4π
Example based on electric flux and Gauss’s
Example: 91 Electric charge is uniformly distributed along a long straight wire of radius 1 mm. The charge per cm length
of the wire is Q coulomb. Another cylindrical surface of radius 50 cm and length 1 m symmetrically
encloses the wire as shown in the figure. The total electric flux passing through the cylindrical surface is
[MP PET 2001]
(a) Q +
ε0 +
+
(b) 100 Q
ε0 1m
+
(c) 10 Q + 50 cm
(πε 0 ) +
(d) 100 Q
(πε 0 )
Solution: (b) Given that charge per cm length of the wire is Q. Since 100 cm length of the wire is enclosed so
Qenc = 100 Q
⇒ Electric flux emerging through cylindrical surface φ = 100 Q .
ε0
Example: 92 A charge Q is situated at the corner A of a cube, the electric flux through the one face of the cube is
[CPMT 2000]
Q
(a) Q (b) Q (c) Q (d) Q
6ε 0 8ε 0 24ε 0 2ε 0
Solution: (c) For the charge at the corner, we require eight cube to symmetrically enclose it in a Gaussian surface. The
Example: 93 total flux φT = Q . Therefore the flux through one cube will be φ cube = Q . The cube has six faces and
Solution: (a) ε0 8ε 0
Example: 94
Solution: (b) flux linked with three faces (through A) is zero, so flux linked with remaining three faces will φ . Now
8ε 0
as the remaining three are identical so flux linked with each of the three faces will be
= 1 × 1 Q = 1 Q .
3 ε0 24 ε0
8
A square of side 20 cm is enclosed by a surface of sphere of 80 cm radius. Square and sphere have the
same centre. Four charges + 2 × 10–6 C, – 5 × 10– 6 C, – 3 × 10– 6 C, +6 × 10– 6 C are located at the
four corners of a square, then out going total flux from spherical surface in N–m2/C will be
(a) Zero (b) (16 π) × 10– 6 (c) (8π) × 10–6 (d) 36π × 10–6
Since charge enclosed by Gaussian surface is
φenc. = (2 × 10−6 − 5 × 10−6 − 3 × 10−6 + 6 × 10−6 ) = 0 so φ = 0
In a region of space, the electric field is in the x-direction and proportional to x, i.e., E = E0 xˆi . Consider
an imaginary cubical volume of edge a, with its edges parallel to the axes of coordinates. The charge
inside this cube is
(a) Zero (b) ε 0 E0a 3 (c) 1 E0 a 3 (d) 1 ε 0 E0 a 2
ε0 6
The field at the face ABCD = E0 x0ˆi.
∴ Flux over the face ABCD = – (E0x0)a2 YC G
The negative sign arises as the field is directed into the cube.
The field at the face EFGH = E0 (x0 + a)ˆi. D H
x0 B a
∴ Flux over the face EFGH = E0 (x0 + a)a 2
The flux over the other four faces is zero as the field is parallel to the Aa F
surfaces. Ea
X
Z
∴ Total flux over the cube = E0a2 = 1 q
2
where q is the total charge inside the cube. ∴ q = ε0 E0a3.
Tricky example: 13
In the electric field due to a point charge + Q a spherical closed surface is drawn as shown by the
dotted circle. The electric flux through the surface drawn is zero by Gauss’s law. A conducting sphere is
inserted intersecting the previously drawn Gaussian surface. The electric flux through the surface
(a) Still remains zero
(b) Non zero but positive +Q
(c) Non-zero but negative
(d) Becomes infinite
Solution: (b) Due to induction some positive charge will lie within the Gaussian surface drawn and hence flux
becomes something positive.
Application of Gauss’s Law.
Gauss’s law is a powerful tool for calculating electric field in case of symmetrical charge distribution by
choosing a Gaussian surface in such away that E is either parallel or perpendicular to it’s various faces.
e.g. Electric field due to infinitely long line of charge : Let us consider a + nˆ
uniformly charged wire of infinite length having a constant linear charge density is 2 +
l +
λ λ = charge . Let P be a point distant r from the wire at which the electric field is to be P
length +a
calculated. +
+
+
Draw a cylinder (Gaussian surface) of radius r and length l around the line charge 1 nˆ
which encloses the charge Q ( Q = λ . l ). Cylindrical Gaussian surface has three surfaces;
two circular and one curved for surfaces (1) and (2) angle between electric field and normal to the surface is
90o i.e., θ = 90o.
So flux linked with these surfaces will be zero. Hence total flux will pass through curved surface and it is
∫φ = E ds cosθ …. (i)
According to Gauss’s law …. (ii)
φ=Q
ε0
∫Equating equation (i) and (ii) E ds = Q
ε0
∫⇒ E ds = Q ⇒ Ex2πrl = Q
ε0 ε0
⇒ E = Q = λ = 2kλ K = 1
2πε0rl 2πε0r r 4πε0
Electrostatics
Capacitance.
(1) Definition : We know that charge given to a conductor increases it’s potential i.e., Q ∝ V ⇒ Q = CV
Where C is a proportionality constant, called capacity or capacitance of conductor. Hence capacitance is the
ability of conductor to hold the charge.
(2) Unit and dimensional formula : S.I. unit is Coulomb = Farad (F)
Volt
Smaller S.I. units are mF, µF, nF and pF ( 1mF = 10−3 F , 1µF = 10−6 F , 1nF = 10 −9 F ,
1pF = 1µµF = 10 −12 F )
C.G.S. unit is Stat Farad 1F = 9 × 1011 Stat Farad . Dimension : [C] = [M −1L−2T 4 A2 ].
(3) Capacity of an isolated spherical conductor : When charge Q is given to a spherical conductor of
radius R, then potential at the surface of sphere is + + + Q
+
Q k 1 +
R 4πε 0 +
V = k . = + R +
+
+
+ O +
+
Q 1 + + + ++
Hence it’s capacity C = V = 4πε 0 R ⇒ C = 4πε0 R = 9 × 109 .R
in C.G.S. C=R
Note : ≅ If earth is assumed to be spherical having radius R = 6400 km. It’s theortical capacitance
C = 1 × 6400 × 103 = 711 µF . But for all practical purpose capacitance of earth is taken
9 × 109
infinity.
(4) Energy of a charged conductor : When a conductor is charged it’s potential increases from 0 to V as
shown in the graph; and work is done against repulsion, between charge stored in the conductor and charge
coming from the source (battery). This work is stored as “electrostatic potential energy”
From graph : Work done = Area of graph = 1 QV V
2
Hence potential energy U= 1 QV ; By using Q = CV, we can write
2
U 1 QV 1 CV 2 Q2 Q
2 2 2C
= = =
(5) Combination of drops : Suppose we have n identical drops each having – Radius – r, Capacitance – c,
Charge – q, Potential – v and Energy – u.
If these drops are combined to form a big drop of – Radius – R, Capacitance – C, Charge – Q, Potential – V
and Energy – U then –
(i) Charge on big drop : Q = nq
(ii) Radius of big drop : Volume of big drop = n × volume of a single drop i.e., 4 πR 3 = n× 4 πr 3 , R = n1/3r
3 3
(iii) Capacitance of big drop : C = n1/3c
(iv) Potential of big drop : V = Q = nq V = n2/3v
C n1/ 3c
(v) Energy of big drop : U = 1 CV 2 = 1 (n1 / 3 c)(n 2 / 3v) 2 U = n5/3u
2 2
(6) Sharing of charge : When two conductors joined together through a conducting wire, charge begins to
flow from one conductor to another till both have the same potential, due to flow of charge, loss of energy also
takes place in the form of heat.
Suppose there are two spherical conductors of radii r1 and r2, having charge Q1 and Q2, potential V1 and
V2, energies U1 and U2 and capacitance C1 and C2 respectively, as shown in figure. If these two spheres are
connected through a conducting wire, then alteration of charge, potential and energy takes place.
Q1 r2 Q2 Q1′ r2 Q2′
C1 r1 Q2=C2V2 C2 C1 r1 Q2′=C2V2 C2
V1 V2 V V
U1 U1′
U2 U2′
Q1=C1V1 Q1′=C1V1
(A) (B)
(i) New charge : According to the conservation of charge Q1 + Q2 = Q1' + Q2' = Q (say), also
Q1' = C1 V = 4πε 0r1 , Q1' = r1 ⇒ 1+ Q1' = 1 + r1 ⇒ Q1' + Q2' = r1 + r2
Q2' C2V 4πε 0r2 Q2' r2 Q2' r2 Q2' r2
⇒ Q2' = Q r1 r2 r2 and similarly Q1' = Q r1 r1
+ + r2
(ii) Common potential : Common potential (V ) = Total charge = Q1 + Q2 = Q1' + Q ' ⇒ V = C1 V1 + C2V2
Total capacity C1 + C2 C1 2 C1 + C2
+ C2
(iii) Energy loss : As electrical energy stored in the system before and after connecting the spheres is
Ui = 1 C1 V12 + 1 C 2 V22 and Uf = 1 (C1 + C2).V 2 = 1 (C1 + C2 ) C1 V1 + C2V2 2
2 2 2 2 C1 + C2
so energy loss ΔU = Ui −Uf = C1 C 2 ) (V1 − V2 )2
2(C1 +
C2
Concept
Capacity of a conductor is a constant term, it does not depend upon the charge Q, and potential (V) and nature of the material of
the conductor.
+ ++ ++ Examples based on sharing of charge, drops and general concept of capacity
+ +
+ + +
++ + ++
Example: 95 Eight drops of mercury of same radius and having same charge coalesce to form a big drop. Capacitance of
big drop relative to that of small drop will be [MP PMT 2002, 1990; MNR 1999, 87; DCE 1998]
(a) 16 times (b) 8 times (c) 4 times (d) 2 times
Solution: (d) By using relation C = n1/ 3 . c ⇒ C = (8)1/ 3 . c = 2c
Example: 96 Two spheres A and B of radius 4 cm and 6 cm are given charges of 80µC and 40µC respectively. If they are
connected by a fine wire, the amount of charge flowing from one to the other is [MP PET 1991]
(a) 20 µ C from A to B (b) 16 µ C from A to B (c) 32µC from B to A (d) 32 µ C from A to B
Solution: (d) Total charge Q = 80 + 40 = 120 µ C . By using the formula Q1' = Q r1 . New charge on sphere A is
Example: 97 +
r1 r2
Q ' = Q rA rA rB = 120 4 4 6 = 48 µ C . Initially it was 80 µ C, i.e., 32 µ C charge flows from A to B.
A + +
Two insulated metallic spheres of 3µF and 5µF capacitances are charged to 300V and 500V respectively.
The energy loss, when they are connected by a wire, is [Pb PMT 1999; CPMT 1999; KCET (Engg.) 2000]
(a) 0.012 J (b) 0.0218 J (c) 0.0375J (d) 3.75 J
Solution: (c) By using ΔU = C1 C2 ) (V1 − V2)2 ; ΔU = 0.375 J
2(C1 +C
2
Example: 98 64 small drops of mercury, each of radius r and charge q coalesce to form a big drop. The ratio of the
surface density of charge of each small drop with that of the big drop is [KCET 2002]
(a) 1 : 64 (b) 64 : 1 (c) 4 : 1 (d) 1 : 4
σ Small q / 4πr 2 q R 2
σ Big Q / 4πR 2 Q r
Solution: (d) = = ; since R = n1/3r and Q = nq
So σ Small 1 ⇒ σ Small 1
σ Big = n1/ 3 σ Big =4
Tricky example: 14
Two hollow spheres are charged positively. The smaller one is at 50 V and the bigger one is at 100 V.
How should they be arranged so that the charge flows from the smaller to the bigger sphere when they
are connected by a wire [Kerala PET 2002]
(a) By placing them close to each other
(b) By placing them at very large distance from each other
(c) By placing the smaller sphere inside the bigger one
(d) Information is insufficient
Solution: (c) By placing the smaller sphere inside the bigger one. The potential of the smaller one will now be 150 V.
So on connecting it with the bigger one charge will flow from the smaller one to the bigger one.
Capacitor.
(1) Definition : A capacitor is a device that stores electric energy. It is also named condenser.
or
A capacitor is a pair of two conductors of any shape, which are close to each other and have equal and
opposite charge.
+Q –Q
+ –
+ –
+ –
+ –
+
+ –
–
(2) Symbol : The symbol of capacitor are shown below variable capacitor
or
(3) Capacitance : The capacitance of a capacitor is defined as the magnitude of the charge Q on the positive
plate divided by the magnitude of the potential difference V between the plates i.e., C = Q
V
(4) Charging : A capacitor get’s charged when a battery is connected across the plates. The plate attached to
the positive terminal of the battery get’s positively charged and the one joined to the negative terminal get’s
negatively charged. Once capacitor get’s fully charged, flow of charge carriers stops in the circuit and in this
condition potential difference across the plates of capacitor is same as the potential difference across the terminals of
battery (say V).
+Q –Q C
+–
+V – ⇒
+V –
(5) Charge on capacitor : Net charge on a capacitor is always zero, but when we speaks of the charge Q on
a capacitor, we are referring to the magnitude of the charge on each plate. Charge distribution in making of parallel
plate capacitor can easily be understand by reading carefully the following sequence of figures –
+Q
+ Q + Q + Q + Q − Q + Q +Q –Q
2 2 2 2 2 2 XY
⇒⇒ ⇒
XX XY
(6) Energy stored : When a capacitor is charged by a voltage source (say battery) it stores the electric
energy. If C = Capacitance of capacitor; Q = Charge on capacitor and V = Potential difference across
capacitor then energy stored in capacitor U = 1 CV 2 = 1 QV = Q2
2 2 2C
Note : ≅ In charging capacitor by battery half the energy supplied is stored in the capacitor and
remaining half energy (1/2 QV) is lost in the form of heat.
(7) Types of capacitors : Capacitors are of mainly three types as described in given table
Parallel Plate Capacitor Spherical Capacitor Cylindrical Capacitor
It consists of two concentric cylinders
It consists of two parallel metallic plates It consists of two concentric of radii a and b (a < b), inner
(may be circular, rectangular, square) conducting spheres of radii a and b cylinder is given charge +Q while
separated by a small distance (a < b). Inner sphere is given charge outer cylinder is earthed. Common
+Q, while outer sphere is earthed length of the cylinders is l then
+Q –Q – –– b a q –q
+ – – + + ––Q–
+ – l
– +
A+ air – +Q
+ + b–
– –+ a
+ –+ +–
+ –
– –+ –
d –– –
A = Effective overlaping area of Capacitance C = 4πε0 . ab Capacitance
each plate b−a
2πε 0l
d = Separation between the in C.G.S. C = ab . In the presence of C =
plates b−a b
dielectric medium (dielectric constant K) log e a
Q = Magnitude of charge on the between the spheres C' = 4πε 0 K ab In the presence of dielectric medium
inner side of each plate b−a (dielectric constant K) capacitance
increases by K times and
σ = Surface density of charge of Special Case :
each plate = Q If outer sphere is given a charge +Q C' 2πε 0 Kl
A while inner sphere is earthed
= b
a
V = Potential difference across the –Q′– – +Q log e
plates a
– –
– –
E = Electric field between the plates – – –b
–
= σ Induced charge on the inner sphere
ε0
Capacitance : C = ε0 A Q' = − a . Q , C' = 4πε0 . b2
d b b−a
A This arrangement is not a capacitor.
4πd But it’s capacitance is equivalent to
in C.G.S. : C = the sum of capacitance of spherical
If a dielectric medium of dielectric capacitor and spherical conductor
constant K is filled completely
between the plates then capacitance i.e.
increases by K times C' = KC
4πε 0 . b2 a = 4πε 0 ab + 4πε 0b
b− b−a
Concepts
It is a very common misconception that a capacitor stores charge but actually a capacitor stores electric energy in the electrostatic
field between the plates.
Two plates of unequal area can also form a capacitor because effective overlapping area is considered.
A
d
If two plates are placed side by side then three capacitors are formed. One between distant earthed bodies and the first face of the
first plate, the second between the two plates and the third between the second face of the second plate and distant earthed
objects. However the capacitances of the first and third capacitors are negligibly small in comparision to that between the plates
which is the main capacitance.
Capacitance of a parallel plate capacitor depends upon the effective overlapping area of plates (C ∝ A) , separation between the
plates C ∝ 1 and dielectric medium filled between the plates. While it is independent of charge given, potential raised or
d
nature of metals and thickness of plates.
The distance between the plates is kept small to avoid fringing or edge effect (non-uniformity of the field) at the bounderies of the
plates.
+–
+–
+–
+–
+–
+–
Spherical conductor is equivalent to a spherical capacitor with it’s outer sphere of infinite radius.
A spherical capacitor behaves as a parallel plate capacitor if it’s spherical surfaces have large radii and are close to each other.
The intensity of electric field between the plates of a parallel plate capacitor (E = σ/ε0) does not depends upon the distance
between them.
The plates of a parallel plate capacitor are being moved away with some velocity. If the plate separation at any instant of time is
‘d’ then the rate of change of capacitance with time is proportional to 1 .
d2
Radial and non-uniform electric field exists between the spherical surfaces of spherical capacitor.
Two large conducting plates X and Y kept close to each other. The plate X is given a charge Q1 while plate Y is given a charge
Q2 (Q1 > Q2 ) , the distribution of charge on the four faces a, b, c, d will be as shown in the following figure.
Q1 X Q2 X − Q1 − Q2 Y
b d 2
a c ⇒ Q1 − Q2 Q1 − Q2
2 2
Q1 + Q2
2
+–
+ – Example based on simple concepts of capacitor
+–
Example: 99 The capacity of pure capacitor is 1 farad. In D.C. circuit, its effective resistance will be [MP PMT 2000]
(a) Zero (b) Infinite (c) 1 ohm (d) 1 ohm
2
Solution: (b) Capacitor does not work in D.C. for D.C. it’s effective resistance is infinite i.e. it blocks the current to flow in
the circuit.
Example: 100 A light bulb, a capacitor and a battery are connected together as shown here, with switch S initially open.
When the switch S is closed, which one of the following is true [MP PMT 1995]
s +–
(a) The bulb will light up for an instant when the capacitor starts charging
(b) The bulb will light up when the capacitor is fully charged
(c) The bulb will not light up at all
(d) The bulb will light up and go off at regular intervals
Solution: (a) Current through the circuit can flow only for the small time of charging, once capacitor get’s charged it blocks
the current through the circuit and bulb will go off.
Example: 101 Capacity of a parallel plate condenser is 10µF when the distance between its plates is 8 cm . If the distance
between the plates is reduced to 4cm, its capacity will be
[CBSE 2001; Similar to CPMT 1997; AFMC 2000]
(a) 10µF (b) 15µF (c) 20µF (d) 40µF
Solution: (c) C = ε0A ∝ 1 ∴ C1 = d2 or C2 = d1 × C1 = 8 × 10 = 20µ F
d d C2 d1 d2 4
Example: 102 What is the area of the plates of a 3F parallel plate capacitor, if the separation between the plates is 5 mm
(a) 1.694 × 109 m2 (b) 4.529 × 109 m2 (c) 9.281 × 109 m2 [BHU 2002; AIIMS 1998]
(d) 12.981 × 109 m2
Solution: (a) By using the relation C = ε0A ⇒ A= Cd = 3 × 5 × 10 −3 = 1.694 × 109 m2 .
d ε0 8.85 × 10 −12
Example: 103 If potential difference of a condenser (6 µ F) is changed from 10 V to 20 V then increase in energy is
[CPMT 1997, 87]
(a) 2 × 10 −4 J (b) 4 × 10−4 J (c) 3 × 10−4 J (d) 9 × 10−4 J
Solution: (d) Initial energy Ui = 1 CV12 ; Final energy Uf = 1 CV22
2 2
∴ Increase in energy ∆U = Uf − Ui = 1 C (V22 − V12 ) = 1 × 6 × 10 −6 (20 2 − 10 2 ) = 9 × 10 −4 J.
2 2
Example: 104 A spherical capacitor consists of two concentric spherical conductors. The inner one of radius R1 maintained
at potential V1 and the outer conductor of radius R2 at potential V2 . The potential at a point P at a distance
x from the centre (where R2 > x > R1 ) is [MP PMT 1997]
(a) V1 − V2 (x − R1 ) (b) V1R1(R2 − x) + V2 R2 (x − R1)
R2 − R1 (R2 − R1) x
(c) V1 + V2 x (d) (V1 + V2 ) x
(R2 − R1 ) (R1 + R2)
Solution: (b) Let Q1 and Q2 be the charges on the inner and the outer sphere respectively. Now V1 is the total potential
on the sphere of radius R1,
So, V1 = Q1 + Q2 …….. (i)
R1 R2
and V2 is the total potential on the surface of sphere of radius R2 ,
So, V2 = Q2 + Q1 …….. (ii)
R2 R2
If V be the potential at point P which lies at a distance x from the common centre then
V = Q1 + Q2 = Q1 + V1 − Q1 = Q1 1 − 1 + V1 = Q1(R1 − x) + V1 ……..(iii)
x R2 x R1 x R1 xR1
Substracting (ii) from (i)
V1 − V2 = Q1 − Q2 ⇒ (V1 − V2 )R1R2 = R2Q1 − R1Q1 ⇒ Q1 = (V1 − V2 )R1 R2
R1 R2 R2 − R1
Now substituting it in equation (iii), we have
V = (R1 − x)(V1 − V2 )R1R2 + V1 ⇒ V = V1 R1(R2 − x) + V2 R2(x − R1 )
xR1(R2 − R1) x(R2 − R1 )
Example: 105 The diameter of each plate of an air capacitor is 4 cm. To make the capacity of this plate capacitor equal to
that of 20 cm diameter sphere, the distance between the plates will be [MP PET 1996]
(a) 4 × 10−3 m (b) 1 × 10−3 m (c) 1 cm (d) 1 × 10−3 cm
Solution: (b) According to the question ε0A = 4πε 0 R ⇒ d= A = π (2 × 10−2 )2 = 1 × 10−3 m.
d 4πR 4π × 10 × 10−2
Example: 106 A spherical condenser has inner and outer spheres of radii a and b respectively. The space between the two is
filled with air. The difference between the capacities of two condensers formed when outer sphere is earthed
and when inner sphere is earthed will be [MP PET 1996]
(a) Zero (b) 4πε 0a (c) 4πε 0b (d) 4πε 0 a b b a
−
Solution: (c) Capacitance when outer sphere is earthed C1 = 4πε 0 . ab and capacitance when inner sphere is earthed
b−a
C2 = 4πε 0 . b2 a . Hence C2 − C1 = 4πε 0 .b
b−
Example: 107 After charging a capacitor of capacitance 4µ F upto a potential 400 V, its plates are connected with a
resistance of 1kΩ . The heat produced in the resistance will be [CBSE PMT 1994]
(a) 0.16 J (b) 1.28 J (c) 0.64 J (d) 0.32 J
Solution: (d) This is the discharging condition of capacitor and in this condition energy released
U = 1 CV 2 = 1 × 4 × 10−6 × (400)2 = 0.32J = 0.32 J.
2 2
Example: 108 The amount of work done in increasing the voltage across the plates of a capacitor from 5V to 10V is W. The
work done in increasing it from 10V to 15V will be
(a) 0.6 W (b) W (c) 1.25 W (d) 1.67 W
Solution: (d) As we know that work done = U final − Uinitial = 1 C (V22 − V12 )
2
When potential difference increases from 5V to 10V then
W = 1 C (10 2 − 52) ……..(i)
2
When potential difference increases from 10V to 15V then
W' = 1 C (15 2 − 10 2 ) ……..(ii)
2
On solving equation (i) and (ii) we get
W' = 1.67W.
Tricky example: 15
In an isolated parallel plate capacitor of capacitance C, the four surface have charges Q1 , Q2 , Q3 and
Q4 as shown. The potential difference between the plates is [UPSEAT 2003; IIT-JEE 1999]
Q1 Q3
Q2 Q4
(a) Q1 + Q2 + Q3 + Q4 (b) Q2 + Q3 (c) Q2 − Q3 (d) Q1 + Q4
2C 2C 2C 2C
Solution: (c) Plane conducting surfaces facing each other must have equal and opposite charge densities. Here as the
plate areas are equal, Q2 = −Q3 .
The charge on a capacitor means the charge on the inner surface of the positive plate (here it is Q2 )
Potential difference between the plates = charge = Q2 = 2Q2
capacitance C 2C
= Q2 − (−Q2 ) = Q2 − Q3 .
2C 2C
Dielectric.
Dielectrics are insulating (non-conducting) materials which transmits electric effect without conducting we
know that in every atom, there is a positively charged nucleus and a negatively charged electron cloud surrounding
it. The two oppositely charged regions have their own centres of charge. The centre of positive charge is the centre
of mass of positively charged protons in the nucleus. The centre of negative charge is the centre of mass of
negatively charged electrons in the atoms/molecules.
(1) Type of Dielectrics : Dielectrics are of two types –
(i) Polar dielectrics : Like water, Alcohol, CO2 , NH3 , (ii) Non polar dielectric : Like N 2 , O2 , Benzene,
HCl etc. are made of polar atoms/molecules.
Methane etc. are made of non-polar atoms/molecules. In
In polar molecules when no electric field is applied non-polar molecules, when no electric field is applied the
centre of positive charge coincides with the centre of
centre of positive charges does not coincide with the centre negative charge in the molecule. Each molecule has zero
dipole moment in its normal state.
of negative charges. O– –
105o
H+ H+ – + + – P = 0
P
A p(opl)arinmtohleecaublesenhcaes permanent electric dipole When electric field is applied, positive charge
moment of electric field also. But a
experiences a force in the direction of electric field and
polar dielectric has net dipole moment is zero in the
absence of electric field because polar molecules are negative charge experiences a force in the direction
randomly oriented as shown in figure.
opposite to the field i.e., molecules becomes induced
electric dipole. E
– – + –
+ + –
–
+ ++
+ –
In the presence of electric field polar molecules tends –
to line up in the direction of electric field, and the
substance has finite dipole moment. In general, any non-conducting, material can be called
as a dielectric but broadly non conducting material having
–+ –+ –+ + non polar molecules referred to as dielectric because
–+ –+ induced dipole moment is created in the non polar
– molecule.
–+
P
(2) Polarization of a dielectric slab : It is the process of inducing equal and opposite charges on the two
faces of the dielectric on the application of electric field. E
Suppose a dielectric slab is inserted between the plates of a capacitor. + –
As shown in the figure. + ++–+ –
Induced electric field inside the dielectric is Ei , hence this induced +–
electric field decreases the main field E to E − Ei i.e., New electric field + –+ –+ –
between the plates will be E' = E − Ei . + Ei –
+ –+ –+ –
+–
+ –+ –+ –
+–
+ –+ –+ –
(3) Dielectric constant : After placing a dielectric slab in an electric
field. The net field is decreased in that region hence
If E = Original electric field and E' = Reduced electric field. Then E = K where K is called dielectric constant
E'
K is also known as relative permittivity (ε r ) of the material or SIC (Specific Inductive Capacitance)
The value of K is always greater than one. For vacuum there is no polarization and hence E = E' and K = 1
(4) Dielectric breakdown and dielectric strength : If a very high electric field is created in a dielectric,
the outer electrons may get detached from their parent atoms. The dielectric then behaves like a conductor. This
phenomenon is known as dielectric breakdown.
The maximum value of electric field (or potential gradient) that a dielectric material can tolerate without it’s
electric breakdown is called it’s dielectric strength.
S.I. unit of dielectric strength of a material is V but practical unit is kV .
m mm
Variation of Different Variables (Q, C, V, E and U) of Parallel Plate Capacitor.
Suppose we have an air filled charged parallel plate capacitor having variables are as follows :
Charge – Q, Surface charge density – σ = Q, Capacitance – C = ε0 A +Q –Q
A d + –
–
+
Potential difference across the plates – V = E .d +–
Air
A +–
Electric field between the plates – E= σ = Q +–
ε0 Aε 0
+–
d
Energy stored – U = 1 CV 2 = Q2 = 1 QV V
2 2C 2
(1) When dielectric is completely filled between plates : If a dielectric slab is fills completely the gap
between the plates, capacitance increases by K times i.e., C' = Kε 0 A ⇒ C' = KC
d
The effect of dielectric on other variables such as charge. Potential difference field and energy associated with a
capacitor depends on the fact that whether the charged capacitor is disconnected from the battery or battery is still connected.
Quantity Battery is Removed Battery Remains connected
AK AK
d d
V
Capacity C′ = KC
Charge Q′ = Q (Charge is conserved) C′ = KC
Potential V′ = V/K Q′ = KQ
V′ = V (Since Battery maintains the potential
Intensity E′ = E/K difference)
Energy U′ = U/K E′ = E
U′ = U/K
Note : ≅ If nothing is said it is to be assumed that battery is disconnected.
(2) When dielectric is partially filled between the plates : If a dielectric slab of thickness t (t < d) is
inserted between the plates as shown below, then E = Main electric field between the plates, Ei = Induced electric
field in dielectric. E' = (E − Ei ) = The reduced value of electric field in the dielectric. Potential difference between the
two plates of capacitor is given by
V'= E (d − t) + E' t = E (d − t) + E .t +– E
K +– +–
+–
⇒ V'= E d − t + t = σ d − t + t = Q d − t + t +– +–
K ε0 K Aε 0 K A + –Ei + –
Now capacitance of the capacitor +– +–
+– +–
Q ε0A
V' −t+ d
C' = ⇒ C' = t
K
d
t1 t2 t2 t1 t2 t3 t4
A K1 K2 K3 A K1 K2 K3 K4
d d
C' = ε0A C' = ε 0 A
d − (t1 + t2 + t3 + ........) + t1 + t2 + t3 + ........ t1 + t2 + t3 + t4
K1 K2 K3 K1 K2 K3 K4
(3) When a metallic slab is inserted between the plates :
t
A K=∞ A K=∞ or A
d dd
Capacitance C' = ε0A C' = ∞ (In this case capacitor is said to be short circuited)
(d − t)
(4) When separation between the plates is changing : If separation between the plates changes then it’s
capacitance also changes according to C∝ 1 . The effect on other variables depends on the fact that whether the
d
charged capacitor is disconnected from the battery or battery is still connected.
(i) Separation is increasing Battery remains connected
Quantity Battery is removed
AA
d′ d′
V
Capacity Decreases because C ∝ 1 i.e., C' < C Decreases i.e., C' < C
Charge d
Potential Remains constant because a battery is not present Decreases because battery is present i.e., Q' < Q
difference i.e., Q' = Q
Electric Remaining charge (Q − Q' ) goes back to the
field battery.
Energy Increases because V = Q ⇒ V ∝ 1 i.e., V' > V V′ = V (Since Battery maintains the potential
C C difference)
Remains constant because E= σ = Q Decrease because E = Q ⇒ E∝Q
ε0 Aε 0 Aε 0
i.e., E' = E i.e., E' < E
Increases because U = Q2 ⇒ U ∝ 1 Decreases because U = 1 CV 2 ⇒ U∝C
2C C 2
i.e., U' > U i.e., U' < U
(ii) Separation is decreasing Battery remains connected
Quantity Battery is removed Increases i.e., C' > C
Capacity Increases because C ∝ 1 i.e., C' > C
Charge d
Remains constant because battery is not present Increases because battery is present i.e., Q' > Q
i.e., Q' = Q Remaining charge (Q' − Q) supplied from the
battery.
Potential Decreases because V = Q ⇒V ∝ 1 i.e., V'< V V′ = V (Since Battery maintains the potential
difference C difference)
Electric C
field
Remains constant because E= σ = Q Increases because E = Q ⇒ E∝Q
Energy ε0 Aε 0 Aε 0
i.e., E' = E i.e., E' > E
Decreases because U = Q2 ⇒ U ∝ 1 Increases because U = 1 CV 2 ⇒ U∝C
2C C 2
i.e., U' < U i.e., U' > U
Force Between the Plates of a Parallel Plate Capacitor.
Field due to charge on one plate on the other is E = σ , hence the force F = QE
2ε 0
+–
σ2 +–
F = −σA × σ = − 2ε 0 A
2ε + Air –
0 A +–
⇒ | F |= σ 2A = Q2 +–
2ε 0 2ε 0 A +–
d
Energy Density Between the Plates of a Parallel Plate Capacitor.
The energy stored in a capacitor is not localised on the charges or the plates but is distributed in the field. And
as in case of a parallel plate capacitor field is only between the plates i.e. in a volume (A× d), the so called energy
density.
Energy 1 CV 2 1 ε0 A V 2 1 V 2 1
Volume 2 2 d Ad 2 d 2
Hence Energy density = = = = ε 0 = ε 0 E 2 .
Ad
Concepts
In the expression of capacitance of parallel plate capacitor filled partially with dielectric term d − t + t is known as effective
K
air separation between the plates.
When dielectric is partially filled between the plates of a parallel plate capacitor then it’s capacitance increases but potential
difference decreases. To maintain the capacitance and potential difference of capacitor as before (i.e., c = ε0A , V = σ d)
d ε0
separation between the plates has to be increased. Suppose separation is increased by d' so in this case
ε0A t = ε0A which gives us K = t t
d + d'−t + K d − d'
Example based on capacitor with dielectric
Example: 109 The mean electric energy density between the plates of a charged capacitor is (here Q = Charge on the
capacitor and A = Area of the capacitor plate)
[MP PET 2002]
(a) Q2 (b) Q (c) Q2 (d) None of these
2ε 0 A 2 2ε 0 A 2 2ε 0 A
Solution: (a) Energy density = 1 ε 0 E 2 = 1 ε 0 Q 2 = Q2 .
2 2 Aε 0 2ε 0 A 2
Example: 110 Plate separation of a 15µ F capacitor is 2 mm. A dielectric slab (K = 2) of thickness 1 mm is inserted between
the plates. Then new capacitance is given by [BHU 1994, Similar to BHU 2000]
(a) 15µ F (b) 20µ F (c) 30µ F (d) 25µ F
Solution: (b) Given C = ε0A = 15µ F ……..(i)
d
Then by using C' = ε0A t = ε0A 10−3 = 2 × ε0 A × 103 ; From equation (i) C' = 20µ F.
d−t+ K 2 3
2 × 10−3 − 10−3 +
Example: 111 There is an air filled 1 pF parallel plate capacitor. When the plate separation is doubled and the space is filled
with wax, the capacitance increases to 2 pF. The dielectric constant of wax is [MNR 1998]
(a) 2 (b) 4 (c) 6 (d) 8
Solution: (b) Given that capacitance C = 1 pF
After doubling the separation between the plates C' = C
2
and when dielectric medium of dielectric constant k filled between the plates then C' = KC
2
According to the question, C' = KC = 2 ⇒ K = 4.
2
Example: 112 If a slab of insulating material 4 × 10−5 m thick is introduced between the plate of a parallel plate capacitor, the
distance between the plates has to be increased by 3.5 × 10−5 m to restore the capacity to original value. Then
the dielectric constant of the material of slab is [AMU 1999]
(a) 10 (b) 12 (c) 6 (d) 8
Solution: (d) By using K = t t d' ; here t = 4 × 10−5 m ; d' = 3.5 × 10−5 m ⇒ K = 4 × 10 −5 = 8
− 4 × 10 −5 − 3.5 × 10 −5
Example: 113 The force between the plates of a parallel plate capacitor of capacitance C and distance of separation of the
plates d with a potential difference V between the plates, is [MP PMT 1999]
(a) CV 2 (b) C2V 2 (c) C2V 2 (d) V 2d
2d 2d 2 d2 C
Solution: (a) Since F = Q2 ⇒ F = C2V 2 = CV 2 .
2ε 0 A 2ε 0 A 2d
Example: 114 A capacitor when filled with a dielectric K = 3 has charge Q0 , voltage V0 and field E0 . If the dielectric is
replaced with another one having K = 9, the new values of charge, voltage and field will be respectively
(a) 3Q0 , 3V0 , 3E0 (b) Q0 , 3V0 , 3E0 (c) Q0 , V0 , 3E0 (d) Q0 , V0 , E0
3 3 3
Solution: (d) Suppose, charge, potential difference and electric field for capacitor without dielectric medium are Q, V and E
respectively
With dielectric medium of K = 3 With dielectric medium of K = 9
Charge Q0 = Q Charge Q' = Q = Q0
Potential difference V0 = V Potential difference V' = V = V0
3 9 3
Electric field E0 = E Electric field E' = E = E0 .
3 9 3
Example: 115 A slab of copper of thickness b is inserted in between the plates of parallel plate capacitor as shown in the
figure. The separation between the plates is d. If b= d then the ratio of capacities of the capacitor after and
2
before inserting the slab will be
[IIT-JEE 1976; Similar to Orissa JEE 2002, KCET 2001, MP PMT 1994]
A = Area
d Cu b
(a) 2 : 1 (b) 2 : 1 (c) 1 : 1 (d) 1 : 2
Solution: (b) Capacitance before inserting the slab C = ε0A and capacitance after inserting the slab C' = ε0A .
d d−t
Where t = b = d so C' = 2ε 0 A hence, C' = 2 .
2 d C 1
Example: 116 The capacity of a parallel plate condenser is C0 . If a dielectric of relative permitivity ε r and thickness equal to
one fourth the plate separation is placed between the plates, then its capacity becomes C. The value of C
C0
will be
(a) 5ε r (b) 4ε r (c) 3ε r (d) 2ε r
4εr + 1 3ε r + 1 2ε r + 1 εr +1
Solution: (b) Initially capacitance C0 = ε0A ……..(i) Finally capacitance C = ε0A ……..(ii)
d d d/4
d − 4 + εr
By dividing equation (ii) by equation (i) C = 4ε r
C0 3ε r + 1
Tricky example: 16
An air capacitor of capacity C = 10µ F is connected to a constant voltage battery of 12 V. Now the
space between the plates is filled with a liquid of dielectric constant 5. The charge that flows now from
battery to the capacitor is [MP PMT 1997]
(a) 120µ C (b) 600µ C (c) 480µ C (d) 24µ C
Solution: (c) Initially charge on the capacitor Qi = 10 × 12 = 120µ C
When dielectric medium is filled, so capacitance becomes K times, i.e. new capacitance
C' = 5 × 10 = 50µ C
Final charge on the capacitor Q f = 50 × 12 = 600µ C
Hence additional charge supplied by the battery = Q f − Qi = 480µ C.
Grouping of Capacitors.
Series grouping Parallel grouping
(1) Charge on each capacitor remains same and equals to (1) Potential difference across +Q1 – –Q1
the main charge supplied by the battery each capacitor remains same + –
C2 C3 and equal to the applied + –
C1 +Q –Q +Q –Q potential difference + –
+Q –Q + – –Q2
+– + –– + –– Q1 –
+ –– ++ – ++ – Q2 +Q2 –
Q ++ – + – +– + –
V = V1 + V2 + V3 +
V1 V2 V3 Q = Q1 + Q2 + Q3 +
+
+– Q Q3 +Q3 – –Q3
+
V +–
(2) Equivalent capacitance +–
+–
1 111
Ceq = C1 + C2 + C3 or Ceq = (C1−1 + C −1 + C −1 ) −1
2 3
V
(2) Ceq = C1 + C2 + C3
(3) In series combination potential difference and energy
distribution in the reverse ratio of capacitance i.e., (3) In parallel combination charge and energy distributes in
the ratio of capacitance i.e. Q ∝ C and U ∝ C
V ∝ 1 and U ∝ 1 .
C C
(4) If two capacitors having capacitances C1 and C2 are
(4) If two capacitors having capacitance C1 and C2
connected in series then Ceq = C1C2 = Multiplication respectively are connected in parallel then
C1 + C2 Addition Ceq = C1 + C2
V1 = C1 C1 . V and V2 = C2 . V Q1 = C1 . Q and Q2 = C2 . Q
+ C2 C1 + C C1 + C2 C1 + C
2 2
(5) If n identical capacitors each having capacitances C are (5) If n identical capacitors are connected in parallel
connected in series with supply voltage V then
Equivalent capacitance Ceq = nC and Charge on each
Equivalent capacitance Ceq = C and Potential
n Q
V capacitor Q' = n
difference across each capacitor V'= n .
Redistribution of Charge Between Two Capacitors.
When a charged capacitor is connected across an uncharged capacitor, then redistribution of charge occur to
equalize the potential difference across each capacitor. Some energy is also
wasted in the form of heat.
Suppose we have two charged capacitors C1 and C2 after disconnecting C1 – – – –– – C2
these two from their respective batteries. These two capacitors are connected to Q1 ++ + ++ + Q2
each other as shown below (positive plate of one capacitor is connected to V2
positive plate of other while negative plate of one is connected to negative plate V1
of other)
Charge on capacitors redistributed and new charge on them will be Q1' = Q C1 C1 , Q ' = Q C1 C2
+ C2 2 +C
2
The common potential V = Q1 + Q2 = C1 V1 + C2V2 and loss of energy ∆U = 2 C1 C 2 ) (V1 − V2 )2
C1 + C2 C1 + C2 (C1 +
C 2
Note : ≅Two capacitors of capacitances C1 and C2 are charged to potential of V1 and V2 respectively. After
disconnecting from batteries they are again connected to each other with reverse polarity i.e., positive plate
of a capacitor connected to negative plate of other. So common potential V = Q1 − Q2 = C1 V1 − C2V2 .
C1 + C2 C1 + C2
Electrostatics- capacitors
Circuit With Resistors and Capacitors.
(1) A resistor may be connected either in series or in parallel with the capacitor as shown below
Series RC Circuit Parallel RC Circuit
CR R
C
S V0 S V0
In this combination capacitor takes longer time to charge. Resistor has no effect on the charging of capacitor.
The charging current is maximum in the beginning; it Resistor provides an alternative path for the electric
decreases with time and becomes zero after a long time. current.
(2) Three states of RC circuits
(i) Initial state : i.e., just after closing the switch or just after opening the switch.
(ii) Transient state : or instantaneous state i.e., any time after closing or opening the switch.
(iii) Steady state : i.e., a long time after closing or opening the switch. In the steady state condition, the
capacitor is charged or discharged.
(3) Charging and discharging of capacitor in series RC circuit : As shown in the following figure (i)
when switch S is closed, capacitor start charging. In this transient state potential difference appears across
capacitor as well as resistor. When capacitor gets fully charged the entire potential difference appeared across the
capacitor and nothing is left for the resistor. [shown in figure (ii)]
(i) C R (ii) C R Steady state
+– V′
Transient state i V +– V0 Q0 (Max charge)
V0 C
=
S
S + – + –
V0 V0
(i) Charging : In transient state of charging charge on the capacitor at any instant Q = Q0 1 − e −t and
RC
potential difference across the capacitor at any instant V = V0 1 − e −t
RC
Q0 V0
Q Q = Q0(1 – e–t/RC) VC VC = V0(1 – e–t/RC)
Ot
Ot
(ii) Discharging : After the completion of charging, if battery is removed capacitor starts discharging. In
transient state charge on the capacitor at any instant Q = Q0e −t / RC and potential difference cross the capacitor at
any instant V = V0e −t / CR .
Q0 V0
Q Q= Q0 e–t/RC VC VC = V0 e–t/RC
Ot Ot
(iii) Time constant (τ) : The quantity RC is called the time constant i.e., τ = RC .
In charging : It is defined as the time during which charge on the capacitor rises to 0.63 times (63%) the
maximum value. That is when t = τ = RC , Q = Q0 (1 − e −1) = 0.639 Q0 or
In discharging : It is defined as the time during which charge on a capacitor falls to 0.37 times (37%) of the
initial charge on the capacitor that is when t = τ = RC , Q = Q0 (e −1) = 0.37Q0
(iv) Mixed RC circuit : In a mixed RC circuit as shown below, when switch S is closed current flows
through the branch containing resistor as well as through the branch contains capacitor and resistor (because
capacitor is in the process of charging)
R1 i R1
C R2 C R2 No
current
iS R iS R
V0 V0
Transient state Steady state
When capacitor gets fully charged (steady state), no current flows through the line in which capacitor is
connected. Therefore the current through resistor R1 is V0 r) , hence potential difference across resistance will be
(R1 +
equal to V0 r ) R1 . The same potential difference will appear across the capacitor, hence charge on capacitor in
(R1 +
steady state Q = CV0 R1
(R1 + r)
Concepts
In series combination equivalent capacitance is always lesser than that of either of the individual capacitors e.g. If two capacitors
having capacitances 3µF and 6µF respectively are connected in series then equivalent capacitance
Ceq = 3×6 = 2µF Which is lesser than the smallest capacitance (3µF) of the network.
3+6
In parallel combination, equivalent capacitance is always greater than the maximum capacitance of either capacitor in network.
If n identical plates arranged as shown below, constitutes (n – 1) capacitors in series. If each capacitors having capacitance ε0A
d
then Ceq = ε0A + +–+– +– + –
(n − 1)d +–+– +– + ––
+–+– +– + –
–
+–+– +– +
In this situation except two extreme plates each plate is common to adjacent capacitors.
If n identical plates are arranged such that even no. of plates are connected together and odd number of plates are connected
together, then (n – 1) capacitors will be formed and they will be in parallel grouping.
246
1 3 57
Equivalent capacitance C' = (n − 1)C Where C = capacitance of a capacitor = ε0A
d
If n identical capacitors are connected in parallel which are charged to a potential V. If these are separated and connected in series then
potential difference of combination will be nV.
Network Solving.
To solve capacitive network for equivalent capacitance following guidelines should be followed.
Guideline 1. Identify the two points across which the equivalent capacitance is to be calculated.
Guideline 2. Connect (Imagine) a battery between these points.
Guideline 3. Solve the network from the point (reference point) which is farthest from the points
between which we have to calculate the equivalent capacitance. (The point is likely to be not a node)
(1) Simple circuits : Suppose equivalent capacitance is to be determined in the following networks
between points A and B
Suppose equivalent capacitance is to be determined in the following networks between points A and B
(i) 3µF 6µF 3 × 6 = 2µF
3µF 6µF ⇒ ⇒A 2µF C AB = 2 + 3 = 5µF
3µF B
B A 3µF B
A 3µF
(ii) 6µF ⇒ 6µF 6µF ⇒ 3 + 3 = 6µF 6 = 3µF CAB = 3 + 2 = 5μF
3µF 3µF 3µF 2 6µF
6µF 6µF 6µF
2µF 3µF ⇒ 6µF
B 6µF
6 = 3µF A A 2µF B A 2µF B
2
A 2µF B
(iii) 9µF 9µF 9µF 9µF 9µF 9µF Series 9µF 9µF
6µF 6µF A 6µF 9µF 6µF
A 9 = 3µF A
9µF 9µF 9µF 9µF 3
9µF + 6µF + 6µF 3µF
–
⇒ –
B B B 9µF Parallel
9µF 9µF 9µF 6 + 3 = 9µF
⇓
9µF 9µF
A
By similar process CAB = 3µF ⇐ + 6µF 9µF
–
B 9µF
9µF
(2) Circuits with extra wire : If there is no capacitor in any branch of a network then every point of this
branch will be at same potential. Suppose equivalent capacitance is to be determine in following cases
(i) A A C
CB
A CC C ⇒ CB C CB ⇒A C
(ii) B AA A B + – CAB = 3C
A BB No p.d. across
C vertical
A A B
C branch so it is
C⇒ C ⇒ C B removed
A C
C A BA CAB = 2C
B
CB
A
⇒
(iii) C C B C
C A C
A A B
C C
BA AC B C
+–
B
CAB = 3C
(iv) C C CC B C
C C B
A A ⇒ CC
B
⇒ C C
⇐
BA Parallel A B
A C + C = 2C +–
AC
⇓
Parallel Series
Hence equivalent capacitance between 2C/3 2C + C = 5C C 2C × C 2C
C 3 3 =
5C 2C 2C + C 3
A and B is 3 A
C
A B B
+– +–
(v) Since there is no capacitor in the path APB, the points A, P and B are electrically same i.e., the input and
CC
output points are directly connected (short circuited).
C ⇒ Thus, entire charge will prefer to flow along path APB. It means
A that the capacitors connected in the circuit will not receive any
B charge for storing. Thus equivalent capacitance of this circuit is
P zero.
(3) Wheatstone bride based circuit : If in a network five capacitors are arranged as shown in following
figure, the network is called wheatstone bridge type circuit. If it is balanced then C1 = C3 hence C5 is removed
C2 C4
and equivalent capacitance between A and B
(i) (ii) D (iii) C1
C1 C2
A D BA A E C5 C2
C1 C2 C5 B
C3 D B
CC55 C3 C4
C3 C4 E C4
E
C AB = C1C2 + C3C4
C1 + C2 C3 + C4
(4) Extended wheatstone bridge : The given figure consists of two wheatstone bridge connected
together. One bridge is connected between points AEGHFA and the other is connected between points EGBHFE.
This problem is known as extended wheatstone bridge problem, it has two branches EF and GH to the left
and right of which symmetry in the ratio of capacities can be seen.
It can be seen that ratio of capacitances in branches AE and EG is same as that between the capacitances of
the branches AF and FH. Thus, in the bridge AEGHFA; the branch EF can be removed. Similarly in the bridge
EGBHFE branch GH can be removed
CE CG C CE CG C
CC ⇒ B
A BA
CF CH C CF CH C C AB = 2C
3
(5) Infinite chain of capacitors : In the following figure equivalent capacitance between A and B
(i)
A C1 C1 Suppose the effective capacitance between A and B is CR.
C1 C2 C2 Since the network is infinite, even if we remove one pair
C2 ⇒ ∞ of capacitors from the chain, remaining network would still
have infinite pair of capacitors, i.e., effective capacitance
B between X and Y would also be CR
C1 X C1 Series
A A
CR C2 Parallel ⇒ CR C1(C2 + CR )
CR (C2 + CR) (C2 + CR) C1 + C2 + CR
B B
Y
Hence equivalent capacitance between A and B
C AB = C1(C2 + CR ) = CR ⇒ C AB = C2 1 + 4 C1 −
C1 + C2 + CR 2 C2 1
(ii) For what value of C0 in the circuit shown below will the net effective capacitance between A and B be
independent of the number of sections in the chain
A C1 C1 C1 C1 C Suppose there are n sections between A and B
C2 C2 C2 C2 and the network is terminated by C0 with
Parallel C0 equivalent capacitance CR. Now if we add one
C2 + C0 more sections to the network between D and C (as
C0
BA C C1 ⇓ shown in the following figure), the equivalent
CR D capacitance of the network CR will be independent
C0 C2
of number of sections if the capacitance between
D and C still remains C0 i.e.,
Hence C0 = C1 × (C2 + C0 ) ⇒ C0 2 + C2C0 − C1C2 = 0
C1 + C2 + C0
BD
On simplification C0 = C2 1 + 4 C1 −
2 C2 1
(6) Network with more than one cell :
(i) E1 E1 – E2
C1 C2 ⇒ C1 Potential difference across C1 is C2 (E1 − E2)
C1 + C2
E2 C2
and potential difference across C2 is
C1 (E1 − E2)
C1 + C2
(ii) Potential difference between the ends of this arrangement is given V. (E – V)
C1 +E – C2
C1 E C2 A B
⇒
A B⇒
+– C1 C2
V
(7) Advance case of compound dielectrics : If several dielectric medium filled between the plates of a
parallel plate capacitor in different ways as shown.
(i) 2 The system can be assumed to be made up of two capacitors C1 and C2 which may
be said to connected in series
1
A K1 K2 C1 = K1ε 0 A , C2 = K2ε 0 A and 1 11 ⇒ Ceq = 2K1 K 2 . ε0A
d d Ceq = C1 + C2 K1 + K2 d
22
d/2 d/2 Also K eq = 2K1 K 2
K1 + K2
(ii) In this case these two capacitors are in parallel and
C1 = K 1ε 0 A , C2 = K2ε 0 A
2d 2d
1
A/2 Hence, Ceq = C1 + C2 ⇒ Ceq = K1 + K2 . ε 0A
A/2 K1 2 d
K2
(iii) Also K eq = K1 + K2
2 2
d
In this case C1 and C2 are in series while this combination is in parallel with C3
C3 C1 = K1ε 0 A = K1ε 0 A , C2 = K1ε 0 A = K2ε 0 A and C3 = K 3ε 0 A = K3ε0 A
A/2 d 2 d d 2 d d 2 2d
K3 22 2
A/2 K1 K2 C1C2 k1ε 0 A × k 2ε 0 A
C22 C1 + C2 d d
C1 Hence, Ceq = + C3 = + k3ε 0 A ⇒ Ceq = k1k 2 + k3 . ε 0A
d/2 d/2 2d k1 + k2 2 d
k1ε 0 A k2ε 0 A
d + d
Also k eq = k3 + k1k 2
2 k1 + k2
Electrostatics- Capacitors
Example based on series and parallel grouping of capacitors
Example: 117 Three capacitors of 2µ f, 3µ f and 6µ f are joined in series and the combination is charged by means of a 24
volt battery. The potential difference between the plates of the 6µ f capacitor is
(a) 4 volts (b) 6 volts (c) 8 volts [MP PMT 2002 Similar to MP PMT 1996]
(d) 10 volts
Solution: (a) Equivalent capacitance of the network is 1 = 1 + 1 + 1
Ceq 2 3 6
2µF 3µF 6µF
Q +24V–
Ceq = 1µF
Charge supplied by battery Q = Ceq.V ⇒ 1 × 24 = 24 µC
Hence potential difference across 6µ F capacitor = 24 = 4 volt.
6
Example: 118 Two capacitors each of 1 µ f capacitance are connected in parallel and are then charged by 200 V D.C.
supply. The total energy of their charges in joules is [MP PMT 2002]
(a) 0.01 (b) 0.02 (c) 0.04 (d) 0.06
Solution: (c) By using formula U = 1 Ceq V 2 1µF
2
Here Ceq = 2µF 1µF
+200–V
∴ U = 1 × 2 × 10 −6 × (200)2
2
= 0.04 J
Example: 119 Five capacitors are connected as shown in the figure. The equivalent capacitance between the point A and B is
2µf [MP PMT 2002; SCRA 1996; Pantnagar 1987]
A
1µf 1µf 2µf
B
2µf
(a) 1 µ f (b) 2 µ f (c) 3 µ f (d) 4 µ f
Solution: (b) A 2µf A Parallel
+ B 1 + 1 = 2µF
– 1µf 1µf Series
2µf 1µf 1µf
2µf 2 = 1µF ⇒ 1µf
2
B 2µf
⇓
A Parallel A Series
1µf 1 + 1 = 2µF ⇐
1µf 2µf 2 = 1µF
1µf 2
B B 2µf
Hence equivalent capacitance between A and B is 2µF.
Example: 120 In the following network potential difference across capacitance of 4.5 µF is [RPET 2001; MP PET 1992]
3µF
4.5µF
6µF
+–
12 V
(a) 8 V (b) 4 V (c) 2 V (d) 6 V Parallel
3 + 6 = 9µF
9 × 4.5 3µF
Solution: (a) Equivalent capacitance Ceq = 9 + 4.5 = 3µF
4.5µF
Charge supplied by battery Q = Ceq × V = 3 × 12 = 36 µC
Hence potential difference across 4.5 µF = 36 = 8V. 6µF
4.5
+–
12 V
Example: 121 A parallel plate capacitor of area A, plate separation d and capacitance C is filled with three different dielectric
materials having dielectric constants K1, K2 and K3 as shown in fig. If a single dielectric material is to be used to
have the same capacitance C in this capacitor, then its dielectric constant K is given by [IIT Screening 2000]
A/2 A/2
K1 K2 d/2
d
K3
A
A = Area of plates
(a) 11 1 1 (b) 11 1
K = K1 + K2 + 2K3 K = K1 + K2 + 2K3
(c) K = K1 K 2 + 2K3 (d) K = K1 + K 2 + 2K 3
K1 + K2
Solution: (b) The effective capacitance is given by 1 = d 1 + (K1 1
Ceq + K 2 )
ε 0 A 2K 3
The capacitance of capacitor with single dielectric of dielectric constant K is C = Kε 0 A
d
According to question Ceq = C i.e., 1 ε0A 1 = Kε 0 A
2K + K1 + d
d
K2
3
⇒ 1 = 1 + K1 1 K2 .
K 2K 3 +
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