The second chapter of Class 12 Physics introduces you to Electrostatic Potential and Capacitance. Different electric fields possess varying electrostatic potential. This chapter informs you about the electric potential and its applications, potential difference, equipotential surfaces, the electrical potential energy of charges in an electrostatic field, etc. This chapter is comprised of a number of formulae and terms associated with the electrostatic potential. There are questions related to this topic at the end of the chapter for you to work out.
There are two charges,
Distance between the two charges, d = 16 cm = 0.16 m
Consider a point P on the line joining the two charges, as shown in the given figure.
r = Distance of point P from charge q1
Let the electric potential (V) at point P be zero.
Potential at point P is the sum of potentials caused by charges q1 and q2 respectively.
Where,
= Permittivity of free space
For V = 0, equation (i) reduces to
Therefore, the potential is zero at a distance of 10 cm from the positive charge between the charges.
Suppose point P is outside the system of two charges at a distance s from the negative charge, where potential is zero,
as shown in the following figure.
For this arrangement, potential is given by,
For V = 0, equation (ii) reduces to
Therefore, the potential is zero at a distance of 40 cm from the positive charge outside the system of charges.
The given figure shows six equal amount of charges, q, at the vertices of a regular hexagon.
Where, Charge, q = 5 µC = 5 × 10- 6 C
Side of the hexagon, l = AB = BC = CD = DE = EF = FA = 10 cm
Distance of each vertex from centre O, d = 10 cm
Electric potential at point O,
Where,
= Permittivity of free space
Therefore, the potential at the centre of the hexagon is 2.7 × 106 V.
(a) The situation is represented in the given figure.
An equipotential surface is the plane on which total potential is zero everywhere. This plane is normal to line AB. The plane is located at the mid-point of line AB because the magnitude of charges is the same.
(b) The direction of the electric field at every point on this surface is normal to the plane in the direction of AB.
(a) Radius of the spherical conductor, r = 12 cm = 0.12 m
Charge is uniformly distributed over the conductor, q = 1.6 × 10 -7 C
Electric field inside a spherical conductor is zero. This is because if there is field inside the conductor, then charges will move to neutralize it.
(b) Electric field E just outside the conductor is given by the relation,
Where,
= Permittivity of free space
Therefore, the electric field just outside the sphere is 105 N C-1
(c) Electric field at a point 18 m from the centre of the sphere = E1
Distance of the point from the centre, d = 18 cm = 0.18 m
Therefore, the electric field at a point 18 cm from the centre of the sphere is .
4.4 x 104 N/C
Capacitance between the parallel plates of the capacitor, C = 8 pF
Initially, distance between the parallel plates was d and it was filled with air. Dielectric constant of air, k = 1
Capacitance, C, is given by the formula,
Where, A = Area of each plate
= Permittivity of free space
If distance between the plates is reduced to half, then new distance, d' = d / 2
Dielectric constant of the substance filled in between the plates,
= 6
Hence, capacitance of the capacitor becomes
Taking ratios of equations (i) and (ii), we obtain
Therefore, the capacitance between the plates is 96 pF.
(a) Capacitance of each of the three capacitors, C = 9 pF
Equivalent capacitance (C') of the combination of the capacitors is given by the relation,
Therefore, total capacitance of the combination is 3µF
(b) Supply voltage, V = 100 V
Potential difference (V') across each capacitor is equal to one-third of the supply voltage.
Therefore, the potential difference across each capacitor is 40 V.
(a) Capacitances of the given capacitors are
C1 = 2 pF,
C2 = 3 pF,
C3 = 4 pF,
For the parallel combination of the capacitors, equivalent capacitoris C' given by the algebraic sum,
C'=2 +3 + 4= 9 pF
Therefore, total capacitance of the combination is 9 pF.
(b) Supply voltage, V = 100 V
The voltage through all the three capacitors is same = V = 100 V
Charge on a capacitor of capacitance C and potential difference V is given by the relation,
q = VC … (i)
For C = 2 pF,
Charge= VC = 100 x 2 = 200 pC = 2 x 10-10 C
For C = 3 pF,
Charge = VC = 100 x 3 = 300 pC = 3 x 10-10 C
For C = 4 pF
Charge = VC = 100 x 4 = 400 pC = 4 x 10-10 C
Area of each plate of the parallel plate capacitor, A = 6 × 10-3 m2
Distance between the plates, d = 3 mm = 3 × 10-3m
Supply voltage, V = 100 V
Capacitance C of a parallel plate capacitor is given by,
Where, 
= Permittivity of free space
= 8.854 × 10-12 N-1 m-2 C-2
Therefore, capacitance of the capacitor is 17.71 pF and charge on each plate is 1.771 × 10-9 C.
(a) Dielectric constant of the mica sheet, k = 6
Initial capacitance, C = 1.771 × 10 -11 F
New Capacitance, C'= kC= 6x1.771x10-11 =106 pF
Supply voltage, V = 100 V
New Charge, q'=C'V=6x1.771x10-9 =1.06x 10-8 C
Potential across the plates remains 100 V
(b) Dielectric constant, k = 6
Initial capacitance, C = 1.771 × 10-11 F
New Capacitance, C'= kC= 6x1.771x10-11 =106 pF
If supply voltage is removed, then there will be no effect on the amount of charge in the plates.
Charge = 1.771 × 10- 9 C
Potential across the plates is given by,
Capacitor of the capacitance, C = 12 pF = 12 × 10 - 12 F
Potential difference, V = 50 V
Electrostatic energy stored in the capacitor is given by the relation,
Therefore, the electrostatic energy stored in the capacitor is 1.5 x10-8 J
Capacitance of the capacitor, C = 600 pF
Potential difference, V = 200 V
Electrostatic energy stored in the capacitor is given by,
If supply is disconnected from the capacitor and another capacitor of capacitance C = 600 pF is connected to it, then equivalent capacitance (C') of the combination is given by,
New electrostatic energy can be calculated as
Loss in electrostatic enegy = E - E'
= 1.2 x 10-5 - 0.6 x 10-5
= 0.6 x 10-5
= 6 x 10-6 J
Therefore, the electrostatic energy lost in the process is 6 x 10-6 J.
Charge located at the origin, q = 8 mC= 8 × 10 - 3 C
Magnitude of a small charge, which is taken from a point P to point R to point Q, q1 = - 2 × 10 - 9 C
All the points are represented in the given figure.
Point P is at a distance, d1 = 3 cm, from the origin along z-axis.
Point Q is at a distance, d2 = 4 cm, from the origin along y-axis.
Potential at point P, 
Potential at point Q, 
Work done (W) by the electrostatic force is independent of the path.
Therefore, work done during the process is 1.27 J.
Length of the side of a cube = b
Charge at each of its vertices = q
A cube of side b is shown in the following figure.
d = Diagonal of one of the six faces of the cube
l = Length of the diagonal of the cube
The electric potential (V) at the centre of the cube is due to the presence of eight charges at the vertices.
Therefore, the potential at the centre of the cube is 
The electric field at the centre of the cube, due to the eight charges, gets cancelled. This is because the charges are distributed symmetrically with respect to the centre of the cube. Hence, the electric field is zero at the centre.
Two charges placed at points A and B are represented in the given figure. O is the mid-point of the line joining the two charges.
Magnitude of charge located at A, q1 = 1.5 μC
Magnitude of charge located at B, q2 = 2.5 μC
Distance between the two charges, d = 30 cm = 0.3 m
(a) Let V1 and E1 are the electric potential and electric field respectively at O
V1 = Potential due to charge at A + Potential due to charge at B
Where, ∈0 = Permittivity of free space
E1 = Electric field due to q2 - Electric field due to q1
Therefore, the potential at mid-point is 2.4 × 105 V and the electric field at mid-point is 4 × 105 V m - 1. The field is directed from the larger charge to the smaller charge.
(b) Consider a point Z such that normal distanceOZ = 10 cm = 0.1 m, as shown in the following figure.
V2 and E2 are the electric potential and electric field respectively at Z.
It can be observed from the figure that distance,
V2= Electric potential due to A + Electric Potential due to B
Electric field due to q at Z,
Electric field due to q2 at Z,
The resultant field intensity at Z,
Where, 2θis the angle, ∠AZ B
From the figure, we obtain
Therefore, the potential at a point 10 cm (perpendicular to the mid-point) is 2.0 × 105 V and electric field is 6.6 ×105 V m -1
(a) Charge placed at the centre of a shell is +q. Hence, a charge of magnitude - q will be induced to the inner surface of the shell. Therefore, total charge on the inner surface of the shell is - q.
Surface charge density at the inner surface of the shell is given by the relation,
A charge of +q is induced on the outer surface of the shell. A charge of magnitude Q is placed on the outer surface of the shell. Therefore, total charge on the outer surface of the shell is Q + q. Surface charge density at the outer surface of the shell,
(b) Yes
The electric field intensity inside a cavity is zero, even if the shell is not spherical and has any irregular shape. Take a closed loop such that a part of it is inside the cavity along a field line while the rest is inside the conductor. Net work done by the field in carrying a test charge over a closed loop is zero because the field inside the conductor is zero. Hence, electric field is zero, whatever is the shape.
Electric field on one side of a charged body is E1 and electric field on the other side of the same body is E2. If infinite plane charged body has a uniform thickness, then electric field due to one surface of the charged body is given by,
Where,
= Unit vector normal to the surface at a point
σ = Surface charge density at that point
Electric field due to the other surface of the charged body,
Electric field at any point due to the two surfaces,
Since inside a closed conductor, 
= 0,
Therefore, the electric field just outside the conductor is 
(b) When a charged particle is moved from one point to the other on a closed loop, the work done by the electrostatic field is zero. Hence, the tangential component of electrostatic field is continuous from one side of a charged surface to the other
Charge density of the long charged cylinder of length L and radius r is λ.
Another cylinder of same length surrounds the pervious cylinder. The radius of this cylinder is R.
Let E be the electric field produced in the space between the two cylinders.
Electric flux through the Gaussian surface is given by Gauss's theorem as,
Where, d = Distance of a point from the common axis of the cylinders
Let q be the total charge on the cylinder.
It can be written as
Where, q = Charge on the inner sphere of the outer cylinder
∈0 = Permittivity of free space
Therefore, the electric field in the space between the two cylinders is 
The distance between electron-proton of a hydrogen atom, d = 0.53 Å
Charge on an electron, q1 = - 1.6 ×10 -19 C
Charge on a proton, q2 = +1.6 ×10-19 C
(a) Potential at infinity is zero.
Potential energy of the system, = Potential energy at infinity - Potential energy at distance d
where,
∈0 is the permittivity of free space
The system of two protons and one electron is represented in the given figure.
Charge on proton 1, q1 = 1.6 ×10 - 19 C
Charge on proton 2, q2 = 1.6 ×10 - 19 C
Charge on electron, q3 = - 1.6 ×10 - 19 C
Distance between protons 1 and 2, d1 = 1.5 ×10 - 10 m
Distance between proton 1 and electron, d2 = 1 ×10 - 10 m
Distance between proton 2 and electron, d3 = 1 × 10 - 10 m
The potential energy at infinity is zero.
Potential energy of the system,
Therefore, the potential energy of the system is - 19.2 eV.
Let a be the radius of a sphere A, QA be the charge on the sphere, and CA be the capacitance of the sphere. Let b be the radius of a sphere B, QB be the charge on the sphere, and CB be the capacitance of the sphere. Since the two spheres are connected with a wire, their potential (V) will become equal.
Let EA be the electric field of sphere A and EB be the electric field of sphere B. Therefore, their ratio,
Putting the value of (2) in (1), we obtain
Therefore, the ratio of electric fields at the surface is 
(a) Zero at both the points
Charge - q is located at (0, 0, - a) and charge + q is located at (0, 0, a). Hence, they form a dipole. Point (0, 0, z) is on the axis of this dipole and point (x, y, 0) is normal to the axis of the dipole. Hence, electrostatic potential at point (x, y, 0) is zero.
Electrostatic potential at point (0, 0, z) is given by,
Where, 
= Permittivity of free space
p = Dipole moment of the system of two charges = 2qa
(b) Distance r is much greater than half of the distance between the two charges. Hence, the potential (V) at a distance r is inversely proportional to square of the distance i.e., 
(c) Zero
The answer does not change if the path of the test is not along the x-axis.
A test charge is moved from point (5, 0, 0) to point ( - 7, 0, 0) along the x-axis. Electrostatic potential (V1) at point (5, 0, 0) is given by.
Four charges of same magnitude are placed at points X, Y, Y, and Z respectively, as shown in the following figure.
A point is located at P, which is r distance away from point Y.
The system of charges forms an electric quadrupole.
It can be considered that the system of the electric quadrupole has three charges.
Charge +q placed at point X
Charge - 2q placed at point Y
Charge +q placed at point Z
XY = YZ = a
YP = r
PX = r + a
PZ = r - a
Electrostatic potential caused by the system of three charges at point P is given by,
Since,
is taken as negligible.
It can be inferred that potential,
However, it is known that for a dipole, 
And, for a monopole,
Total required capacitance, C = 2 µF
Potential difference, V = 1 kV = 1000 V
Capacitance of each capacitor, C1 = 1 µF
Each capacitor can withstand a potential difference, V1 = 400 V
Suppose a number of capacitors are connected in series and these series circuits are connected in parallel (row) to each other. The potential difference across each row must be 1000 V and potential difference across each capacitor must be 400 V. Hence, number of capacitors in each row is given as
Hence, there are three capacitors in each row.
Capacitance of each row 
Let there are n rows, each having three capacitors, which are connected in parallel. Hence, equivalent capacitance of the circuit is given as
Hence, 6 rows of three capacitors are present in the circuit. A minimum of 6 × 3 i.e., 18 capacitors are required for the given arrangement.
Capacitance of a parallel capacitor, V = 2 F
Distance between the two plates, d = 0.5 cm = 0.5 × 10 - 2 m
Capacitance of a parallel plate capacitor is given by the relation,
Where,
= Permittivity of free space = 8.85 × 10 - 12 C2 N - 1 m - 2
Hence, the area of the plates is too large. To avoid this situation, the capacitance is taken in the range of µF.
Capacitance of capacitor C1 is 100 pF.
Capacitance of capacitor C2 is 200 pF.
Capacitance of capacitor C3 is 200 pF.
Capacitance of capacitor C4 is 100 pF.
Supply potential, V = 300 V
Capacitors C2 and C3 are connected in series. Let their equivalent capacitance be C' .
Capacitors C1 and C' are in parallel. Let their equivalent capacitance be
C" and C4 are connected in series. Let their equivalent capacitance be C.
Hence, the equivalent capacitance of the circuit is 
Potential difference across C" = V"
Potential difference across C4 = V4
Charge on C4 is given by
Q4= CV
Hence, potential difference, V1, across C1 is 100 V.
Charge on C1 is given by,
Q1 = C1V1
= 100 x 10-12 x 100
= 10-8 C
C2 and C3 having same capacitances have a potential difference of 100 V together. Since C2 and C3 are in series,
the potential difference across C2 and C3 is given by,
V2= V3 = 50 V
Therefore, charge on C2 is given by,
Q2 = C2V2
= 200 x 10-12 x 50
= 10-8 C
And charge on C3 is given by,
Q3 = C3V3
= 200 x 10-12 x 50
= 10-8 C
Hence, the equivalent capacitance of the given circuit is
Area of the plates of a parallel plate capacitor, A = 90 cm2 = 90 × 10 - 4 m2
Distance between the plates, d = 2.5 mm = 2.5 × 10 - 3 m
Potential difference across the plates, V = 400 V
(a) Capacitance of the capacitor is given by the relation,
Electrostatic energy stored in the capacitor is given by the relation,
Where,
= Permittivity of free space = 8.85 × 10 - 12 C2 N - 1 m - 2
Hence, the electrostatic energy stored by the capacitor is 2.55 x 10-6 J
(b) Volume of the given capacitor,
Energy stored in the capacitor per unit volume is given by,
Where,
= Electric intensity = E
Capacitance of a charged capacitor, C1=4µF = 4 x 10-6 F
Supply voltage, V1 = 200 V
Electrostatic energy stored in C1 is given by,
Capacitance of an uncharged capacitor, C2=2µF = 2 x 10-6 F
When C2 is connected to the circuit, the potential acquired by it is V2.
According to the conservation of charge, initial charge on capacitor C1 is equal to the final charge on capacitors, C1 and C2.
Electrostatic energy for the combination of two capacitors is given by,
Hence, amount of electrostatic energy lost by capacitor C1
= E1 - E2
= 0.08 - 0.0533 = 0.0267
=2.67 × 10 - 2 J
Let F be the force applied to separate the plates of a parallel plate capacitor by a distance of Δx.
Hence, work done by the force to do so = FΔx
As a result, the potential energy of the capacitor increases by an amount given as uAΔx.
Where,
u = Energy density
A = Area of each plate
d = Distance between the plates
V = Potential difference across the plates
The work done will be equal to the increase in the potential energy i.e.,
Electric intensity is given by,
However, capacitance,
Charge on the capacitor is given by,
Q = CV
The physical origin of the factor,
, in the force formula lies in the fact that just outside the conductor, field is E and inside it is zero. Hence, it is the average value,
, of the field that contributes to the force.
Radius of the outer shell = r1
Radius of the inner shell = r2
The inner surface of the outer shell has charge +Q.
The outer surface of the inner shell has induced charge - Q.
Potential difference between the two shells is given by,
Hence, proved.
Radius of the inner sphere, r2 = 12 cm = 0.12 m
Radius of the outer sphere, r1= 13 cm = 0.13 m
Charge on the inner sphere, q = 2.5 μC = 2.5 x 10-6 C
Dielectric constant of a liquid, ∈r = 32
(a) Capacitance of the capacitor is given by the relation,
Where,
∈0 = Permittivity of free space = 8.85 x 10-12 C2 N-1 m-2
Hence, the capacitance of the capacitor is approximately 5.5 x 10-9 F.
(b) Potential of the inner sphere is given by,
Hence, the potential of the inner sphere is 4.5 x 102 V.
(c) Radius of an isolated sphere, r = 12 x 10-2 m
Capacitance of the sphere is given by the relation,
The capacitance of the isolated sphere is less in comparison to the concentric spheres. This is because the outer sphere of the concentric spheres is earthed. Hence, the potential difference is less and the capacitance is more than the isolated sphere.
(a) The force between two conducting spheres is not exactly given by the expression, Q1Q2/4π∈0r2, because there is a non-uniform charge distribution on the spheres.
(b) Gauss's law will not be true, if Coulomb's law involved 1/r3 dependence, instead of1/r2, on r.
(c) Yes,
If a small test charge is released at rest at a point in an electrostatic field configuration, then it will travel along the field lines passing through the point, only if the field lines are straight. This is because the field lines give the direction of acceleration and not of velocity.
(d) Whenever the electron completes an orbit, either circular or elliptical, the work done by the field of a nucleus is zero.
(e) No
Electric field is discontinuous across the surface of a charged conductor. However, electric potential is continuous.
(f) The capacitance of a single conductor is considered as a parallel plate capacitor with one of its two plates at infinity.
(g) Water has an unsymmetrical space as compared to mica. Since it has a permanent dipole moment, it has a greater dielectric constant than mica.
Length of a co-axial cylinder, l = 15 cm = 0.15 m
Radius of outer cylinder, r1 = 1.5 cm = 0.015 m
Radius of inner cylinder, r2 = 1.4 cm = 0.014 m
Charge on the inner cylinder, q = 3.5 µC = 3.5 x 10 - 6 C
Capacitance of a co-axial cylinder of radii r1 and r2 is
given by the relation,
Where,
∈0 = Permittivity of free space = 8.85 x 10-12 x N-1m-2C2
Potential difference of the inner cylinder is given by,
Potential rating of a parallel plate capacitor, V = 1 kV = 1000 V
Dielectric constant of a material, ∈r=3
Dielectric strength = 107 V/m
For safety, the field intensity never exceeds 10% of the dielectric strength.
Hence, electric field intensity, E = 10% of 107 = 106 V/m
Capacitance of the parallel plate capacitor, C = 50 pF = 50 × 10-12 F
Distance between the plates is given by,
Capacitance is given by the relation,
Where,
A = Area of each plate
∈0 = Permittivity of free space = 8.85 x10-12 N-1C2m-2
Hence, the area of each plate is about 19 cm2.
(a) Equidistant planes parallel to the x-y plane are the equipotential surfaces.
(b) Planes parallel to the x-y plane are the equipotential surfaces with the exception that when the planes get closer, the field increases.
(c) Concentric spheres centered at the origin are equipotential surfaces.
(d) A periodically varying shape near the given grid is the equipotential surface. This shape gradually reaches the shape of planes parallel to the grid at a larger distance.
Potential difference, V = 15 x 106 V
Dielectric strength of the surrounding gas = 5 x 107 V/m
Electric field intensity, E = Dielectric strength = 5 x 107 V/m
Minimum radius of the spherical shell required for the purpose is given by,
Hence, the minimum radius of the spherical shell required is 30 cm.
According to Gauss's law, the electric field between a sphere and a shell is determined by the charge q1 on a small sphere. Hence, the potential difference, V, between the sphere and the shell is independent of charge q2.For positive charge q1, potential difference V is always positive.
(a) We do not get an electric shock as we step out of our house because the original equipotential surfaces of open air changes, keeping our body and the ground at the same potential.
(b) Yes, the man will get an electric shock if he touches the metal slab next morning. The steady discharging current in the atmosphere charges up the aluminium sheet. As a result, its voltage rises gradually. The raise in the voltage depends on the capacitance of the capacitor formed by the aluminium slab and the ground.
(c) The occurrence of thunderstorms and lightning charges the atmosphere continuously. Hence, even with the presence of discharging current of 1800 A, the atmosphere is not discharged completely. The two opposing currents are in equilibrium and the atmosphere remains electrically neutral.
(d) During lightning and thunderstorm, light energy, heat energy, and sound energy are dissipated in the atmosphere.
is from side 1 to side 2.) Hence show that just outside a conductor, the electric field is σ 
