NCERT Solutions Class 12 Physics Chapter 2 Electrostatic Potential And Capacitance

Question 2.1: Two charges 5 × 10⁻⁸ C and −3 × 10⁻⁸ C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.
Answer There are two charges,
q₁ = 5 x 10^-8 C
q₂ = - 3 x 10^-8 C
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.

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.

Therefore, the potential is zero at a distance of 40 cm from the positive charge outside the system of charges.

Question 2.2: A regular hexagon of side 10 cm has a charge 5 5C at each of its vertices. Calculate the potential at the centre of the hexagon.
Answer The given figure shows six equal amount of charges, q, at the vertices of a regular hexagon.

Where,
Charge, q = 5 5C = 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,

Therefore, the potential at the centre of the hexagon is 2.7 × 10⁶ V.

Question 2.3: Two charges 2 μC and −2 5C are placed at points A and B 6 cm apart.
(a) Identify an equipotential surface of the system.
(b) What is the direction of the electric field at every point on this surface?
Answer (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.

Question 2.4: A spherical conductor of radius 12 cm has a charge of 1.6 × 10−7C distributed uniformly on its surface. What is the electric field
(a) Inside the sphere
(b) Just outside the sphere
(c) At a point 18 cm from the centre of the sphere?
Answer (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,

Therefore, the electric field just outside the sphere is .
(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

4.4 x 10⁴ N/C
Therefore, the electric field at a point 18 cm from the centre of the sphere is 4.4 x 10⁴ N/C.

Question5. A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF = 10−12 F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?
Solution : Given: Capacitance of capacitor when medium between two plates is air, C = 8 pF = 8×10^–12 F

Question 2.6: Three capacitors each of capacitance 9 pF are connected in series.
(a) What is the total capacitance of the combination?
(b) What is the potential difference across each capacitor if the combination is connected to a 120 V supply?
Answer (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 3μF is .
(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.

Question7. Three capacitors of capacitances 2 pF, 3 pF and 4 pF are connected in parallel.
(a) What is the total capacitance of the combination?
(b) Determine the charge on each capacitor if the combination is connected to a 100 V supply.

Solution :Given, C₁ = 2 pF
 C₂ = 3 pF
 C₃ = 4 pF , V = 100 volt
Total capacitance of the parallel combination is
C = C₁ + C₂ + C₃ = 2 + 3 + 4 = 9 pF
Let q₁ , q₂ and q₃ be that charges on the capacitor C_{1 ,} C₂ and C₃ respectively.
In the parallel combination the potential difference across each capacitor will be equal to the supply voltage i.e., 100 V
⇒ q₁ = C₁V = 2 x 10^-12×100 = 2× 10^-10 C
⇒ q₂ = C₂V = 3 x 10^-12×100 = 3× 10^-10 C
⇒ q₃ = C₃V = 4 x 10^-12×100 = 4× 10^-10 C

Question8. In a parallel plate capacitor with air between the plates, each plate has an area of 6 × 10⁻³ m² and the distance between the plates is 3 mm. Calculate the capacitance of the capacitor. If this capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor?
Solution :

Question9. Explain what would happen if in the capacitor given in Exercise 2.8, a 3 mm thick mica sheet (of dielectric constant = 6) were inserted between the plates,
(a) While the voltage supply remained connected.
(b) After the supply was disconnected.
Solution :(a) When the voltage supply remains connected:
The capacitance of the capacitor will become K times.
Therefore, C’ = kC
Where k = dielectric constant = 6×17.7pF = 106.2 pF
The potential difference across the two plates of the capacitor will remain equal to the supply voltage i.e. 100 V
The charge on the capacitor,
q’ = C’V = 160.2 x 10^-12 x 100
 = 1.602 x 10^-8 C
 (b) After the voltage supply is disconnected:
As calculated above, the capacitance of the capacitor, C’ = 106.2 pF
The potential difference will decrease on introducing mica sheet by a factor of K,
⇒ V = K/V 100 = 100/6 = 16.67 V
The charge on the capacitor, 
q' = C'V' = 106.2 x 10^-12 x 16.67 
q' = 1.77 x 10-9 C

Question10. A 12 pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor?
Solution :Given, C = 12 pF = 12 x 10^-12 F
 V = 50 V
The electrostatic energy stored in the capacitor,
W = (½) CV² = (½) × 12 × 10^-12× (50)² = 1.5 × 10^-8 J

Question11. A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process?
Solution :Given, C₁ = 600 pF = 600 x 10^-12 F
 V₁ = 200 V
Energy stored in the capacitor,
U₁ = (½) C₁ (V₁)² = (½)×600× 10^-12× (200)²
 = 12×10^-6 J
When this charged capacitor is connected to another uncharged capacitor C₂ ( = 600 pF) ,they will share charges, till potential differences across their plates become equal.
Total charge on the two capacitors,
q = C₁V₁ + C₂V₂ = 600 × 10^-12× 200 + 0
 = 12 ×10^-8 C
Total capacitance of the two capacitors,
C = C₁ + C₂ = 600 pF + 600 pF
 = 1200 pF
 = 1200 x 10^-12 F

Question 2.12: A charge of 8 mC is located at the origin. Calculate the work done in taking a small charge of −2 × 10⁻⁹ C from a point P (0, 0, 3 cm) to a point Q (0, 4 cm, 0), via a point R
(0, 6 cm, 9 cm). 
Answer 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.

Therefore, work done during the process is 1.27 J.

Question 2.13: A cube of side b has a charge q at each of its vertices. Determine the potential and electric field due to this charge array at the centre of the cube.
Answer 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.

The electric potential (V) at the centre of the cube is due to the presence of eight charges at the vertices.

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.

Question 2.14: Two tiny spheres carrying charges 1.5 μC and 2.5 μC are located 30 cm apart. Find the potential and electric field:
(a) at the mid-point of the line joining the two charges, and
(b) at a point 10 cm from this midpoint in a plane normal to the line and passing through the mid-point.
Answer Two charges placed at points A and B are represented in the given figure. O is the midpoint 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

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.

Where, 2θ is 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.

Question15. A spherical conducting shell of inner radius r1 and outer radius r2 has a charge Q.
(a) A charge q is placed at the centre of the shell. What is the surface charge density on the inner and outer surfaces of the shell?
(b) Is the electric field inside a cavity (with no charge) zero, even if the shell is not spherical, but has any irregular shape? Explain.
Solution :

Question 2.16: (a) Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by

Where n is a unit vector normal to the surface at a point and σ is the surface charge density at that point. (The direction of n is from side 1 to side 2.) Hence show that just outside a conductor, the electric field is σ n/∈₀
(b) Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another. [Hint: For (a), use Gauss’s law. For, (b) use the fact that work done by electrostatic field on a closed loop is zero.]
Answer (a) 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,

Therefore, the electric field just outside the conductor is σ (n/∈₀)
(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.

Question 2.17: A long charged cylinder of linear charged density λ is surrounded by a hollow co-axial conducting cylinder. What is the electric field in the space between the two cylinders? 
Answer 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,
ϕ = E (2πd)L
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
∴ ϕ = E (2πdL) q/∈0

Question 2.18: In a hydrogen atom, the electron and proton are bound at a distance of about 0.53 Å:
(a) Estimate the potential energy of the system in eV, taking the zero of the potential energy at infinite separation of the electron from proton.
(b) What is the minimum work required to free the electron, given that its kinetic energy in the orbit is half the magnitude of potential energy obtained in (a)?
(c) What are the answers to (a) and (b) above if the zero of potential energy is taken at 1.06 Å separation?
Answer The distance between electron-proton of a hydrogen atom, d = 0.53A
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, p-e = Potential energy at infinity − Potential energy at distance, d

Therefore, the potential energy of the system is −27.2 eV.
(b) Kinetic energy is half of the magnitude of potential energy.

Question 2.19: If one of the two electrons of a H₂ molecule is removed, we get a hydrogen molecular ion H₂. In the ground state of an H₂, the two protons are separated by roughly 1.5 Å, and the electron is roughly 1 Å from each proton. Determine the potential energy of the system. Specify your choice of the zero of potential energy.
Answer

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.

Question 2.20: Two charged conducting spheres of radii a and b are connected to each other by a wire. What is the ratio of electric fields at the surfaces of the two spheres? Use the result obtained to explain why charge density on the sharp and pointed ends of a conductor is higher than on its flatter portions.
Answer 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 EAbe the electric field of sphere A and EB be the electric field of sphere B. Therefore, their ratio,

Question21. Two charges −q and +q are located at points (0, 0, − a) and (0, 0, a), respectively.
(a) What is the electrostatic potential at the points?
(b) Obtain the dependence of potential on the distance r of a point from the origin when r/a >> 1.
(c) How much work is done in moving a small test charge from the point (5, 0, 0) to (−7, 0, 0) along the x-axis? Does the answer change if the path of the test charge between the same points is not along the x-axis?
Solution : (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,
∈₀ = 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.,V ∝ 1/r²
(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,

Hence, no work is done in moving a small test charge from point (5, 0, 0) to point (−7,0, 0) along the x-axis.
The answer does not change because work done by the electrostatic field in moving a test charge between the two points is independent of the path connecting the two points.

Question 2.22: Figure 2.34 shows a charge array known as an electric quadrupole. For a point on the axis of the quadrupole, obtain the dependence of potential on r for r/a >> 1, and contrast your results with that due to an electric dipole, and an electric monopole (i.e., a single charge).

Answer 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,

Question23. An electrical technician requires a capacitance of 2 µF in a circuit across a potential difference of 1 kV. A large number of 1 µF capacitors are available to him each of which can withstand a potential difference of not more than 400 V. Suggest a possible arrangement that requires the minimum number of capacitors.
Solution :Potential difference across the circuit = 1kV = 1000V
Capacitance of each capacitor = 1 μF
Potential difference each capacitor can withstand = 400V
Capacitance required across the circuit = 2 μF
Assume n number of capacitors are connected in series and further m number of such series circuits are connected in parallel to each other.
As the potential difference in the circuit is 1000V so the potential difference across each row of n capacitors is 1000V, as the potential difference each capacitor can withstand is 400V,
Therefore, 400V × n = 1000V
⇒ n = 1000V/400V = 2.5~3capacitors in each row.
Now,

Question24. What is the area of the plates of a 2 F parallel plate capacitor, given that the separation between the plates is 0.5 cm? [You will realize from your answer why ordinary capacitors are in the range of µF or less. However, electrolytic capacitors do have a much larger capacitance (0.1 F) because of very minute separation between the conductors.]
Solution : Let the area of the plates of capacitor be A.
Capacitance of a parallel capacitor (C) = 2 F
Distance between the two plates (d) = 0.5cm = 0.005m
Capacitance of a parallel plate capacitor is given by the relation,
C = ∈₀A/d
A = Cd/∈₀
Eg = Absolute Permittivity of free space = 8.85 x 10^-12 C²N^-1m^-2

As the area of the plates required becomes too large so the capacitance is taken in the range of µF

 

 

 

 

Question34. Describe schematically the equipotential surfaces corresponding to
(a) a constant electric field in the z-direction,
(b) a field that uniformly increases in magnitude but remains in a constant (say, z) direction,
(c) a single positive charge at the origin, and
(d) a uniform grid consisting of long equally spaced parallel charged wires in a plane.
Solution : (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.

Question35. In a Van de Graaff type generator a spherical metal shell is to be a 15 × 106 V electrode. The dielectric strength of the gas surrounding the electrode is 5 × 107 Vm−1. What is the minimum radius of the spherical shell required? (You will learn from this exercise why one cannot build an electrostatic generator using a very small shell which requires a small charge to acquire a high potential.)
Solution : Potential difference, V = 15 × 10⁶ V
Dielectric strength of the surrounding gas = 5 × 10⁷ V/m
Electric field intensity, E = Dielectric strength = 5 × 10⁷ V/m
Minimum radius of the spherical shell required for the purpose is given by,
r = V/E
∴ r = 15 x 10⁶ V/5 x 10⁷ Vm^-1 = 0.3m
Hence, the minimum radius of the spherical shell required is 30 cm.

Question36. A small sphere of radius r₁ and charge q₁ is enclosed by a spherical shell of radius r₂ and charge q2. Show that if q₁ is positive, charge will necessarily flow from the sphere to the shell (when the two are connected by a wire) no matter what the charge q₂ on the shell is.
Solution : 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.

Question37. Answer the following:
(a) The top of the atmosphere is at about 400 kV with respect to the surface of the earth, corresponding to an electric field that decreases with altitude. Near the surface of the earth, the field is about 100 Vm⁻¹. Why then do we not get an electric shock as we step out of our house into the open? (Assume the house to be a steel cage so there is no field inside!)
(b) A man fixes outside his house one evening a two metre high insulating slab carrying on its top a large aluminium sheet of area 1m². Will he get an electric shock if he touches the metal sheet next morning?
(c) The discharging current in the atmosphere due to the small conductivity of air is known to be 1800 A on an average over the globe. Why then does the atmosphere not discharge itself completely in due course and become electrically neutral? In other words, what keeps the atmosphere charged?
(d) What are the forms of energy into which the electrical energy of the atmosphere is dissipated during a lightning? (Hint: The earth has an electric field of about 100 Vm⁻¹ at its surface in the downward direction, corresponding to a surface charge density = −10⁻⁹ C m⁻². Due to the slight conductivity of the atmosphere up to about 50 km (beyond which it is good conductor), about + 1800 C is pumped every second into the earth as a whole. The earth, however, does not get discharged since thunderstorms and lightning occurring continually all over the globe pump an equal amount of negative charge on the earth.)
Solution : (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.