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NCERT · Class 12 · Physics

Chapter 2: Electrostatic Potential and Capacitance

Practice each question, then open the answer or worked solution.

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Questions

35 questions
Q2.32Written answer

A cylindrical capacitor has two co-axial cylinders of length 15 cm and radii 1.5 cm and 1.4 cm. The outer cylinder is earthed, and the inner cylinder is given a charge of 3.5 µC. Determine the capacitance of the system and the potential of the inner cylinder. Neglect end effects (i.e., bending of field lines at the ends).

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Worked solution

Length of the coaxial cylinders,l = 15 cm = 0.15 m

Radius of the outer cylinder, r1 = 1.5 cm = 0.015 m

Radius of the inner cylinder, r2 = 1.4 cm = 0.014 m

Charge of the inner cylinder, q = 3.5 μC = 3.5 x 10-6 C

The outer cylinder is earthed

The capacitance of the co-axial cylinder of radii r1 and r2 is given by the relation

\(\begin{array}{l}C = \frac{2\pi \epsilon _{0}l}{log_{e}\frac{r_{1}}{r_{2}}}\end{array} \)

Here,

ε0 is the permittivity of free space = 8.85 x 10-12 N-1 m-2 C2

\(\begin{array}{l}C = \frac{2\times 3.14\times 8.85\times 10^{-12}\times 0.15}{2.3026log_{10}\frac{0.15}{0.14}}\end{array} \)

 

\(\begin{array}{l}C = \frac{2\times 3.14\times 8.85\times 10^{-12}\times 0.15}{2.3026\times 0.0299}=1.2 \times 10^{-10}F\end{array} \)

Therefore, the potential difference of the inner cylinder is given by

V = q/C

= 3.5 x 10-6/1.2 x 10-10 = 2.92 x 104 V

Q2.33Written answer

A parallel plate capacitor is to be designed with a voltage rating of 1 kV, using a material of dielectric constant 3 and dielectric strength of about 107Vm–1. (Dielectric strength is the maximum electric field a material can tolerate without breakdown, i.e. without starting to conduct electricity through partial ionisation.) For safety, we should like the field never to exceed, say, 10% of the dielectric strength. What minimum area of the plates is required to have a capacitance of 50 pF?

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Worked solution

Voltage rating of the parallel plate capacitor, V = 1kV= 1000 V.

Dielectric constant, ε = 3

Dielectric strength = 107 V/m

For safety, the field should never exceed, say, 10% of the dielectric strength, E = 10% of dielectric strength = (10/100) x 107 = 106  V/m

Capacitance of capacitor , C = 50pF = 50 × 10-12 F

Distance between the plates, d = V/E = 10³/106 = 10-3 m

Capacitance, 

\(\begin{array}{l}C = \frac{\epsilon_{0}\epsilon _{r}A}{d}\end{array} \)

Area of the plate, A = 

\(\begin{array}{l}A = \frac{d}{\epsilon_{0}\epsilon _{r}C}\end{array} \)

= (50 × 10-12 × 10-3)/(3 × 8.85 × 10-12)

= 1.9 × 10-3 m²

= 19 cm²

Therefore, the minimum area of the plates, A = 19 cm²

Q2.34Written answer

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
(d) a uniform grid consisting of long, equally spaced parallel charged wires in a plane

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Worked solution

(a) Planes parallel to the x-y plane.
(b) Same as in (a), except that planes differing by a fixed potential get closer as the field increases.
(c) Concentric spheres centred at the origin.
(d) A periodically varying shape near the grid, which gradually reaches the shape of planes parallel to the grid at far distances.

Q2.35Written answer

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, the 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

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Worked solution

By Gauss’s law, the electric field between the sphere and the shell is determined only by the charge q1. Hence, the potential difference between the sphere and the shell is independent of q2. If q1 is positive, then the potential difference is also positive.

Q2.36Written answer

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 the next morning?
(c) The discharging current in the atmosphere due to the small conductivity of air is known to be 1800 A on 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 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 a 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.

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Worked solution

(a) Our body and the ground form an equipotential surface. As we step out into the open, the original equipotential surfaces of open-air change, keeping our heads and the ground at the same potential.
(b) Yes, the steady discharging current in the atmosphere charges up the aluminium sheet gradually and raises its voltage to an extent depending on the capacitance of the capacitor (formed by the sheet, slab and the ground).
(c) The atmosphere is continually being charged by thunderstorms and lightning all over the globe and discharged through regions of ordinary weather. The two opposing currents are, on average, in equilibrium.
(d) Light energy involved in lightning; heat and sound energy in the accompanying thunder.

NCERT Class 12 Physics Solutions for Electrostatic Potential and Capacitance

Chapter 2, Electrostatic Potential and Capacitance of Class 12 Physics, is prepared as per the current CBSE Syllabus 2023-24. Also, this chapter provides good marks weightage to derivations and numerical problems for the board exam. The derivation of topics like potential energy of the system of charges, potential due to electric dipole and energy stored in the capacitor is frequently asked in exams. Numerical problems based on the effective capacitance of a complex combination of capacitors are asked in exams regularly. All the topics in the chapter are covered in the