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Electrostatic Potential and Capacitance — NEET & JEE MCQs with answers
This chapter develops the concept of electrostatic potential energy and potential, equipotential surfaces, and their relation to the electric field. It then covers potential energy of charge systems and dipoles, electrostatics of conductors and dielectrics, and the theory of capacitors including parallel-plate capacitors, the effect of dielectrics, series/parallel combinations, and energy stored in a capacitor.
Topics in this chapter
- 2.1 Introduction
- 2.2 Electrostatic Potential
- 2.3 Potential Due to a Point Charge
- 2.4 Potential Due to an Electric Dipole
- 2.5 Potential Due to a System of Charges
- 2.6 Equipotential Surfaces
- 2.7 Potential Energy of a System of Charges
- 2.8 Potential Energy in an External Field
- 2.9 Electrostatics of Conductors
- 2.10 Dielectrics and Polarisation
- 2.11 Capacitors and Capacitance
- 2.12 The Parallel Plate Capacitor
- 2.13 Effect of Dielectric on Capacitance
- 2.14 Combination of Capacitors
- 2.15 Energy Stored in a Capacitor
Free practice questions
When a non-polar molecule is placed in an external field, it develops
- A. an induced dipole moment
- B. a permanent dipole moment
- C. a net negative charge
- D. no change at all
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Answer: A — In an external field the positive and negative charges of a non-polar molecule are displaced in opposite directions, so the non-polar molecule develops an induced dipole moment.
NCERT Class 12 Physics Ch 2.10
The work done in bringing charge $q_2$ from infinity to a point $r_2$ (where a charge $q_1$ already exists) equals:
- A. zero, since $q_1$ is already present
- B. $q_1$ times the field at $r_2$ due to $q_2$
- C. $q_2$ times the potential at $r_2$ due to $q_1$
- D. the sum of the two charges divided by the distance
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Answer: C — NCERT: work done on $q_2$ = $q_2$ times the potential at $r_2$ due to $q_1$, i.e. $\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{r_{12}}$.
NCERT Class 12 Physics Ch 2.7
A $4\ \mu F$ and a $12\ \mu F$ capacitor are connected in series. The equivalent capacitance is:
- A. $16\ \mu F$
- B. $3\ \mu F$
- C. $6\ \mu F$
- D. $8\ \mu F$
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Answer: B — Series: $C=\frac{4\times 12}{4+12}=\frac{48}{16}=3\ \mu F$.
NCERT Class 12 Physics Ch 2.14
The electrostatic potential at a point due to a given charge configuration is denoted by the symbol:
- A. $E$
- B. $V$
- C. $q$
- D. $U$
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Answer: B — NCERT denotes the electrostatic potential due to a given charge configuration by the symbol V.
NCERT Class 12 Physics Ch 2.2
The SI unit of the electrostatic energy density $u=\tfrac{1}{2}\varepsilon_0E^2$ is:
- A. $\text{J/m}^2$
- B. $\text{J/m}^3$
- C. $\text{J}$
- D. $\text{J·m}^3$
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Answer: B — Energy density is energy per unit volume, so its SI unit is joule per cubic metre, $\text{J/m}^3$.
NCERT Class 12 Physics Ch 2.15
The dipole potential expression $V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{p\cos\theta}{r^2}$ is exact for
- A. distances comparable to $a$
- B. any dipole at any distance
- C. only points in the equatorial plane
- D. a point dipole (all higher-order terms in $a/r$ vanish exactly)
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Answer: D — NCERT: Eq. (2.15) is approximately true only for $r \gg a$; however, for a point dipole $p$ at the origin the expression is exact.
NCERT Class 12 Physics Ch 2.4
For a conservative force, when the external force is removed and the body moves, the body gains kinetic energy and
- A. loses twice as much potential energy
- B. gains an equal amount of potential energy
- C. keeps its potential energy unchanged
- D. loses an equal amount of potential energy
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Answer: D — NCERT states that when the external force is removed the body moves, gaining kinetic energy and losing an equal amount of potential energy, so the sum is conserved.
NCERT Class 12 Physics Ch 2.1
When a conductor is placed in an external electric field, in the final static situation the field due to the induced charges:
- A. is always zero
- B. adds to the external field inside
- C. only partly reduces the external field inside
- D. exactly cancels the external field inside the conductor
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Answer: D — The free charges rearrange so that the field due to the induced charges opposes and exactly cancels the external field within the conductor. This continues until the net electrostatic field inside the conductor becomes zero.
NCERT Class 12 Physics Ch 2.9
The potential at a point P due to a point charge $q_1$ at distance $r_{1P}$ is given by:
- A. $4\pi\varepsilon_0\dfrac{q_1}{r_{1P}}$
- B. $\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1^2}{r_{1P}}$
- C. $\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1}{r_{1P}}$
- D. $\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1}{r_{1P}^2}$
Show answer
Answer: C — The potential due to a single point charge is $V_1 = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1}{r_{1P}}$, where $r_{1P}$ is the distance between the charge $q_1$ and the point P. This is the basic building block that is summed for a system of charges.
NCERT Class 12 Physics Ch 2.5
Two points A and B are at distances r and 2r from a point charge Q. On moving a unit positive test charge from B (farther) to A (nearer) for Q > 0, the potential:
- A. Increases
- B. Remains constant
- C. Decreases
- D. Becomes zero at A
Show answer
Answer: A — V = kQ/r increases as r decreases (for Q > 0). Since V(A) = kQ/r > V(B) = kQ/2r, moving from B to A the potential increases.
NCERT Class 12 Physics Ch 2.3
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