JEE Physics Electrostatic Potential and Capacitance MCQs Set 31

Physics Objective Questions and Answers: Electrostatic Potential and Capacitance

Review structured MCQ sets for JEE Physics Electrostatic Potential and Capacitance. Built according to official JEE guidelines, these downloadable questions support daily revision and core concept reinforcement.

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Multiple Choice Questions

Question 1: A positive point charge is released from rest at a distance \( r_0 \) from a positive line charge with uniform density. The speed v of the point charge, as a function of instantaneous distance r from the line charge, is proportional to
(a) \( v\propto\dfrac r{r_0} \)
(b) \( \sqrt{\ln\left(\dfrac r{r_0}\right)} \)
(c) \( v\propto e^{+r/r_0} \)
(d) \( v\propto\ln\left(\dfrac r{r_0}\right) \)
Show Answer & Explanation

Answer: (b) \( \sqrt{\ln\left(\dfrac r{r_0}\right)} \)

Explanation:
1. Line charge field \( E=\dfrac\lambda{2\pi\epsilon_0r} \), so the potential difference is \( \dfrac\lambda{2\pi\epsilon_0}\ln\dfrac r{r_0} \).
2. \( \tfrac12mv^2=q\Delta V\propto\ln\dfrac r{r_0}\Rightarrow v\propto\sqrt{\ln(r/r_0)} \).

Teacher's Note:
1. Potential is a scalar: add potentials of point charges, and use \( U = k q1 q2 / r \) summed over pairs for the energy of a system.
2. The work done by the field is \( q \) times the fall in potential, and equipotential surfaces carry zero work.

Question 2: Three charges Q, +q and +q are placed at the vertices of a right-angled isosceles triangle (Q at the vertex of the acute angle, the right angle being at one of the +q charges, the two equal sides having length a). The net electrostatic energy of the configuration is zero if the value of Q is
(a) \( \dfrac{-q}{1+\sqrt2} \)
(b) \( -2q \)
(c) \( +q \)
(d) \( \dfrac{-\sqrt2\,q}{\sqrt2+1} \)
Show Answer & Explanation

Answer: (d) \( \dfrac{-\sqrt2\,q}{\sqrt2+1} \)

Explanation:
1. Pairs: \( Q\text{-}q_1 \) at a, \( q_1\text{-}q_2 \) at a, \( Q\text{-}q_2 \) at \( a\sqrt2 \). \( U=k\left[\dfrac{Qq}a+\dfrac{q^2}a+\dfrac{Qq}{a\sqrt2}\right]=0 \).
2. \( Q\left(1+\dfrac1{\sqrt2}\right)=-q\Rightarrow Q=\dfrac{-\sqrt2q}{\sqrt2+1} \).

Teacher's Note:
1. Potential is a scalar: add potentials of point charges, and use \( U = k q1 q2 / r \) summed over pairs for the energy of a system.
2. The work done by the field is \( q \) times the fall in potential, and equipotential surfaces carry zero work.

Question 3: A charge Q is distributed over three concentric spherical shells of radii a, b, c (\( a\lt b\lt c \)) such that their surface charge densities are equal to one another. The total potential at a point at distance r from their common centre, where \( r\lt a \), would be
(a) \( \dfrac{Q(a^2+b^2+c^2)}{4\pi\epsilon_0(a^3+b^3+c^3)} \)
(b) \( \dfrac{Q(a+b+c)}{4\pi\epsilon_0(a^2+b^2+c^2)} \)
(c) \( \dfrac Q{4\pi\epsilon_0(a+b+c)} \)
(d) \( \dfrac Q{12\pi\epsilon_0}\cdot\dfrac{ab+bc+ca}{abc} \)
Show Answer & Explanation

Answer: (b) \( \dfrac{Q(a+b+c)}{4\pi\epsilon_0(a^2+b^2+c^2)} \)

Explanation:
1. \( Q=4\pi\sigma(a^2+b^2+c^2) \), and inside all the shells \( V=\dfrac\sigma{\epsilon_0}(a+b+c) \).
2. \( V=\dfrac{Q(a+b+c)}{4\pi\epsilon_0(a^2+b^2+c^2)} \).

Teacher's Note:
1. Potential is a scalar: add potentials of point charges, and use \( U = k q1 q2 / r \) summed over pairs for the energy of a system.
2. The work done by the field is \( q \) times the fall in potential, and equipotential surfaces carry zero work.

Question 4: An electric field \( \vec E=30x^2\hat i \) V/m exists in space. Then the potential difference \( V_A-V_O \), where \( V_O \) is the potential at the origin and \( V_A \) the potential at \( x=2 \) m, is
(a) 120 V
(b) -120 V
(c) 80 V
(d) -80 V
Show Answer & Explanation

Answer: (d) -80 V

Explanation:
1. \( V_A-V_O=-\int_0^2E\,dx=-\int_0^230x^2dx \).
2. \( =-10x^3\Big|_0^2=-80 V \).

Teacher's Note:
1. Potential is a scalar: add potentials of point charges, and use \( U = k q1 q2 / r \) summed over pairs for the energy of a system.
2. The work done by the field is \( q \) times the fall in potential, and equipotential surfaces carry zero work.

Question 5: Positive and negative point charges of equal magnitude are kept at \( (0,0,a/2) \) and \( (0,0,-a/2) \) respectively. The work done by the electric field when another positive point charge is moved from \( (-a,0,0) \) to \( (0,a,0) \) is
(a) positive
(b) zero
(c) negative
(d) depends on the path connecting the initial and final positions
Show Answer & Explanation

Answer: (b) zero

Explanation:
1. Both points lie in the plane \( z=0 \), the perpendicular bisector plane of the dipole, which is an equipotential \( (V = 0) \).
2. The work \( q(V_i-V_f)=0 \), independent of path.

Teacher's Note:
1. Potential is a scalar: add potentials of point charges, and use \( U = k q1 q2 / r \) summed over pairs for the energy of a system.
2. The work done by the field is \( q \) times the fall in potential, and equipotential surfaces carry zero work.

Question 6: A uniform electric field pointing in the positive x-direction exists in a region. Let A be the origin, B the point on the x-axis at \( x=+1 \) cm and C the point on the y-axis at \( y=+1 \) cm. Then the potentials at the points A, B and C satisfy
(a) \( V_A\lt V_B \)
(b) \( V_A\lt V_C \)
(c) \( V_A\gt V_B \)
(d) \( V_A\gt V_C \)
Show Answer & Explanation

Answer: (c) \( V_A\gt V_B \)

Explanation:
1. Potential falls in the direction of the field, so \( V_B\lt V_A \).
2. A and \( C \) lie on the same equipotential line (perpendicular to the field), so \( V_A=V_C \).

Teacher's Note:
1. Potential is a scalar: add potentials of point charges, and use \( U = k q1 q2 / r \) summed over pairs for the energy of a system.
2. The work done by the field is \( q \) times the fall in potential, and equipotential surfaces carry zero work.

Question 7: A non-conducting ring of radius 0.5 m carries a total charge of \( 1.11\times10^{-10} \) C distributed non-uniformly on its circumference producing an electric field E everywhere in space. The value of the integral \( \int_{l=\infty}^{l=0}-\vec E\cdot d\vec l \) (l = 0 being the centre of the ring) in volts is
(a) +2
(b) -1
(c) -2
(d) zero
Show Answer & Explanation

Answer: (a) +2

Explanation:
1. The integral equals \( V(0)-V(\infty)=V_{centre} \).
2. Whatever the distribution on the ring, \( V_{centre}=\dfrac{kQ}R=\dfrac{9\times10^9\times1.11\times10^{-10}}{0.5}\approx2 V \).

Teacher's Note:
1. Potential is a scalar: add potentials of point charges, and use \( U = k q1 q2 / r \) summed over pairs for the energy of a system.
2. The work done by the field is \( q \) times the fall in potential, and equipotential surfaces carry zero work.

Question 8: The electric potential V at any point (x, y, z), all in metres, in space is given by \( V=4x^2 \) volt. The electric field at the point (1, 0, 2) m is
(a) \( +8 \) V/m along the x-axis
(b) \( -16 \) V/m along the x-axis
(c) \( -8 \) V/m along the x-axis
(d) zero
Show Answer & Explanation

Answer: (c) \( -8 \) V/m along the x-axis

Explanation:
1. \( E_x=-\dfrac{\partial V}{\partial x}=-8x \), and \( E_y=E_z=0 \).
2. At \( x=1 \): \( E=-8 V/m \), i.e. \( 8 V/m \) directed along the negative x-axis.

Teacher's Note:
1. Potential is a scalar: add potentials of point charges, and use \( U = k q1 q2 / r \) summed over pairs for the energy of a system.
2. The work done by the field is \( q \) times the fall in potential, and equipotential surfaces carry zero work.

Question 9: Eight point charges (alternately \( +q \) and \( -q \) on neighbouring vertices, so that every edge joins unlike charges) are placed at the corners of a cube of edge a. The work done in disassembling this system of charges (taking all charges to infinity) is
(a) \( 5.82\,\dfrac{q^2}{4\pi\epsilon_0a} \)
(b) \( -5.82\,\dfrac{q^2}{4\pi\epsilon_0a} \)
(c) \( 3\,\dfrac{q^2}{4\pi\epsilon_0a} \)
(d) \( 12\,\dfrac{q^2}{4\pi\epsilon_0a} \)
Show Answer & Explanation

Answer: (a) \( 5.82\,\dfrac{q^2}{4\pi\epsilon_0a} \)

Explanation:
1. For one corner: 3 neighbours at a (unlike): \( -3 \); 3 face-diagonal at \( a\sqrt2 \) (like): \( +\dfrac3{\sqrt2} \); 1 body-diagonal at \( a\sqrt3 \) (unlike): \( -\dfrac1{\sqrt3} \). Sum \( =-1.456 \).
2. \( U=\dfrac12\times8\times(-1.456)\dfrac{kq^2}a=-5.82\dfrac{kq^2}a \). The work needed to disassemble is \( +5.82\dfrac{q^2}{4\pi\epsilon_0a} \).

Teacher's Note:
1. Potential is a scalar: add potentials of point charges, and use \( U = k q1 q2 / r \) summed over pairs for the energy of a system.
2. The work done by the field is \( q \) times the fall in potential, and equipotential surfaces carry zero work.

Question 10: Two identical parallel plate capacitors, each of capacitance C (plates of area A, separation d), are filled with three dielectrics of equal thickness and dielectric constants \( K_1,K_2,K_3 \). In the first capacitor the dielectrics are stacked in layers parallel to the plates; in the second they sit side by side, each with one third of the plate area. If both are charged to the same potential V, the ratio of the energy stored, \( E_1/E_2 \), is
(a) \( \dfrac{(K_1+K_2+K_3)(K_2K_3+K_3K_1+K_1K_2)}{K_1K_2K_3} \)
(b) \( \dfrac{9K_1K_2K_3}{(K_1+K_2+K_3)(K_2K_3+K_3K_1+K_1K_2)} \)
(c) \( \dfrac{K_1K_2K_3}{(K_1+K_2+K_3)(K_1K_2+K_2K_3+K_3K_1)} \)
(d) \( \dfrac{(K_1+K_2+K_3)(K_2K_3+K_3K_1+K_1K_2)}{9K_1K_2K_3} \)
Show Answer & Explanation

Answer: (b) \( \dfrac{9K_1K_2K_3}{(K_1+K_2+K_3)(K_2K_3+K_3K_1+K_1K_2)} \)

Explanation:
1. Layers (series): \( C_1=\dfrac{3C}{1/K_1+1/K_2+1/K_3}=\dfrac{3CK_1K_2K_3}{K_1K_2+K_2K_3+K_3K_1} \). Side by side (parallel): \( C_2=\dfrac{C(K_1+K_2+K_3)}3 \).
2. At the same \( V \), \( E\propto C \): \( \dfrac{E_1}{E_2}=\dfrac{9K_1K_2K_3}{(K_1+K_2+K_3)(K_1K_2+K_2K_3+K_3K_1)} \).

Teacher's Note:
1. For capacitors use \( C = K epsilon_0 A / d \), add in parallel and add reciprocals in series, and decide first whether the battery stays connected (\( V \) fixed) or is removed (\( Q \) fixed).
2. Energy stored is \( Q^2/2C = CV^2/2 \); compare before and after to get the work done.

Question 11: The parallel combination of two air-filled parallel plate capacitors of capacitance C and nC is connected to a battery of voltage V. When the capacitors are fully charged, the battery is removed and after that a dielectric material of dielectric constant K is placed between the two plates of the first capacitor. The new potential difference of the combined system is
(a) \( \dfrac{nV}{K+n} \)
(b) \( V \)
(c) \( \dfrac V{K+n} \)
(d) \( \dfrac{(n+1)V}{K+n} \)
Show Answer & Explanation

Answer: (d) \( \dfrac{(n+1)V}{K+n} \)

Explanation:
1. Total charge \( Q=(C+nC)V=(n+1)CV \), conserved after the battery is removed.
2. New capacitance \( KC+nC \): \( V'=\dfrac{(n+1)CV}{(K+n)C}=\dfrac{(n+1)V}{K+n} \).

Teacher's Note:
1. For capacitors use \( C = K epsilon_0 A / d \), add in parallel and add reciprocals in series, and decide first whether the battery stays connected (\( V \) fixed) or is removed (\( Q \) fixed).
2. Energy stored is \( Q^2/2C = CV^2/2 \); compare before and after to get the work done.

Question 12: A capacitor with capacitance 5 \(\mu\)F is charged to 5 \(\mu\)C. If the plates are pulled apart to reduce the capacitance to 2 \(\mu\)F, how much work is done?
(a) \( 3.75\times10^{-6}\ \text{J} \)
(b) \( 6.25\times10^{-6}\ \text{J} \)
(c) \( 2.16\times10^{-6}\ \text{J} \)
(d) \( 2.55\times10^{-6}\ \text{J} \)
Show Answer & Explanation

Answer: (a) \( 3.75\times10^{-6}\ \text{J} \)

Explanation:
1. Charge is constant: \( W=\dfrac{Q^2}2\left(\dfrac1{C_2}-\dfrac1{C_1}\right) \).
2. \( W=\dfrac{25\times10^{-12}}2\left(\dfrac1{2\times10^{-6}}-\dfrac1{5\times10^{-6}}\right)=3.75\times10^{-6} J \).

Teacher's Note:
1. For capacitors use \( C = K epsilon_0 A / d \), add in parallel and add reciprocals in series, and decide first whether the battery stays connected (\( V \) fixed) or is removed (\( Q \) fixed).
2. Energy stored is \( Q^2/2C = CV^2/2 \); compare before and after to get the work done.

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Multiple Choice Questions (MCQs) for JEE Physics Electrostatic Potential and Capacitance

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