JEE Physics Electrostatic Potential and Capacitance MCQs Set 24

Answer: (a) \( \dfrac{a^2V}{2r^2} \)

Explanation:
1. Charge on one drop: \( q=4\pi\varepsilon_0rV \). Number of drops when half full: \( n\cdot\tfrac43\pi r^3=\tfrac12\cdot\tfrac43\pi a^3\Rightarrow n=\dfrac{a^3}{2r^3} \).
2. Total charge \( Q=nq=\dfrac{2\pi\varepsilon_0a^3V}{r^2} \), so the sphere's potential is \( \dfrac{Q}{4\pi\varepsilon_0a}=\dfrac{a^2V}{2r^2} \).

Teacher's Note:
1. A droplet of radius r at potential V carries charge 4 pi epsilon naught r V. Count how many drops fill half the sphere.

Answer: (a) 11

Explanation:
1. \( \lambda=3\times10^{-6}\ \text{C/m} \). Force per unit length at separation d: \( f=\dfrac{\lambda^2}{2\pi\varepsilon_0d} \).
2. Work per unit length from 2 cm to 1 cm: \( W=\dfrac{\lambda^2}{2\pi\varepsilon_0}\ln2=(9\times10^{-12})(1.8\times10^{10})(0.693)=0.112\ \text{J/m} \).
3. So x = 11.2, nearest integer 11.

Teacher's Note:
1. Force per unit length falls as 1 over d, so the work involves a natural logarithm. Convert the density to C/m first.

Answer: (a) 30

Explanation:
1. The potential difference between the sphere and the shell depends only on the charge Q of the inner sphere: \( \Delta V=kQ\left(\dfrac1a-\dfrac1b\right) \).
2. The charge given to the shell shifts the potential of both surfaces equally, so the difference stays 30 V.

Teacher's Note:
1. A uniform shift in potential does not change a difference. Only the inner charge produces a field between the surfaces.

Answer: (a) 3

Explanation:
1. The field inside the cavity is zero, so A is at the potential of the inner surface r = a.
2. In the shell: \( E(r)=\dfrac{\rho(r^3-a^3)}{3\varepsilon_0r^2} \). \( V_A-V_B=\dfrac{\rho}{3\varepsilon_0}\left[\dfrac{r^2}2+\dfrac{a^3}r\right]_a^{2a}=\dfrac{\rho}{3\varepsilon_0}\left(2.5a^2-1.5a^2\right)=\dfrac{\rho a^2}{3\varepsilon_0} \).
3. So n = 3.

Teacher's Note:
1. Find the field in the shell with Gauss's law and integrate it from the inner to the outer surface. The cavity itself is an equipotential region.

Answer: (a) 4

Explanation:
1. At closest approach both move with the same speed \( V/2 \) (momentum conservation).
2. Energy conservation: \( \tfrac12mV^2=\tfrac12(2m)\left(\dfrac V2\right)^2+\dfrac{Q^2}{4\pi\varepsilon_0d} \) so \( \dfrac{Q^2}{4\pi\varepsilon_0d}=\tfrac14mV^2 \).
3. \( d=\dfrac{4Q^2}{4\pi\varepsilon_0mV^2} \), x = 4.

Teacher's Note:
1. The free second particle moves away, so the closest approach occurs at common velocity, not at zero velocity.

Answer: (a) 16

Explanation:
1. Total energy of a circular orbit: \( E=-\dfrac{qq_0}{8\pi\varepsilon_0r} \).
2. Change to radius 2r: \( \Delta E=-\dfrac{qq_0}{16\pi\varepsilon_0r}+\dfrac{qq_0}{8\pi\varepsilon_0r}=\dfrac{qq_0}{16\pi\varepsilon_0r} \), so n = 16.

Teacher's Note:
1. For an inverse square attraction the total energy of an orbit is minus half of the magnitude of the potential energy.

Answer: (a) 1

Explanation:
1. Angular momentum about A: \( mud=mvr \) so \( v=ud/r \). Energy: \( \tfrac12mu^2=\tfrac12mv^2+\dfrac{Qq}{4\pi\varepsilon_0r}=\tfrac12mv^2+\dfrac{mu^2d}{r} \).
2. Substituting v: \( 1=\dfrac{d^2}{r^2}+\dfrac{2d}{r}\Rightarrow r^2-2dr-d^2=0 \) so \( r=d(1+\sqrt2) \).
3. With \( d=\sqrt2-1 \): \( r=(\sqrt2-1)(\sqrt2+1)=1 \) m.

Teacher's Note:
1. At the closest point the velocity is perpendicular to the line to the charge. Use conservation of angular momentum and energy together.

Answer: (a) 2

Explanation:
1. When the first ball is released it gains kinetic energy from its interaction with all the other 2018 balls.
2. When the adjacent ball is released later, the first ball is already far away, so it has one interaction fewer: the difference in the kinetic energies is the interaction energy between these two neighbours, \( K=\dfrac{kq^2}{a} \).
3. \( q^2=\dfrac{Ka}{k}=\dfrac{9\times10^9\times4\times10^{-6}}{9\times10^9}=4\times10^{-6}\Rightarrow q=2\times10^{-3}\ \text{C}=2\ \text{mC} \).

Teacher's Note:
1. The two kinetic energies differ by exactly one pair interaction. Everything else cancels.

Answer: (a) 2

Explanation:
1. \( \vec E=-\nabla V=-(3\hat i+4\hat j) \), magnitude 5 N/C, so the force on 1 C is 5 N along \( -(3\hat i+4\hat j)/5 \).
2. Starting from rest it moves in a straight line. To reach y = 0 it must go 3.2 m in y, i.e. a path length \( 3.2\times\tfrac54=4 \) m.
3. Work = 5 × 4 = 20 J, so \( v=\sqrt{2\times20/10}=2\ \text{m/s} \). (Check: the potential drops by \( 3\times(2-(-0.4))+4\times3.2=20 \) V.)

Teacher's Note:
1. Potential difference times charge gives the kinetic energy gained. Field is the negative gradient of potential, so the path is a straight line along it.

Answer: (a) 10

Explanation:
1. Charges: \( Q_1=\sigma4\pi R^2 \) and \( Q_2=\sigma4\pi(2R)^2=4Q_1 \); total = 5Q_1.
2. After connection the potentials are equal: \( \dfrac{Q_1'}{R}=\dfrac{Q_2'}{2R}\Rightarrow Q_2'=2Q_1' \), so \( Q_2'=\tfrac{10}3Q_1 \).
3. \( \sigma_2'=\dfrac{Q_2'}{4\pi(2R)^2}=\dfrac{10}{3}\cdot\dfrac{4\pi R^2\sigma}{16\pi R^2}=\dfrac{5\sigma}{6}\cdot... \) so with \( \sigma=12 \): \( \sigma_2'=\dfrac{10}{3}\times\dfrac{12}{4}=10 \) μC/m².

Teacher's Note:
1. After connection the potentials are equal, so charge divides in proportion to radius. Convert back to density using the new surface area.

Answer: (b) (B) and (D)

Explanation:
1. Potential energy of Q at distance x: \( U=-\dfrac{2kqQ}{\sqrt{x^2+a^2}} \), which has its minimum at x = 0, so Q is bound and oscillates about the origin with maximum speed there: (B) true, (C) false.
2. The restoring force is not proportional to x for a displacement of 2a, so the motion is not simple harmonic: (A) false, (D) true.

Teacher's Note:
1. Check the potential energy: its minimum at the origin gives maximum speed there. SHM would need a force linear in displacement, which holds only for small displacements.

Answer: (c) \( R_1=0\ \text{and}\ R_2\lt (R_4-R_3) \)

Explanation:
1. Inside: \( V=\dfrac{V_0}{2}\left(3-\dfrac{r^2}{R^2}\right) \). \( V=\tfrac32V_0 \) at r = 0 so \( R_1=0 \); \( V=\tfrac54V_0 \) gives \( R_2=R/\sqrt2=0.71R \).
2. Outside: \( V=V_0R/r \). \( \tfrac34V_0 \) gives \( R_3=\tfrac43R \) and \( \tfrac14V_0 \) gives \( R_4=4R \).
3. \( R_4-R_3=\tfrac83R=2.67R\gt R_2 \), so (c) holds; and \( R_4=4R\gt 2R \), so the last option is false.

Teacher's Note:
1. Use the inside and outside potential formulas of a uniformly charged sphere. Calculate each radius and compare.

Answer: (d) \( \dfrac{Q\ln2}{4\pi\varepsilon_0L} \)

Explanation:
1. Linear density \( \lambda=Q/L \). An element at distance x from O gives \( dV=\dfrac{\lambda\,dx}{4\pi\varepsilon_0x} \).
2. \( V=\dfrac{\lambda}{4\pi\varepsilon_0}\int_L^{2L}\dfrac{dx}x=\dfrac{Q}{4\pi\varepsilon_0L}\ln2 \).

Teacher's Note:
1. Integrate dq over distance from O. The limits are L and 2L, so the logarithm gives ln 2.

Answer: (c) -80 V

Explanation:
1. \( V_A-V_O=-\int_0^2E\,dx=-\int_0^2 30x^2dx=-10x^3\Big|_0^2=-80\ \text{V} \).

Teacher's Note:
1. Potential difference is minus the line integral of the field. The potential falls along the direction of the field.

Answer: (d) \( \dfrac\sigma{\varepsilon_0}\left[\dfrac{a^2-b^2}{b}+c\right] \)

Explanation:
1. Charges: \( Q_A=4\pi a^2\sigma,\ Q_B=-4\pi b^2\sigma,\ Q_C=4\pi c^2\sigma \).
2. \( V_B=k\left[\dfrac{Q_A}b+\dfrac{Q_B}b+\dfrac{Q_C}c\right]=\dfrac\sigma{\varepsilon_0}\left[\dfrac{a^2}{b}-b+c\right]=\dfrac\sigma{\varepsilon_0}\left[\dfrac{a^2-b^2}{b}+c\right] \).

Teacher's Note:
1. At a shell the potential from an inner shell is kQ over the shell's radius, and from an outer shell kQ over that shell's radius. Add them all.

Answer: (c) \( v\propto\sqrt{\ln\left(\dfrac r{r_0}\right)} \)

Explanation:
1. Potential of a line charge: \( V(r)=-\dfrac{\lambda}{2\pi\varepsilon_0}\ln r \).
2. Energy conservation: \( \tfrac12mv^2=q[V(r_0)-V(r)]=\dfrac{q\lambda}{2\pi\varepsilon_0}\ln\dfrac r{r_0} \).
3. So \( v\propto\sqrt{\ln(r/r_0)} \).

Teacher's Note:
1. The potential of a line charge is logarithmic. Kinetic energy gained is charge times the potential drop.