JEE Physics Electrostatic Potential and Capacitance MCQs Set 17

Answer: (a) \( \frac{5\rho R^2}{12\varepsilon_0} \)

Explanation:
1. Use superposition: full sphere of density \( \rho \) plus a sphere of density \( -\rho \) filling the cavity.
2. Full sphere, at its centre: \( \frac{\rho R^2}{2\varepsilon_0} \). Cavity sphere (radius R/2): C lies on its surface, where the potential is \( \frac{\rho(R/2)^2}{3\varepsilon_0}=\frac{\rho R^2}{12\varepsilon_0} \).
3. Hence \( V_C=\frac{\rho R^2}{2\varepsilon_0}-\frac{\rho R^2}{12\varepsilon_0}=\frac{5\rho R^2}{12\varepsilon_0} \).

Teacher's Note:
1. Use superposition of potentials (scalars) and energy conservation; avoid vector work where a potential function exists.

Answer: (b) \( \frac{L}{e^{2\pi\varepsilon_0mv^2/(\lambda q)}-1} \)

Explanation:
1. Potential on the line of the rod at distance x from end A: \( V=\frac{\lambda}{4\pi\varepsilon_0}\ln\frac{x+L}{x} \).
2. At closest approach the kinetic energy is zero: \( \frac12mv^2=qV \), so \( \ln\frac{x+L}{x}=\frac{2\pi\varepsilon_0mv^2}{\lambda q} \).
3. Solving, \( x=\frac{L}{e^{2\pi\varepsilon_0mv^2/(\lambda q)}-1} \).

Teacher's Note:
1. Use superposition of potentials (scalars) and energy conservation; avoid vector work where a potential function exists.

Answer: (c) \( \frac{4a}{3} \)

Explanation:
1. Potential on the axis: \( V(z)=\frac{\sigma}{2\varepsilon_0}\left(\sqrt{z^2+a^2}-z\right) \), so \( U(z)=mgz+qV(z) \) with \( \frac{q\sigma}{2\varepsilon_0m}=2g \).
2. Just reaching the disc: \( U(H)=U(0) \), i.e. \( H+2\left(\sqrt{H^2+a^2}-H\right)=2a \), so \( 2\sqrt{H^2+a^2}=2a+H \).
3. Squaring: \( 3H^2=4aH \) and \( H=\frac{4a}{3} \).

Teacher's Note:
1. Use superposition of potentials (scalars) and energy conservation; avoid vector work where a potential function exists.

Answer: (d) \( \frac{a}{\sqrt3} \)

Explanation:
1. \( U(y)=2mg\left(\sqrt{y^2+a^2}-y\right)+mgy \).
2. Equilibrium: \( \frac{dU}{dy}=2mg\left(\frac{y}{\sqrt{y^2+a^2}}-1\right)+mg=0 \), so \( \frac{y}{\sqrt{y^2+a^2}}=\frac12 \).
3. Hence \( 4y^2=y^2+a^2 \) and \( y=\frac{a}{\sqrt3} \) (the minimum of U, hence stable).

Teacher's Note:
1. Use superposition of potentials (scalars) and energy conservation; avoid vector work where a potential function exists.

Answer: (a) 3 m/s

Explanation:
1. Potential energy on the x-axis: \( U(x)=qk\left[\frac{16\times10^{-6}}{\sqrt{x^2+13.5}}-\frac{2\times10^{-6}}{\sqrt{x^2+1.5}}\right] \). Numerically it is largest not at the origin but at \( x\approx1.58 \) m, where \( U_{max}=2.70\times10^{-3} \) J (it is zero at infinity).
2. The particle must cross this barrier: \( \frac12mv_0^2=2.70\times10^{-3} \), so \( v_0=\sqrt{\frac{2\times2.7\times10^{-3}}{6\times10^{-4}}}=3 \) m/s.

Teacher's Note:
1. Use superposition of potentials (scalars) and energy conservation; avoid vector work where a potential function exists.

Answer: (b) \( 2.5\times10^{-4} \) J

Explanation:
1. Energy conservation: \( K_0=\frac12mv_0^2-U(0) \) with \( U(\infty)=0 \).
2. Here \( \frac12mv_0^2=2.7\times10^{-3} \) J and \( U(0)=qk\left[\frac{16\times10^{-6}}{\sqrt{13.5}}-\frac{2\times10^{-6}}{\sqrt{1.5}}\right]=2.449\times10^{-3} \) J.
3. So \( K=(27-10\sqrt6)\times10^{-4}=2.5\times10^{-4} \) J.

Teacher's Note:
1. Use superposition of potentials (scalars) and energy conservation; avoid vector work where a potential function exists.

Answer: (c) 5.86 m/s

Explanation:
1. The Coulomb force \( F=\frac{kq^2}{L^2}=\frac{9\times10^9\times10^{-12}}{0.64}=0.0141 \) N is always radial (outward), so it does no work and only gravity changes the speed.
2. At the top, with tension just zero: \( \frac{mv_t^2}{L}=mg-F \), so \( v_t^2=0.8\left(10-7.03\right)=2.375 \).
3. At the bottom: \( v^2=v_t^2+4gL=2.375+32=34.375=\frac{275}{8} \) and \( v=5.86 \) m/s.

Teacher's Note:
1. Use superposition of potentials (scalars) and energy conservation; avoid vector work where a potential function exists.

Answer: (d) 10/3 cm

Explanation:
1. The extreme charges 2Q and 8Q go to the two ends (10 cm apart), and Q lies between them at distance x from 2Q.
2. Energy \( U\propto\frac{2}{x}+\frac{8}{10-x}+\text{const} \); \( \frac{dU}{dx}=0 \) gives \( \frac{2}{x^2}=\frac{8}{(10-x)^2} \), so \( 10-x=2x \) and \( x=\frac{10}{3} \) cm. (The electric field at Q is then zero.)

Teacher's Note:
1. Use superposition of potentials (scalars) and energy conservation; avoid vector work where a potential function exists.

Answer: (a) \( \frac{Q^2R}{8\pi\varepsilon_0r^2} \)

Explanation:
1. In the limit contact no longer transfers charge: both spheres are at the same potential, equal to that of S\( _1 \) carrying Q, namely \( V=\frac{Q}{4\pi\varepsilon_0r} \).
2. S\( _2 \) then has charge \( q_\infty=4\pi\varepsilon_0RV=\frac{QR}{r} \).
3. Its energy is \( \frac{q_\infty^2}{8\pi\varepsilon_0R}=\frac{Q^2R}{8\pi\varepsilon_0r^2} \).

Teacher's Note:
1. Use superposition of potentials (scalars) and energy conservation; avoid vector work where a potential function exists.

Answer: (b) a circle of radius 4 m with its centre on the line joining the charges, 2 m from q on the side away from -2q

Explanation:
1. Take q at the origin and -2q at (6, 0). Zero potential: \( \frac{q}{d_1}=\frac{2q}{d_2} \), i.e. \( d_2=2d_1 \).
2. Then \( (x-6)^2+y^2=4(x^2+y^2) \), giving \( (x+2)^2+y^2=16 \).
3. This is a circle of radius 4 m centred at (-2, 0).

Teacher's Note:
1. Use superposition of potentials (scalars) and energy conservation; avoid vector work where a potential function exists.

Answer: (c) \( \frac{R_1}{R_2} \)

Explanation:
1. Energy \( U=\frac{Q_1^2}{8\pi\varepsilon_0R_1}+\frac{Q_2^2}{8\pi\varepsilon_0R_2} \) with \( Q_1+Q_2 \) fixed.
2. Minimising: \( \frac{Q_1}{R_1}=\frac{Q_2}{R_2} \), which means equal potentials (potential difference zero).
3. So \( \frac{Q_1}{Q_2}=\frac{R_1}{R_2} \).

Teacher's Note:
1. Use superposition of potentials (scalars) and energy conservation; avoid vector work where a potential function exists.

Answer: (d) \( -\frac{8Q}{11} \)

Explanation:
1. Let the final charges be x, y, z on A, B, C. A and C are connected, so \( x+z=Q \) and \( V_A=V_C \).
2. Potentials (units of \( \frac{1}{4\pi\varepsilon_0a} \)): \( V_B=\frac x2+\frac y2+\frac z3=0 \) and \( V_A=x+\frac y2+\frac z3=V_C=\frac x3+\frac y3+\frac z3 \).
3. The last gives \( y=-4x \); then \( z=\frac{9x}{2} \), so \( x=\frac{2Q}{11} \) and \( y=-\frac{8Q}{11} \).

Teacher's Note:
1. Use superposition of potentials (scalars) and energy conservation; avoid vector work where a potential function exists.

Answer: (a) \( \frac{kq^2}{2}\left(\frac1b-\frac1a\right) \), with \( k=\frac1{4\pi\varepsilon_0} \)

Explanation:
1. Induced charges: \( -q \) on the inner surface (radius a) and \( +q \) on the outer surface (radius b).
2. Potential at the centre due to the shell charges: \( V=-\frac{kq}{a}+\frac{kq}{b} \).
3. Work done is \( W=\frac12qV=\frac{kq^2}{2}\left(\frac1b-\frac1a\right) \) (negative, since the shell attracts the charge on balance).

Teacher's Note:
1. Use superposition of potentials (scalars) and energy conservation; avoid vector work where a potential function exists.