JEE Physics Electrostatic Potential and Capacitance MCQs Set 16

Answer: (a) \( \sqrt{2mQV} \)

Explanation:
1. \( \frac{p^2}{2m}=QV \), so \( p=\sqrt{2mQV} \).

Teacher's Note:
1. Kinetic energy gained equals charge times potential difference.
2. Express momentum through kinetic energy.

Answer: (d) \( \frac{q^2}{\pi\epsilon_0a}\left[\frac{\sqrt3}{2}-\frac{15}{8}\right] \)

Explanation:
1. Sum over the 15 pairs: 6 pairs at distance a (opposite signs, −1 each), 6 pairs at \( \sqrt3a \) (like signs, +1 each) and 3 pairs at 2a (opposite signs, −1 each).
2. \( U=\frac{q^2}{4\pi\epsilon_0a}\left[-6+\frac{6}{\sqrt3}-\frac32\right]=\frac{q^2}{4\pi\epsilon_0a}\left[2\sqrt3-\frac{15}{2}\right] \).
3. \( =\frac{q^2}{\pi\epsilon_0a}\left[\frac{\sqrt3}{2}-\frac{15}{8}\right] \).

Teacher's Note:
1. Count pairs by their separation: a, root-3 a and 2a.
2. Signs alternate around the hexagon.

Answer: (b) \( \frac{2V}{3} \)

Explanation:
1. Inside a shell the potential is constant and equals the surface value \( V=\frac{kq}{0.10} \).
2. At 15 cm: \( \frac{kq}{0.15}=\frac{0.10}{0.15}V=\frac{2V}{3} \).

Teacher's Note:
1. The potential inside a charged shell is the same everywhere.
2. Outside it behaves like a point charge.

Answer: (b) \( -6a\epsilon_0 \)

Explanation:
1. Field \( E=-\frac{d\phi}{dr}=-2ar \).
2. Gauss's law in differential form: \( \rho=\epsilon_0\frac{1}{r^2}\frac{d(r^2E)}{dr}=\epsilon_0\frac{1}{r^2}(-6ar^2)=-6a\epsilon_0 \).

Teacher's Note:
1. The charge density is epsilon naught times the divergence of E.
2. The result is uniform inside.

Answer: (a) −500 N

Explanation:
1. Slope between 0 and 2 cm is \( +\frac{5}{2}=2.5\ \text{V/cm}=250\ \text{V/m} \), so \( E=-250\ \text{V/m} \).
2. \( F=qE=2\times(-250)=-500\ \text{N} \).

Teacher's Note:
1. Convert V/cm into V/m.
2. The sign shows the force points towards the origin.

Answer: (c) 90 m/s

Explanation:
1. Energy gained \( =kQq\left(\frac11-\frac1{10}\right)=9\times10^9\times10^{-3}\times10^{-6}\times0.9=8.1\ \text{J} \).
2. \( v=\sqrt{\frac{2\times8.1}{2\times10^{-3}}}=90\ \text{m/s} \).

Teacher's Note:
1. Electrostatic potential energy is converted into kinetic energy.
2. Convert 2 g to 0.002 kg.

Answer: (d) A, B and C

Explanation:
1. At A the +q and −q on the left side are equidistant, so the potentials cancel; likewise at C.
2. At B all four distances are equal and the charges add to zero.
3. At P the nearby +q charges dominate, so P and Q are not zero.

Teacher's Note:
1. Zero potential needs equal-distance positive and negative charges.
2. P is closer to the positive pair than to the negative pair.

Answer: (b) \( -\frac{q}{2\pi^2\epsilon_0r^2}\hat j \)

Explanation:
1. Linear density \( \lambda=\frac{q}{\pi r} \). Horizontal components cancel; vertical components add: \( E=\int_0^\pi\frac{\lambda r\,d\theta}{4\pi\epsilon_0r^2}\sin\theta=\frac{\lambda}{2\pi\epsilon_0r}=\frac{q}{2\pi^2\epsilon_0r^2} \).
2. A positive ring above O pushes the field at O downwards, along \( -\hat j \).

Teacher's Note:
1. Use symmetry to drop the horizontal part.
2. Positive charge produces a field pointing away from itself.

Answer: (a) decreases and then increases

Explanation:
1. The potential is infinite at each charge and has a minimum at the mid-point, so it falls and then rises.

Teacher's Note:
1. Potential of a positive charge is large near the charge.
2. The mid-point is where the field is zero.

Answer: (c) \( \frac{\sigma R}{2\epsilon_0} \)

Explanation:
1. Every element is at distance R from the centre, so \( V=\frac{kq}{R} \) with \( q=\sigma\cdot2\pi R^2 \).
2. \( V=\frac{\sigma2\pi R^2}{4\pi\epsilon_0R}=\frac{\sigma R}{2\epsilon_0} \).

Teacher's Note:
1. All points of the shell are equidistant from the centre.
2. Surface area of a hemisphere is 2 pi R squared.

Answer: (c) \( n^{5/3}U \)

Explanation:
1. Charge becomes nq and radius \( R'=n^{1/3}R \) (volume conservation).
2. Energy \( U\propto\frac{q^2}{R} \): \( U'=\frac{n^2}{n^{1/3}}U=n^{5/3}U \).

Teacher's Note:
1. Volume is conserved when the drops merge.
2. Self energy is proportional to q squared over R.

Answer: (a) \( \frac{q\ln2}{4\pi\epsilon_0x_0} \)

Explanation:
1. \( V=\frac{q}{4\pi\epsilon_0x_0}\left(1-\frac12+\frac13-\frac14+\dots\right) \).
2. The series equals \( \ln2 \), so \( V=\frac{q\ln2}{4\pi\epsilon_0x_0} \).

Teacher's Note:
1. The alternating harmonic series sums to ln 2.
2. Potentials add as scalars.

Answer: (c) 6 m

Explanation:
1. For a point charge \( V=Er \), so \( r=\frac{3000}{500}=6\ \text{m} \).

Teacher's Note:
1. V = kq/r and E = kq/r^2.
2. Their ratio is r.

Answer: (d) +2 V

Explanation:
1. The integral from infinity to the centre of \( -\vec E\cdot d\vec\ell \) is the potential at the centre.
2. Every element is at distance R from the centre: \( V=\frac{kq}{R}=\frac{9\times10^9\times1.11\times10^{-10}}{0.5}\approx2\ \text{V} \), independent of the distribution.

Teacher's Note:
1. Potential is the line integral of -E dot dl from infinity.
2. Scalar addition does not depend on the angular distribution.

Answer: (d) 8 V

Explanation:
1. Work \( =2000\times0.02=40\ \text{J} \), so \( \Delta V=\frac{W}{q}=\frac{40}{5}=8\ \text{V} \).

Teacher's Note:
1. Potential difference is work per unit charge.
2. Use the distance in metres.

Answer: (c) 10 V

Explanation:
1. Field \( E=\frac{3000}{3}=1000\ \text{N/C} \).
2. \( \Delta V=Ed=1000\times0.01=10\ \text{V} \).

Teacher's Note:
1. Find E first.
2. Convert the distance into metres.

Answer: (c) \( \frac{V}{6r} \)

Explanation:
1. With charge q: \( V=\frac{kq}{r}-\frac{kq}{3r}=\frac{2kq}{3r} \), so \( kq=\frac{3rV}{2} \).
2. Field at 3r \( =\frac{kq}{9r^2}=\frac{3rV/2}{9r^2}=\frac{V}{6r} \).

Teacher's Note:
1. Express kq through V first.
2. Field outside is kq / r^2.

Answer: (b) may increase or decrease

Explanation:
1. Potential energy is \( U=qV \): for a positive charge it decreases, for a negative charge it increases.

Teacher's Note:
1. The sign of the charge decides the answer.
2. Positive charges move to lower potential on their own.

Answer: (b) −20 V/m

Explanation:
1. \( E=-\frac{dV}{dx}=-(10x+10) \), so at x = 1: \( E=-20\ \text{V/m} \).

Teacher's Note:
1. The field is the negative gradient of potential.
2. Differentiate before substituting x.

Answer: (a) \( -\frac23 \)

Explanation:
1. With equal separations the energy is proportional to \( q(2q)+q(xq)+(2q)(xq)=q^2(2+3x) \).
2. Zero when \( x=-\frac23 \).

Teacher's Note:
1. Sum the products of charges over all pairs.
2. The distance is common to all pairs.