JEE Physics Electric Charges and Fields MCQs Set 14

Answer: (a) \( \tan\theta=\frac{\sigma q}{2\varepsilon_0 mg} \)

Explanation:
1. Field of a large sheet: \( E=\frac{\sigma}{2\varepsilon_0} \), so the electric force is \( qE \) horizontally.
2. Equilibrium: \( T\sin\theta=qE,\ T\cos\theta=mg \).
3. \( \tan\theta=\frac{qE}{mg}=\frac{\sigma q}{2\varepsilon_0 mg} \)

Teacher's Note:
1. The field of an infinite sheet is σ/2ε0, independent of distance; tan θ is the ratio of the horizontal force to the weight.

Answer: (d) \( \frac{Q(R+r)}{4\pi\varepsilon_0(R^2+r^2)} \)

Explanation:
1. Equal densities σ: \( q_1=4\pi r^2\sigma,\ q_2=4\pi R^2\sigma \), so \( Q=4\pi\sigma(r^2+R^2) \).
2. \( V=\frac{1}{4\pi\varepsilon_0}\left(\frac{q_1}{r}+\frac{q_2}{R}\right)=\frac{\sigma}{\varepsilon_0}(r+R) \).
3. \( V=\frac{Q(R+r)}{4\pi\varepsilon_0(R^2+r^2)} \)

Teacher's Note:
1. Potential at the centre is the sum of kq/radius for each shell.

Answer: (c) The number of flux lines entering the surface must be equal to the number of flux lines leaving it

Explanation:
1. Zero net flux means the net charge enclosed is zero.
2. Field lines entering equal field lines leaving, so (c) is true.
3. The field itself need not vanish or be uniform, since charges may lie outside, and (d) is wrong because equal positive and negative charges can lie inside.

Teacher's Note:
1. Gauss's law only tells the net enclosed charge, not the field at each point.

Answer: (c) Q

Explanation:
1. When joined, A and B form one conductor.
2. All the charge resides on the outer surface of the outer shell, so the whole of Q moves from A to B.

Teacher's Note:
1. Charge on a conductor always sits on its outer surface; connecting inner and outer conductors transfers everything outward.

Answer: (d) \( \frac{Q\ln 2}{4\pi\varepsilon_0 L} \)

Explanation:
1. Charge per unit length \( \lambda=Q/L \); take an element at distance x from O.
2. \( V=\frac{1}{4\pi\varepsilon_0}\int_L^{2L}\frac{(Q/L)\,dx}{x} \).
3. \( V=\frac{Q}{4\pi\varepsilon_0 L}\ln 2 \)

Teacher's Note:
1. Integrate dq/x over the rod, from distance L to 2L.

Answer: (c) \( \frac{\sqrt2 Q^2}{\pi\varepsilon_0 a} \)

Explanation:
1. Distance from centre to each corner is \( a/\sqrt2 \), so \( V_{centre}=4\cdot\frac{Q}{4\pi\varepsilon_0}\cdot\frac{\sqrt2}{a}=\frac{\sqrt2 Q}{\pi\varepsilon_0 a} \).
2. \( W=q(V_\infty-V)=(-Q)(0-V)=\frac{\sqrt2 Q^2}{\pi\varepsilon_0 a} \)

Teacher's Note:
1. Work by the external agent equals the change in potential energy, q(V final minus V initial).

Answer: (b) −y

Explanation:
1. At displacement y each charge is at distance \( \sqrt{a^2+y^2} \); the x-components cancel.
2. Net force along y: \( F=-2\frac{q q_0}{4\pi\varepsilon_0}\frac{y}{(a^2+y^2)^{3/2}}\approx-\frac{q q_0 y}{2\pi\varepsilon_0 a^3} \).
3. So F is proportional to −y (restoring, SHM).

Teacher's Note:
1. For y much smaller than a, expand (a² + y²) as a² and the force becomes linear in y.

Answer: (c) A torque as well as a translational force

Explanation:
1. Because the dipole is not parallel to the field, the torque p × E is non-zero.
2. In a non-uniform field the two charges feel unequal forces, so a net translational force also acts.

Teacher's Note:
1. Torque needs a non-zero angle; net force needs a non-uniform field. Here both are present.

Answer: (c) \( \frac{q}{6\varepsilon_0} \)

Explanation:
1. Place the charge at the centre of a cube of edge a; the square is one face of that cube.
2. Total flux \( q/\varepsilon_0 \) is shared equally by 6 faces, so each face gets \( \frac{q}{6\varepsilon_0} \).

Teacher's Note:
1. Imagine the square as one face of a cube with the charge at its centre and use symmetry.

Answer: (b) \( \frac{Q}{8\pi\varepsilon_0 R} \)

Explanation:
1. \( V_A=\frac{1}{4\pi\varepsilon_0}\left(\frac{Q}{R}+\frac{2Q}{2R}\right)=\frac{2Q}{4\pi\varepsilon_0R} \)
2. \( V_B=\frac{1}{4\pi\varepsilon_0}\frac{3Q}{2R} \)
3. \( V_A-V_B=\frac{Q}{8\pi\varepsilon_0 R} \)

Teacher's Note:
1. The potential difference between two concentric shells depends only on the charge on the inner shell: kQ(1/R − 1/2R).

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

Explanation:
1. Charge within radius r: \( Q+\int_a^r\frac{A}{r'}4\pi r'^2dr'=Q+2\pi A(r^2-a^2) \).
2. \( E=\frac{1}{4\pi\varepsilon_0}\frac{Q-2\pi Aa^2+2\pi Ar^2}{r^2} \).
3. For E to be independent of r, the constant part must vanish: \( Q-2\pi Aa^2=0\Rightarrow A=\frac{Q}{2\pi a^2} \)

Teacher's Note:
1. Apply Gauss's law, and demand that the 1/r² term cancels so that E does not depend on r.

Answer: (c) \( \rho(r)\propto\frac{1}{r} \)

Explanation:
1. Equal ΔV for equal Δr means the field magnitude is constant: \( E=-dV/dr=\text{const} \).
2. \( E=\frac{q(r)}{4\pi\varepsilon_0 r^2}=\text{const}\Rightarrow q(r)\propto r^2 \).
3. \( \frac{dq}{dr}=4\pi r^2\rho\propto r\Rightarrow\rho\propto\frac1r \)

Teacher's Note:
1. Constant E means enclosed charge proportional to r²; differentiate to get ρ.

Answer: (c) \( \frac{\sigma}{4a\varepsilon_0} \)

Explanation:
1. Field on the axis of an annulus: \( E=\frac{\sigma}{2\varepsilon_0}\left[\frac{h}{\sqrt{h^2+a^2}}-\frac{h}{\sqrt{h^2+b^2}}\right] \).
2. For small h: \( E\approx\frac{\sigma h}{2\varepsilon_0}\left(\frac1a-\frac1{2a}\right)=\frac{\sigma h}{4a\varepsilon_0} \).

Teacher's Note:
1. Treat the annulus as a full disc of radius b minus a disc of radius a, then expand for small h.

Answer: (c) \( \sigma_1\neq0,\ Q_1=0;\ \sigma_2=0,\ Q_2=0 \)

Explanation:
1. The charges are close to the cavity wall, so induced charge appears on the inner surface with non-zero density; its net value is −(Q − Q) = 0, so Q1 = 0.
2. A neutral shell with zero net charge inside the cavity needs no charge on its outer surface, so Q2 = 0 and the outer surface is uniformly neutral, σ2 = 0.

Teacher's Note:
1. The net enclosed charge is zero, so the outer surface stays uncharged, but the inner surface still shows the induced distribution.

Answer: (b) \( \frac{q}{2\pi\varepsilon_0}\left[\frac1R-\frac1{\sqrt{R^2+d^2}}\right] \)

Explanation:
1. \( V_1=\frac{q}{4\pi\varepsilon_0}\left[\frac1R-\frac1{\sqrt{R^2+d^2}}\right] \) at the centre of the +q ring.
2. By symmetry \( V_2=-V_1 \).
3. \( V_1-V_2=2V_1=\frac{q}{2\pi\varepsilon_0}\left[\frac1R-\frac1{\sqrt{R^2+d^2}}\right] \)

Teacher's Note:
1. Potential at the centre of a ring due to its own charge is kq/R; the other ring contributes kq'/sqrt(R²+d²).

Answer: (c) 5q/6R

Explanation:
1. Induced charge −q spreads on the inner surface (radius 2R) and +q on the outer surface (radius 3R).
2. Potentials at the centre: \( V=k\left[\frac{q}{R}-\frac{q}{2R}+\frac{q}{3R}\right]=kq\cdot\frac{5}{6R} \).

Teacher's Note:
1. Add the point charge, the induced inner charge and the outer charge; each uniform shell contributes kQ/radius at the centre.

Answer: (d) A is false, but R is true.

Explanation:
1. A negative charge is pushed by the field towards higher potential, so the assertion is false.
2. The reason, that an electron is negatively charged, is true.

Teacher's Note:
1. Negative charges move from lower to higher potential if they are free to move.

Answer: (a) −80 V

Explanation:
1. \( V_A-V_0=-\int_0^2 30x^2dx=-10x^3\Big|_0^2=-80\ \text{V} \)

Teacher's Note:
1. Potential difference is the negative line integral of E.

Answer: (b) 2

Explanation:
1. Force is proportional to \( q(Q-q) \).
2. Maximising: \( \frac{d}{dq}[q(Q-q)]=Q-2q=0\Rightarrow q=\frac Q2 \), so Q/q = 2.

Teacher's Note:
1. A product with a fixed sum is greatest when the two parts are equal.

Answer: (a) −8

Explanation:
1. \( E_x=-\frac{\partial V}{\partial x}=-8x \)
2. At x = 1: E = −8 î V/m, so n = −8.

Teacher's Note:
1. Take the negative partial derivative of V with respect to x.

Answer: (a) −110

Explanation:
1. \( V=-\vec E\cdot\vec r=-(25\cdot2+30\cdot2)=-110\ \text{V} \)

Teacher's Note:
1. For a uniform field V = −E.r when V(0) = 0.

Answer: (d) Zero

Explanation:
1. The field is conservative, \( \vec E=\nabla(xy) \), so the work depends only on the end points.
2. \( W=\int\vec E\cdot d\vec r=xy\Big|_A^B=(4)(1)-(2)(2)=0 \)

Teacher's Note:
1. A field of the form (y, x) is the gradient of xy, so the work is the difference of xy between the end points.

Answer: (b) 2

Explanation:
1. At equilibrium the potentials are equal: \( \frac{q_A}{R_A}=\frac{q_B}{R_B} \).
2. \( E=\frac{V}{R}\Rightarrow\frac{E_A}{E_B}=\frac{R_B}{R_A}=2 \)

Teacher's Note:
1. Surface field on a sphere equals V/R, so the smaller sphere has the stronger surface field.

Answer: (b) −18

Explanation:
1. \( W=qV=(2\times1.6\times10^{-19})(25)=8\times10^{-18}\ \text{J} \)
2. So x = −18.

Teacher's Note:
1. An alpha particle has charge 2e.

Answer: (c) 32

Explanation:
1. Hypotenuse = \( \sqrt2\,l \).
2. \( U=\frac{1}{4\pi\varepsilon_0}\left[-\frac{Q_0q}{l}-\frac{Q_0q}{l}+\frac{q^2}{\sqrt2\,l}\right]=0 \).
3. \( Q_0=\frac{q}{2\sqrt2}=\frac{2q}{\sqrt{32}} \), so n = 32.

Teacher's Note:
1. Sum the three pair energies and set the total to zero.