JEE Physics Electric Charges And Fields MCQs Set 11

Answer: (b) \( \frac\phi2 \)

Explanation:
1. By Gauss's law the flux depends only on the enclosed charge: \( \phi=\frac{q}{\varepsilon_0} \).
2. The size of the cube does not matter; halving the charge halves the flux: \( \frac\phi2 \).

Teacher's Note:
1. Flux through a closed surface depends only on the charge inside it.
2. Changing the size of the surface does not change the flux.

Answer: (a) 99 N

Explanation:
1. Force \( F\propto q_1q_2 \).
2. New force \( =100\times1.1\times0.9=99\ \text{N} \).

Teacher's Note:
1. The force is proportional to the product of the charges.
2. The distance stays the same.

Answer: (d) E

Explanation:
1. Field at the surface of a charged conductor \( E=\frac{\sigma}{\varepsilon_0} \).
2. It depends only on σ, which is unchanged, so the field remains E.

Teacher's Note:
1. Surface field of a conductor is σ/ε0 whatever the radius.
2. The total charge increases, but so does the area.

Answer: (c) Stable when moved along the X-axis

Explanation:
1. Along X: if the middle charge moves towards one fixed charge, the repulsion from that charge grows and pushes it back, so the force is restoring: stable.
2. Along Y or Z: the repulsions from the two outer charges both have components pointing away from the axis, which push the charge further away: unstable.

Teacher's Note:
1. Stable equilibrium needs a restoring force for displacements in the direction considered.
2. Like charges repel, so moving closer increases the repulsion.

Answer: (c) \( \vec E/3 \)

Explanation:
1. Field at Q (due to −3Q): \( E=\frac{k(3Q)}{d^2} \), directed from Q towards −3Q.
2. Field at −3Q (due to Q): \( E'=\frac{kQ}{d^2}=\frac E3 \), also directed from Q towards −3Q (away from the positive charge).
3. So the field is \( \vec E/3 \) in the same direction.

Teacher's Note:
1. Each charge produces a field only at the other's location.
2. Both fields point from the positive towards the negative charge.

Answer: (b) \( Q=-2\sqrt2\,q \)

Explanation:
1. Let the side be d. Field on a Q from the other Q (distance \( \sqrt2d \)) is \( \frac{kQ}{2d^2} \) along the diagonal.
2. Fields from the two q charges (each at distance d, at right angles) add along the diagonal to \( \frac{\sqrt2kq}{d^2} \).
3. For zero resultant these must be opposite: \( \frac Q2+\sqrt2q=0\Rightarrow Q=-2\sqrt2\,q \).

Teacher's Note:
1. Resolve the two fields from the q charges along the diagonal.
2. The charges must have opposite signs to cancel.

Answer: (a) \( \pi R^2E \)

Explanation:
1. The flux entering through the flat face is \( E\times\pi R^2 \).
2. No charge is enclosed, so the same flux must leave through the curved surface: \( \pi R^2E \).

Teacher's Note:
1. For a closed surface with no charge inside, flux in equals flux out.
2. The projected area of a hemisphere is a circle of area πR².

Answer: (d) \( \pi,\ \frac{kQ^2}{4L^2} \)

Explanation:
1. With no gravity, the only force on each ball (besides the string tension) is the mutual repulsion, so the balls move as far apart as possible: the strings are in a straight line, \( \theta=\pi \).
2. Distance between the balls \( 2L \): \( T=\frac{kQ^2}{(2L)^2}=\frac{kQ^2}{4L^2} \).

Teacher's Note:
1. Without gravity, the repulsion alone balances the tension.
2. The distance between the balls is twice the string length.

Answer: (d) 16E

Explanation:
1. Field of a sphere outside it is \( \frac{kq}{r^2} \). At the midpoint the distance is half, so the field of A is \( 4E \) (directed towards B).
2. The field of the −3Q sphere is \( \frac{k(3Q)}{(d/2)^2}=3\times4E=12E \), also directed towards B.
3. The fields add: \( 4E+12E=16E \).

Teacher's Note:
1. Halving the distance quadruples the field.
2. The fields of Q and −3Q point the same way between the spheres.

Answer: (c) \( \frac{2k\lambda}{R} \)

Explanation:
1. Use the field of a small element and cancel the components along the axis of symmetry's perpendicular: \( dE=\frac{k\lambda R\,d\theta}{R^2} \).
2. Components perpendicular to the axis cancel; along the axis: \( E=\int_{-\pi/2}^{\pi/2}\frac{k\lambda}{R}\cos\theta\,d\theta=\frac{2k\lambda}{R} \).

Teacher's Note:
1. Use symmetry to drop the perpendicular components.
2. The integral of cos θ from −π/2 to π/2 is 2.

Answer: (b) 90°

Explanation:
1. Torque \( \tau=pE\sin\theta \).
2. It is maximum when \( \sin\theta=1 \), that is \( \theta=90^\circ \).

Teacher's Note:
1. The torque is the cross product of p and E.
2. At 0° and 180° the dipole is in equilibrium and the torque is zero.

Answer: (a) \( \frac\pi3+\tan^{-1}\left(\frac{\sqrt3}{2}\right) \)

Explanation:
1. For a dipole: radial component \( E_r=\frac{2kp\cos\alpha}{r^3} \), tangential component \( E_\perp=\frac{kp\sin\alpha}{r^3} \), with \( \alpha=\frac\pi3 \).
2. The field makes an angle \( \phi \) with OP where \( \tan\phi=\frac{E_\perp}{E_r}=\frac{\tan\alpha}{2}=\frac{\sqrt3}2 \).
3. So \( \theta=\alpha+\phi=\frac\pi3+\tan^{-1}\left(\frac{\sqrt3}2\right) \).

Teacher's Note:
1. The field of a dipole is not along the radius vector.
2. Add the angle of OP to the angle between E and OP.

Answer: (d) 2 : 1

Explanation:
1. In contact the spheres reach the same potential: \( \frac{kq_1}{R}=\frac{kq_2}{2R}\Rightarrow q_2=2q_1 \).
2. Field at the surface \( E=\frac{kq}{r^2} \): \( \frac{E_1}{E_2}=\frac{q_1}{q_2}\cdot\frac{(2R)^2}{R^2}=\frac12\times4=2 \).
3. Ratio \( 2:1 \) (small sphere : large sphere).

Teacher's Note:
1. After contact the potentials are equal, not the charges.
2. Surface charge density is larger on the smaller sphere.