JEE Physics Electric Charges and Fields MCQs Set 30

Answer: (c) \( (63\hat i-27\hat j)\times10^2 \)

Explanation:
1. Charge \( q_1 \): distance \( \sqrt{1+9}=\sqrt{10} \) m, field \( \dfrac{9\times10^9\sqrt{10}\times10^{-6}}{10}=9\sqrt{10}\times10^2 \) along \( (-1,3)/\sqrt{10} \), i.e. \( (-9,27)\times10^2 \).
2. Charge \( q_2 \): distance 5 m, field \( \dfrac{9\times10^9\times25\times10^{-6}}{25}=9\times10^3 \) V/m towards \( q_2 \), along \( (4,-3)/5 \), i.e. \( (72,-54)\times10^2 \). Sum \( =(63,-27)\times10^2 \).

Teacher's Note:
1. Use Coulomb's law and the superposition principle with vectors, and Gauss's law (flux = enclosed charge / epsilon_0) for symmetric charge distributions.
2. Check the direction of the field or force by symmetry before computing the magnitude.

Answer: (a) \( -\dfrac Q4 \)

Explanation:
1. Force on the charge at the origin: \( \dfrac{kQq}{(d/2)^2}+\dfrac{kQ^2}{d^2}=0 \) (the signs must be opposite).
2. \( 4q+Q=0\Rightarrow q=-\dfrac Q4 \).

Teacher's Note:
1. Use Coulomb's law and the superposition principle with vectors, and Gauss's law (flux = enclosed charge / epsilon_0) for symmetric charge distributions.
2. Check the direction of the field or force by symmetry before computing the magnitude.

Answer: (c) \( y \)

Explanation:
1. The y-components add: \( F_y=\dfrac{2kqq_0\,y}{(a^2+y^2)^{3/2}} \), directed away from the axis for like charges.
2. For \( y\ll a \), \( F_y\approx\dfrac{2kqq_0}{a^3}y\propto y \) (so the origin is unstable along y).

Teacher's Note:
1. Use Coulomb's law and the superposition principle with vectors, and Gauss's law (flux = enclosed charge / epsilon_0) for symmetric charge distributions.
2. Check the direction of the field or force by symmetry before computing the magnitude.

Answer: (a) \( (m_p/m_e)^{1/2} \)

Explanation:
1. \( s=\tfrac12at^2=\dfrac{qE}{2m}t^2 \), same s, E and \( |q| \), so \( t\propto\sqrt m \).
2. \( \dfrac{t_2}{t_1}=\sqrt{\dfrac{m_p}{m_e}}\approx43 \).

Teacher's Note:
1. Use Coulomb's law and the superposition principle with vectors, and Gauss's law (flux = enclosed charge / epsilon_0) for symmetric charge distributions.
2. Check the direction of the field or force by symmetry before computing the magnitude.

Answer: (c) execute oscillatory but not simple harmonic motion about the origin

Explanation:
1. The net force on Q is directed towards the origin along the x-axis, \( F=\dfrac{2kqQx}{(a^2+x^2)^{3/2}} \), and Q passes through the origin to the other side.
2. The force is proportional to x only for \( x\ll a \); with the amplitude 2a it is not, so the motion is oscillatory but not simple harmonic.

Teacher's Note:
1. Use Coulomb's law and the superposition principle with vectors, and Gauss's law (flux = enclosed charge / epsilon_0) for symmetric charge distributions.
2. Check the direction of the field or force by symmetry before computing the magnitude.

Answer: (a) \( \dfrac{3Ze}{\pi R^3} \)

Explanation:
1. For \( a=0 \): \( \rho=d\left(1-\dfrac rR\right) \), \( Ze=\int_0^R4\pi r^2d\left(1-\dfrac rR\right)dr=4\pi dR^3\left(\dfrac13-\dfrac14\right)=\dfrac{\pi dR^3}3 \).
2. \( d=\dfrac{3Ze}{\pi R^3} \).

Teacher's Note:
1. Use Coulomb's law and the superposition principle with vectors, and Gauss's law (flux = enclosed charge / epsilon_0) for symmetric charge distributions.
2. Check the direction of the field or force by symmetry before computing the magnitude.

Answer: (c) 6

Explanation:
1. Solid cylinder (no cavity): \( E_1=\dfrac{\rho R^2}{2\epsilon_0(2R)}=\dfrac{\rho R}{4\epsilon_0} \).
2. The cavity acts as charge \( -\rho\cdot\tfrac43\pi\left(\tfrac R2\right)^3=-\dfrac{\pi\rho R^3}6 \) at distance 2R: \( E_2=\dfrac{\rho R}{96\epsilon_0} \) opposite. Net \( =\dfrac{\rho R}{\epsilon_0}\left(\dfrac14-\dfrac1{96}\right)=\dfrac{23\rho R}{96\epsilon_0} \), so \( 16k=96\Rightarrow k=6 \).

Teacher's Note:
1. Use Coulomb's law and the superposition principle with vectors, and Gauss's law (flux = enclosed charge / epsilon_0) for symmetric charge distributions.
2. Check the direction of the field or force by symmetry before computing the magnitude.

Answer: (a) \( \dfrac{q^2}{4\pi\epsilon_0L^2}\ \text{N} \)

Explanation:
1. Six equal +q charges would give zero force; five of them equal minus the force of the missing one.
2. Distance from the centre to a vertex is L, so \( F=\dfrac{q^2}{4\pi\epsilon_0L^2} \).

Teacher's Note:
1. Use Coulomb's law and the superposition principle with vectors, and Gauss's law (flux = enclosed charge / epsilon_0) for symmetric charge distributions.
2. Check the direction of the field or force by symmetry before computing the magnitude.

Answer: (c) \( 3.2\times10^{-9}\ \text{C} \)

Explanation:
1. Each string makes angle \( \theta \) with the vertical, \( \sin\theta=\dfrac{a/\sqrt3}{l}=\dfrac{0.03/\sqrt3}{1}=0.0173 \), so \( \tan\theta\approx0.0173 \).
2. Net horizontal force on a particle \( \sqrt3\dfrac{kq^2}{a^2}=mg\tan\theta \): \( q^2=\dfrac{10^{-3}\times10\times0.0173\times9\times10^{-4}}{\sqrt3\times9\times10^9}=10^{-17} \), so \( q=3.2\times10^{-9} \) C.

Teacher's Note:
1. Use Coulomb's law and the superposition principle with vectors, and Gauss's law (flux = enclosed charge / epsilon_0) for symmetric charge distributions.
2. Check the direction of the field or force by symmetry before computing the magnitude.

Answer: (b) \( V \)

Explanation:
1. The potential difference between the inner sphere and the shell is set by the field in the gap, which depends only on the inner charge Q: \( \dfrac{Q}{4\pi\epsilon_0}\left(\dfrac1a-\dfrac1b\right) \).
2. Giving the shell more charge raises both by the same amount (and the shell is an equipotential), so the difference stays V.

Teacher's Note:
1. Use Coulomb's law and the superposition principle with vectors, and Gauss's law (flux = enclosed charge / epsilon_0) for symmetric charge distributions.
2. Check the direction of the field or force by symmetry before computing the magnitude.

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

Explanation:
1. Enclosed charge at \( r \) (\( a\lt r\lt b \)): \( Q+\int_a^r4\pi r'^2\dfrac A{r'}dr'=Q+2\pi A(r^2-a^2) \).
2. \( E=\dfrac{k[Q-2\pi Aa^2+2\pi Ar^2]}{r^2} \) is constant only if \( Q=2\pi Aa^2 \), so \( A=\dfrac Q{2\pi a^2} \).

Teacher's Note:
1. Use Coulomb's law and the superposition principle with vectors, and Gauss's law (flux = enclosed charge / epsilon_0) for symmetric charge distributions.
2. Check the direction of the field or force by symmetry before computing the magnitude.

Answer: (b) \( E_0a^2 \)

Explanation:
1. Area vector from the edges \( (a,0,a) \) and \( (0,a,0) \): \( \vec A=(a,0,a)\times(0,a,0)=(-a^2,0,a^2) \).
2. Flux \( =\vec E\cdot\vec A=E_0(-a^2) \), magnitude \( E_0a^2 \).

Teacher's Note:
1. Use Coulomb's law and the superposition principle with vectors, and Gauss's law (flux = enclosed charge / epsilon_0) for symmetric charge distributions.
2. Check the direction of the field or force by symmetry before computing the magnitude.

Answer: (d) \( \dfrac{-2\text{ C}}{\epsilon_0} \)

Explanation:
1. The disc is centred on the face \( x=-a/2 \), so half of it (3 C) is inside. The rod from \( a/4 \) to \( 5a/4 \) has only the part up to \( x=a/2 \) inside: \( \tfrac14 \) of 8 C = 2 C. The charge \( -7 \) C is inside; 3 C at \( (-3a/4,\dots) \) is outside.
2. Enclosed charge \( =3+2-7=-2 \) C, so the flux is \( -\dfrac{2}{\epsilon_0} \) C.

Teacher's Note:
1. Use Coulomb's law and the superposition principle with vectors, and Gauss's law (flux = enclosed charge / epsilon_0) for symmetric charge distributions.
2. Check the direction of the field or force by symmetry before computing the magnitude.

Answer: (b) \( V \)

Explanation:
1. The field in the gap depends only on the charge Q of the inner sphere, so the potential difference between the sphere and the shell is unaffected.
2. The shell is a conductor (one potential), so the difference remains V.

Teacher's Note:
1. Use Coulomb's law and the superposition principle with vectors, and Gauss's law (flux = enclosed charge / epsilon_0) for symmetric charge distributions.
2. Check the direction of the field or force by symmetry before computing the magnitude.

Answer: (d) \( -\sqrt3\,ql\,\hat j \)

Explanation:
1. \( \vec p=\sum q\vec r=q(l,0)+(-2q)\left(\dfrac l2,\dfrac{\sqrt3l}2\right) \).
2. \( =q(l-l,\ -\sqrt3l)=-\sqrt3ql\,\hat j \).

Teacher's Note:
1. Use Coulomb's law and the superposition principle with vectors, and Gauss's law (flux = enclosed charge / epsilon_0) for symmetric charge distributions.
2. Check the direction of the field or force by symmetry before computing the magnitude.

Answer: (b) \( 27F \)

Explanation:
1. On the equatorial line \( E\propto\dfrac1{y^3} \), so \( F=QE\propto y^{-3} \).
2. At \( y/3 \): \( F'=3^3F=27F \).

Teacher's Note:
1. Use Coulomb's law and the superposition principle with vectors, and Gauss's law (flux = enclosed charge / epsilon_0) for symmetric charge distributions.
2. Check the direction of the field or force by symmetry before computing the magnitude.

Answer: (d) \( \dfrac R{\sqrt2} \)

Explanation:
1. \( E=\dfrac{kQh}{(R^2+h^2)^{3/2}} \).
2. \( \dfrac{dE}{dh}=0\Rightarrow R^2+h^2=3h^2\Rightarrow h=\dfrac R{\sqrt2} \).

Teacher's Note:
1. Use Coulomb's law and the superposition principle with vectors, and Gauss's law (flux = enclosed charge / epsilon_0) for symmetric charge distributions.
2. Check the direction of the field or force by symmetry before computing the magnitude.

Answer: (b) zero

Explanation:
1. The sphere was neutral and is isolated, so its net charge stays zero.
2. Induction only redistributes charge: negative nearer the point charge, positive farther away.

Teacher's Note:
1. Use Coulomb's law and the superposition principle with vectors, and Gauss's law (flux = enclosed charge / epsilon_0) for symmetric charge distributions.
2. Check the direction of the field or force by symmetry before computing the magnitude.

Answer: (d) 6

Explanation:
1. The line is at height \( h=\dfrac{\sqrt3a}2 \) above the centre. The half-width \( a/2 \) subtends \( \tan\alpha=\dfrac{a/2}{h}=\dfrac1{\sqrt3} \), \( \alpha=30^\circ \), so the rectangle subtends \( 60^\circ \) at the line.
2. The flux through length L of a cylinder around the line is \( \dfrac{\lambda L}{\epsilon_0} \); the fraction through the rectangle is \( \dfrac{60^\circ}{360^\circ}=\dfrac16 \), so \( n=6 \).

Teacher's Note:
1. Use Coulomb's law and the superposition principle with vectors, and Gauss's law (flux = enclosed charge / epsilon_0) for symmetric charge distributions.
2. Check the direction of the field or force by symmetry before computing the magnitude.