JEE Physics Electric Charges and Fields MCQs Set 32

Answer: (d) zero

Explanation:
1. The sphere was neutral and no charge was added or removed, so its net charge stays zero.
2. The external positive charge only induces a redistribution: negative charge on the near side, an equal positive charge on the far side.
3. Net charge on the sphere remains zero.

Teacher's Note:
1. Induction redistributes charge but does not change the total.
2. Conservation of charge applies to an isolated conductor.

Answer: (b) \( \frac{q}{\epsilon_0} \)

Explanation:
1. By Gauss's law, the flux through a closed surface depends only on the charge enclosed.
2. The enclosed charge is \( +q \).
3. The outside charge contributes equal inward and outward flux, so net flux is \( q/\epsilon_0 \).

Teacher's Note:
1. Only enclosed charge enters in the net flux, though all charges contribute to the field.
2. Flux entering equals flux leaving for an outside charge.

Answer: (a) 8 V-m

Explanation:
1. Flux \( \phi=\vec E\cdot\vec A \).
2. Compute the dot product: \( 4\hat i\cdot(2\hat i+3\hat j)=4\times2=8 V-m \).

Teacher's Note:
1. Only the component of \( A \) along \( E \) matters.
2. The dot product of \( i \) with \( j \) is zero.

Answer: (a) A potential difference appears between the two cylinders when a charge density is given to the inner cylinder

Explanation:
1. Charge on the inner cylinder produces a field in the region between the cylinders (induced charge of opposite sign on the inner face of the outer cylinder), hence a potential difference.
2. Charge on the outer cylinder alone produces no field inside it, so no potential difference.
3. A line charge on the axis also produces a field between the cylinders (so \( (c) \) is wrong), and equal charge densities on both give a field in the gap (so \( (d) \) is wrong).

Teacher's Note:
1. Field inside a charged hollow conductor is zero; between nested conductors only inner charge matters.
2. Gauss's law with a cylindrical surface.

Answer: (c) 100 V-m

Explanation:
1. By Gauss's law, flux depends only on the enclosed charge, not on the radius.
2. Both spheres enclose all the charge, so the flux is the same: \( 100 V-m \).

Teacher's Note:
1. Flux is independent of the size of the Gaussian surface if it encloses the same charge.
2. Field falls as \( 1/r^2 \) but area grows as \( r^2 \).

Answer: (c) \( \frac{Q+q}{4\pi b^2} \)

Explanation:
1. The charge \( q \) induces \( -q \) on the inner surface of the shell.
2. The shell's net charge is \( Q \), so the outer surface has \( Q+q \).
3. Surface charge density on the outer surface \( \sigma=\frac{Q+q}{4\pi b^2} \).

Teacher's Note:
1. Inner surface: \( -q \) (to cancel the field inside the conductor). Outer: total minus inner.
2. Outer-surface charge spreads uniformly on a sphere.

Answer: (a) \( \frac{4\phi}{\pi d^2} \)

Explanation:
1. Maximum flux occurs when the plane of the loop is perpendicular to the field: \( \phi=EA \).
2. Area \( A=\pi d^2/4 \).
3. So \( E=\frac{4\phi}{\pi d^2} \).

Teacher's Note:
1. Flux = \( E A \cos(\theta) \), maximum at \( \theta = 0 \).
2. Area of a circle: \( \pi d^2/4 \).

Answer: (c) gravitational and nuclear

Explanation:
1. Neutrons are uncharged, so electrostatic force is absent.
2. They have mass, so the gravitational force is attractive (though very weak).
3. The strong nuclear force acts between nucleons including neutrons and is attractive at short range.

Teacher's Note:
1. Neutral particles: no Coulomb force.
2. Nuclear force is charge independent.

Answer: (a) \( 180^\circ,\ \frac{1}{4\pi\epsilon_0}\frac{Q^2}{(2L)^2} \)

Explanation:
1. With no gravity, only the electric repulsion and thread tension act on each sphere.
2. Equilibrium needs the thread along the line of the force, so both threads are collinear: angle \( 180^\circ \), separation \( 2L \).
3. Tension equals the Coulomb force: \( T=\frac{1}{4\pi\epsilon_0}\frac{Q^2}{(2L)^2} \).

Teacher's Note:
1. Without gravity the repulsion is the only force, so the strings must be straight along it.
2. Distance between the spheres is \( 2L \).

Answer: (b) \( -\frac{Q}{4} \)

Explanation:
1. Let separation be \( d \). Equilibrium of one end charge \( Q \): repulsion from the other \( Q \) is \( kQ^2/d^2 \).
2. Attraction to the centre charge \( q \) (opposite sign) at distance \( d/2 \) is \( kQ|q|/(d/2)^2=4kQ|q|/d^2 \).
3. Equate: \( 4|q|=Q \), so \( q=-Q/4 \).

Teacher's Note:
1. Check the equilibrium of an end charge; the centre charge is automatically in balance by symmetry.
2. The middle charge must be opposite in sign.

Answer: (b) \( \frac{mg}{e} \)

Explanation:
1. Force on the electron is \( eE \).
2. Set it equal to the weight: \( eE=mg \).
3. So \( E=mg/e \).

Teacher's Note:
1. Equate electrical force to the gravitational force.
2. Units of mg/e are \( N/C \).

Answer: (b) \( E/3 \)

Explanation:
1. Field at \( Q \) due to \( -3Q \) is \( E=\frac{3kQ}{r^2} \).
2. Field at \( -3Q \) due to \( Q \) is \( \frac{kQ}{r^2}=\frac{E}{3} \).
3. Both fields point along the line from the positive towards the negative charge, so the field has the same direction: \( E/3 \).

Teacher's Note:
1. Field at a point depends only on the other charge.
2. Newton's third law: the two forces are equal in magnitude.

Answer: (c) \( 2.5\times10^5 \)

Explanation:
1. Field just outside: \( E=\frac{kq}{R^2} \), so \( q=\frac{ER^2}{k}=\frac{0.036\times0.01}{9\times10^9}=4\times10^{-14}\ \text{C} \).
2. Number of electrons \( n=\frac{q}{e}=\frac{4\times10^{-14}}{1.6\times10^{-19}}=2.5\times10^5 \).

Teacher's Note:
1. Use \( E = kq/R^2 \) just outside a charged conducting sphere.
2. Divide the charge by \( e \).

Answer: (c) \( \text{varies as }1/r \), where \( r \) is the distance from the axis

Explanation:
1. By Gauss's law for a coaxial cylinder of radius \( r \) and length \( l \): \( E\cdot2\pi rl=\lambda l/\epsilon_0 \).
2. So \( E=\frac{\lambda}{2\pi\epsilon_0 r}\propto\frac1r \).
3. The field is stronger near the inner cylinder.

Teacher's Note:
1. Line charge field varies as \( 1/r \); point charge as \( 1/r^2 \).
2. Gauss's law in cylindrical symmetry.

Answer: (c) \( qEy \)

Explanation:
1. Force \( qE \) is constant, so work done over distance \( y \) is \( qEy \).
2. By the work-energy theorem, KE gained = \( qEy \).

Teacher's Note:
1. Work-energy theorem.
2. The force does not depend on distance in a uniform field.

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

Explanation:
1. Acceleration \( a=eE/m \) (same charge magnitude).
2. Distance \( s=\tfrac12 at^2 \), so \( t=\sqrt{2s/a}\propto\sqrt m \).
3. Hence \( t_2/t_1=\sqrt{m_p/m_e} \) (about 43).

Teacher's Note:
1. Time is proportional to the square root of mass for the same force and distance.
2. Use \( s = (1/2) a t^2 \).

Answer: (b) \( \tan\theta \)

Explanation:
1. Field near a large conducting sheet is \( E=\sigma/\epsilon_0 \), uniform and horizontal.
2. Equilibrium of ball: \( T\cos\theta=mg \), \( T\sin\theta=qE \).
3. So \( \tan\theta=qE/mg\propto\sigma \).

Teacher's Note:
1. Resolve tension horizontally and vertically.
2. \( \tan(\theta) \) = electric force / weight.

Answer: (c) \( 2L \)

Explanation:
1. Zero field must lie outside the pair, on the side of the smaller charge (\( x>L \)).
2. Set \( \frac{8q}{x^2}=\frac{2q}{(x-L)^2} \), so \( 2(x-L)=x \).
3. Thus \( x=2L \).

Teacher's Note:
1. Between charges of opposite sign the fields add; null point is outside, near the smaller charge.
2. Take square roots to solve.

Answer: (c) the total flux through the surface is zero

Explanation:
1. The integral is the total (net) flux through the closed surface.
2. It being zero means the net charge enclosed is zero and inward flux equals outward flux.
3. It does not mean \( E \) is zero at every point.

Teacher's Note:
1. Zero net flux does not require zero field.
2. Outward flux equals inward flux.