JEE Physics Electrostatic Potential And Capacitance MCQs Set 27

Answer: (d) \( 4\sqrt{35}\ \text{N} \)

Explanation:
1. E = −∇V gives E_x = −(6 − 8y) = 2, E_y = −(−8x − 8 + 6z) = 10 and E_z = −6y = −6 at (1, 1, 1).
2. |E| = √140 = 2√35, so F = qE = 4√35 N.

Teacher's Note:
1. E = −∇V.

Answer: (b) charge will flow from B to A

Explanation:
1. V_A = 4πkσR_A and V_B = 4πkσ(R_B + R_A) because B also feels the potential of A's shell.
2. V_B > V_A, so charge flows from B to A.

Teacher's Note:
1. Charge flows from higher to lower potential.

Answer: (d) zero

Explanation:
1. W = pE(cos θ₁ − cos θ₂) with θ₁ = 90° and θ₂ = 270°.
2. cos 90° = cos 270° = 0, so W = 0.

Teacher's Note:
1. Initial and final energies are equal.

Answer: (a) \( \frac{4\pi\varepsilon_0r_1r_2}{r_2-r_1} \)

Explanation:
1. With the outer shell earthed, the system is a spherical capacitor.
2. C = 4πε₀r₁r₂/(r₂ − r₁).

Teacher's Note:
1. Spherical capacitor.

Answer: (d) I, II and III

Explanation:
1. An external field separates the charge centres of a non-polar molecule until the restoring force balances the electric force.
2. The molecule then has an induced dipole moment, so all three are true.

Teacher's Note:
1. Induced dipole.

Answer: (c) zero along all the paths AB, AC, AD and AE

Explanation:
1. All the points lie on a sphere centred on the charge, which is an equipotential surface.
2. So the work done is zero along every path.

Teacher's Note:
1. Equipotential means zero work.

Answer: (b) 10 V

Explanation:
1. Inside a hollow conductor the potential is constant and equals the surface potential, 10 V.

Teacher's Note:
1. Potential is uniform inside.

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

Explanation:
1. U = ½CV² with V fixed and C → kC, so the energy becomes k times.
2. The charge, and hence the surface density, increases k times, so the reason is false.

Teacher's Note:
1. V fixed, Q increases.

Answer: (d) −12 V

Explanation:
1. The positive terminal is at 0 V.
2. The negative terminal is 12 V lower, that is −12 V.

Teacher's Note:
1. Battery gives a 12 V difference.

Answer: (c) both (a) and (b)

Explanation:
1. Charge is sprayed on a belt by sharp points, and carried to the inside of a hollow sphere.
2. It then moves to the outer surface and spreads uniformly.

Teacher's Note:
1. Corona action and surface charge.

Answer: (a) a force and a torque

Explanation:
1. The two forces are unequal in a non-uniform field, so there is a net force.
2. They act at different points, so there is also a torque.

Teacher's Note:
1. Net force and torque.

Answer: (b) \( V_a \gt V_b=V_d \gt V_c \)

Explanation:
1. The potential on the axis near +q is positive and near −q is negative.
2. The equatorial points have zero potential, so V_a > V_b = V_d = 0 > V_c.

Teacher's Note:
1. V = 0 on the equator.

Answer: (c) A-2, B-1, C-4, D-3

Explanation:
1. V = kq/r is positive near a positive charge (A-2) and negative near a negative charge (B-1).
2. At the charge's own location r = 0 and V is infinite (C-4).
3. Outside a uniformly charged sphere V ∝ 1/r (D-3).

Teacher's Note:
1. Use V = kq/r.

Answer: (b) decreased, proportional to 1/2

Explanation:
1. E = E₀/K for a fixed charge.
2. The field decreases by a factor 1/2.

Teacher's Note:
1. E = E₀/K.

Answer: (b) the decrease in potential is steepest

Explanation:
1. The field points down the potential gradient.
2. It is directed along the steepest decrease in potential.

Teacher's Note:
1. E = −∇V.

Answer: (c) \( \frac{C_1V}{C_1+C_2} \)

Explanation:
1. Charge is conserved: C₁V = (C₁ + C₂)V′.
2. So V′ = C₁V/(C₁ + C₂).

Teacher's Note:
1. Common potential.

Answer: (c) electric field is normal to the surface of the cube

Explanation:
1. The surface of a conductor is an equipotential, so the field there is normal to it.
2. The field inside is zero and the potential is constant (not zero).

Teacher's Note:
1. Field is normal at a conductor surface.

Answer: (c) zero

Explanation:
1. Both points are at distance 2 from the origin.
2. They are at the same potential, so the difference is zero.

Teacher's Note:
1. Equal distances.

Answer: (a) \( \frac WQ\ \text{V} \)

Explanation:
1. Potential difference = work done per unit charge = W/Q.

Teacher's Note:
1. V = W/q.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. The field is conservative, so work per unit charge depends only on the end points.
2. The reason explains why the potential difference is path independent.

Teacher's Note:
1. Conservative field.

Answer: (b) increases K times

Explanation:
1. C = KC₀.
2. The capacity increases K times.

Teacher's Note:
1. C_medium = K C_air.

Answer: (b) potential difference between two points

Explanation:
1. Potential and potential energy shift when the reference is changed.
2. A potential difference stays the same.

Teacher's Note:
1. Only differences are physical.

Answer: (d) the potential energy is the same in both cases and is positive

Explanation:
1. Both points are at the same distance (9) from +Q.
2. U = kQq/9 is the same and positive for like charges.

Teacher's Note:
1. U depends on distance only.

Answer: (c) the field between the plates

Explanation:
1. The energy is held in the electric field between the plates.

Teacher's Note:
1. u = ½ε₀E² per unit volume.

Answer: (a) \( Q=-q \)

Explanation:
1. The corners are equidistant from the centre.
2. V ∝ −Q − q + 2q + 2Q = Q + q = 0, so Q = −q.

Teacher's Note:
1. Sum the charges.