A Gaussian surface inside the conductor material must enclose zero net charge. The charge \( +q \) induces \( -q \) on the inner cavity wall. Since the sphere is neutral, the remaining \( +q \) must migrate to the outer surface.
Correct Answer: C

Charge tends to accumulate at sharp points where the radius of curvature is small, leading to a higher surface charge density \( \sigma \) and a stronger local electric field.
Correct Answer: B

Inside the conductor, \( E = 0 \), so \( V \) is constant and equal to its surface value \( \displaystyle\frac{kQ}{R} \). Outside, it behaves like a point charge: \( V \propto \displaystyle\frac{1}{r} \).
Correct Answer: B

In equilibrium, the electric field must be perpendicular to the surface of a conductor. Field lines from the positive charge will terminate on negative induced charges on the plane’s surface.
Correct Answer: B

Regardless of the net charge on a conductor, the electric field inside the material of a conductor in electrostatic equilibrium is always zero.
Correct Answer: C

Equilibrium requires \( V_1 = V_2 \), so \( \displaystyle\frac{kq_1}{R} = \displaystyle\frac{kq_2}{2R} \Rightarrow q_2 = 2q_1 \). Since \( q_1 + q_2 = Q \), we have \( q_1 + 2q_1 = Q \Rightarrow q_1 = Q/3 \).
Correct Answer: A

\( V \propto \displaystyle\frac{q}{R} \) and \( q = \sigma (4\pi R^2) \), so \( V \propto \sigma R \). If \( V_1 = V_2 \), then \( \sigma_1 R_1 = \sigma_2 R_2 \). This implies \( \sigma \propto 1/R \).
Correct Answer: B

When touched to the inside, the small sphere becomes part of the “inner surface.” Since excess charge must reside on the outermost boundary of a conductor, all charge moves to the container’s exterior.
Correct Answer: C

Charge is conserved, but energy is usually lost as heat (Joule heating) in the wire or as electromagnetic radiation as the charges accelerate to reach equilibrium.
Correct Answer: B

The charge distributes such that \( q_A / q_B = r / R \). If \( R \) is very large, almost all the charge moves to sphere B, leaving sphere A with nearly zero charge.
Correct Answer: C

Disconnected means charge \( Q \) is constant. \( C = \epsilon_0 A / d \), so doubling \( d \) halves \( C \). Since \( U = \displaystyle\frac{Q^2}{2C} \), halving \( C \) doubles the energy \( U \).
Correct Answer: C

\( \displaystyle\frac{1}{C_{eq}} = \displaystyle\frac{1}{2} + \displaystyle\frac{1}{6} = \displaystyle\frac{4}{6} \Rightarrow C_{eq} = 1.5 \text{ \(\mu\)F} \). Total charge \( Q_{tot} = C_{eq} V = 1.5 \times 12 = 18 \text{ \(\mu\)C} \). In series, each capacitor holds the same total charge.
Correct Answer: C

Battery connected means \( V \) is constant. \( C = \epsilon_0 A / d \), so doubling \( A \) doubles \( C \). Since \( U = \displaystyle\frac{1}{2}CV^2 \), doubling \( C \) doubles the energy \( U \).
Correct Answer: B

\( C_p = C + C + C = 3C \).
\( \displaystyle\frac{1}{C_s} = \displaystyle\frac{1}{C} + \displaystyle\frac{1}{C} + \displaystyle\frac{1}{C} = \displaystyle\frac{3}{C} \Rightarrow C_s = C/3 \).
Ratio \( C_p / C_s = 3C / (C/3) = 9 \).
Correct Answer: C

In a uniform field between plates, \( V = Ed \). Therefore, \( E = V/d \).
Correct Answer: C

Charge \( Q \) is constant. \( C = \kappa C_0 \) increases. Since \( V = Q/C \), as \( C \) increases, \( V \) must decrease.
Correct Answer: C

\( V \) is constant. \( C_{new} = \kappa C_0 \). Energy \( U = \displaystyle\frac{1}{2}CV^2 \). If \( C \) increases by \( \kappa \), energy \( U \) also increases by \( \kappa \).
Correct Answer: A

The dielectric polarizes, creating an internal “induced” field that opposes the external field, effectively reducing the net field by the factor \( \kappa \).
Correct Answer: B

Dielectrics are insulators; they have no free charges. The surface charge comes from polarization—the reorientation of molecular dipoles.
Correct Answer: C

The fringing fields of the capacitor attract the polarized dielectric slab. To remove it, an external agent must do positive work against this attractive electrostatic force.
Correct Answer: B