JEE Physics Electrostatic Potential And Capacitance MCQs Set 28

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

Explanation:
1. At an intersection the potential would have two values, so equipotential surfaces never cut, and the assertion is true.
2. Equipotentials need not be parallel (for a point charge they are concentric spheres), so the reason is false.

Teacher's Note:
1. One potential at each point.

Answer: (d) decreasing the distance between the plates

Explanation:
1. C = ε₀A/d, so decreasing d increases C.
2. Plate thickness has no effect.

Teacher's Note:
1. C ∝ 1/d.

Answer: (b) \( V_A-V_B=0 \)

Explanation:
1. All points of an equipotential surface have the same potential.
2. So V_A − V_B = 0.

Teacher's Note:
1. Same surface, same V.

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

Explanation:
1. V = Q/C and C depends on the size.
2. With the same Q but different capacitances the potentials differ.

Teacher's Note:
1. V = Q/C.

Answer: (a) 30 μC

Explanation:
1. 1/C = 1/6 + 1/6 + 1/3 = 2/3, so C = 1.5 μF.
2. In series each has the same charge, Q = CV = 1.5 × 20 = 30 μC.

Teacher's Note:
1. Series: same Q.

Answer: (a) is a conservative field

Explanation:
1. Potential is defined through the work done, which is independent of the path only in a conservative field.

Teacher's Note:
1. Conservative field gives a potential.

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

Explanation:
1. The potentials of +q and −q cancel at every equatorial point.
2. The potential is zero over the whole plane, so it is an equipotential surface.

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

Answer: (c) I and II

Explanation:
1. Field lines are normal to equipotential surfaces and these never intersect.
2. The field lines are not tangent to them, so III is false.

Teacher's Note:
1. E is normal to equipotentials.

Answer: (c) the same along both paths

Explanation:
1. The field is conservative, so the work depends only on the potential difference between S₁ and S₂.
2. It is the same for every path.

Teacher's Note:
1. W = qΔV.

Answer: (c) capacitor

Explanation:
1. Two conductors separated by an insulator form a capacitor.

Teacher's Note:
1. Definition.

Answer: (c) the charge on the capacitor is not conserved

Explanation:
1. The capacitor is disconnected, so its charge stays constant and (c) is incorrect.
2. V falls to V/K and the energy to U/K, so the other three statements are correct.

Teacher's Note:
1. Isolated: Q is conserved.

Answer: (d) dielectric strength

Explanation:
1. This is the definition of dielectric strength.

Teacher's Note:
1. Breakdown field.

Answer: (a) decreases K times

Explanation:
1. F = F₀/K in a medium of dielectric constant K.

Teacher's Note:
1. F = F₀/K.

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

Explanation:
1. The net field on the plane is perpendicular to it, so no work is done in moving the charge along the plane from infinity.
2. Hence the potential, equal to this work, is zero.

Teacher's Note:
1. V equals work per unit charge.

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

Explanation:
1. Energy is stored where the field exists.
2. The field inside the conductor is zero, so all the energy is in the field outside.

Teacher's Note:
1. u = ½ε₀E².

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

Explanation:
1. Equipotential surfaces never intersect, so they cannot be orthogonal and the assertion is false.
2. The field lines are normal to them, so the reason is true.

Teacher's Note:
1. Field is normal to equipotentials.

Answer: (b) the electric field is zero

Explanation:
1. E = −dV/dr.
2. If V is constant, its gradient is zero, so the electric field is zero.

Teacher's Note:
1. E is the negative gradient of V.

Answer: (b) \( \frac{\varepsilon_0A}{t}\cdot\frac{k_1+k_2}{2} \)

Explanation:
1. The two halves are in parallel.
2. C = ε₀k₁(A/2)/t + ε₀k₂(A/2)/t = (ε₀A/t)(k₁ + k₂)/2.

Teacher's Note:
1. Parallel capacitors add.

Answer: (b) a negative charge should be placed at the centroid

Explanation:
1. A negative charge at the centroid attracts all three, adding negative interaction energy.
2. A positive charge or a smaller separation raises the energy, and rotation does nothing.

Teacher's Note:
1. Add attraction.

Answer: (b) less than that of the first conductor

Explanation:
1. The charge is shared until the potentials are equal.
2. Some energy is lost as heat, so the total energy is less than the initial energy.

Teacher's Note:
1. Energy is lost on sharing.

Answer: (d) decrease

Explanation:
1. For a fixed charge, E = E₀/K, and kerosene has K > 1.
2. The field decreases.

Teacher's Note:
1. E = E₀/K.

Answer: (c) 2 times

Explanation:
1. Volume conservation gives R = 2r for the big drop.
2. C = 4πε₀R is proportional to the radius, so it is 2 times.

Teacher's Note:
1. C ∝ R.

Answer: (d) I, II and III

Explanation:
1. All three statements are standard properties of an equipotential surface.

Teacher's Note:
1. Properties of equipotentials.

Answer: (b) II only

Explanation:
1. The two gaps on either side of the plate form two capacitors in series.
2. A conductor behaves like K → ∞, not zero, and the combination is not parallel.

Teacher's Note:
1. Series of two gaps.

Answer: (b) \( \frac{Q}{4\pi\varepsilon_0R}\text{ and zero} \)

Explanation:
1. The potential inside a conductor equals its surface value, Q/4πε₀R.
2. The field inside is zero.

Teacher's Note:
1. V constant, E = 0 inside.