Electrostatic Potential and Capacitance MCQs

Potential V = k q / r.

Substitute k = 9×10⁹, q = 2×10⁻⁹ C and r = 0.3 m: V = (9×10⁹ × 2×10⁻⁹) / 0.3.

The numerator equals 18 V·m, dividing by 0.3 m gives 60 V.

Hence the correct answer is 60 V.

Capacitance of a parallel‑plate capacitor is C = ε₀A/d.

With a dielectric fully filling the space, ε₀ is replaced by ε = Kε₀, so C = Kε₀A/d.

Thus the new capacitance is K times the original value.

Hence the capacitance increases K times.

When the capacitor is isolated, its charge Q remains constant.

With a dielectric of constant K, the capacitance becomes C′ = K C.

Since V = Q/C, the new voltage V′ = Q/(K C) = V/K, which is smaller than the original.

Thus the potential difference across the plates decreases.

Potential energy of a dipole in a uniform field: U = –p·E = –pE cosθ, where θ is the angle between dipole moment p and field direction.

Energy is lowest when cosθ is maximum, i.e., cosθ = 1.

Thus θ = 0°, meaning the dipole aligns parallel to the field.

Hence the minimum potential energy occurs when it is aligned parallel to the field.

Capacitance C = Q/V and for a parallel‑plate capacitor V = Ed = σd/ε₀ where σ = Q/A.

Substituting V gives C = Q/(σd/ε₀) = ε₀ Qd/(Q A) = ε₀ A/d.

Thus the expression depends directly on plate area and inversely on separation.

The correct answer is ε₀ A / d.

When the capacitor is disconnected, the circuit is open so no path exists for charge to flow.

Inserting a dielectric changes the capacitance (C → κC) but the total charge Q on the isolated plates cannot change.

Since Q = CV and V adjusts to the new capacitance, the charge remains unchanged.

Thus the charge on the plates remains the same.

The torque on a dipole is τ = p E sinθ, so the work done in rotating it from θ = 0 (aligned) to a general angle θ is the negative of the integral of τ dθ.

Integrating τ gives the change in potential energy U = –∫₀^θ pE sinθ dθ = –pE cosθ + constant.

Choosing the reference U = 0 when the dipole is perpendicular (θ = 90°) sets the constant to zero, yielding U = –pE cosθ.

Thus the correct expression is –pE cosθ.

Electric field magnitude equals the rate of change of potential with distance: E = –dV/dx.

Thus the numerical value of the field (in V m⁻¹) is the same as the potential gradient (V per metre).

A field of 200 V m⁻¹ corresponds to a potential change of 200 V over each metre, i.e. a gradient of 200 V/m.

Hence the correct answer is 200 V/m.

Work done to bring a unit charge from infinity equals the potential difference, so V = W/q.

Since work W is measured in joule and charge q in coulomb, the unit of V is joule / coulomb.

One joule per coulomb is defined as a volt.

Hence the SI unit of electric potential is volt.

For series connection, reciprocal of total C equals sum of reciprocals: 1/C_eq = 1/3 µF + 1/6 µF = (2+1)/6 = 3/6 = 1/2.

Invert to get C_eq = 2 µF.

Thus the equivalent capacitance is 2 µF.

When the capacitor remains connected to a battery, the voltage across its plates is fixed at V.

Inserting a dielectric raises the capacitance to C′ = κC (κ>1).

Since Q = CV and V is unchanged, the new charge becomes Q′ = C′V = κCV, which is larger than the original charge.

Thus the charge stored increases.

Electric field direction is tangent to field lines, while the potential does not change along an equipotential surface, so the gradient of potential (which gives the field) is normal to that surface.

Since the field vector is perpendicular to the equipotential surface at every point, the angle between a field line and the corresponding equipotential is 90 degrees.

Thus the correct answer is 90 degrees.

U = k q₁q₂ / r = (9×10⁹)(4×10⁻⁶)(2×10⁻⁶) / 0.6

Multiply numerators: 9×4×2 = 72, and 10⁹×10⁻⁶×10⁻⁶ = 10⁻³, so numerator = 72×10⁻³ = 0.072.

Divide by 0.6 gives 0.072 / 0.6 = 0.12 J.

Thus the potential energy is 0.12 J.

For capacitors in parallel, the equivalent capacitance is the sum of individual capacitances.

Add the given values: 2 µF + 3 µF + 5 µF = 10 µF.

Thus the combined capacitance equals 10 µF, which matches the stated answer.

Capacitance C = Q/V for an isolated conductor.

For a sphere of radius R, the potential V = Q/(4π ε₀ R).

Thus C = Q / [Q/(4π ε₀ R)] = 4π ε₀ R.

Hence the capacitance is 4 π ε₀ R.

For series capacitors, charge Q on each is same: Q = C_eq V_total.

Equivalent capacitance 1/C_eq = 1/2 µF + 1/4 µF → C_eq = (2·4)/(2+4) = 4/3 µF.

Thus Q = C_eq·V = (4/3 µF)(60 V) = 80 µC.

Voltage on 2 µF = Q/C = 80 µC / 2 µF = 40 V.

Work done = q ΔV where ΔV is the potential difference between the initial and final points.

On an equipotential surface the potential is the same everywhere, so ΔV = 0.

Hence W = q·0 = 0, which matches the answer zero.

Energy stored in a capacitor: U = ½ C V².

Convert capacitance: 4 µF = 4 × 10⁻⁶ F.

U = ½ (4 × 10⁻⁶)(200)² = 0.5 × 4 × 10⁻⁶ × 4 × 10⁴ = 0.08 J.

Thus the correct answer is 0.08 J.

Work done = charge × potential difference.

W = Q V = 3 C × 12 V.

Multiplying gives 36 joules.

Hence the work done is 36 J.

In a uniform field **E** = E x̂ the potential difference between two points is V = –∫E·dl = –E Δx, so V changes only with the x‑coordinate.

Hence all points having the same x‑value have the same potential, forming a plane of constant x.

These planes are orthogonal to the direction of the field (the x‑axis).

Therefore the equipotential surfaces are perpendicular to the x‑axis.