Electric Potential and Capacitors

0.02 J

Step 1: Formula for energy stored: U = ½CV². Step 2: Substitute: C = 4 × 10⁻⁶ F, V = 100 V. Step 3: U = ½ × 4 × 10⁻⁶ × (100)² = ½ × 4 × 10⁻⁶ × 10⁴ = ½ × 0.04 = 0.02 J. Option B (0.04 J) is a common mistake where students forget the factor of ½. Option C halves the correct answer again incorrectly. Option D results from a power-of-ten error.

400 V/m

Step 1: The relation between electric field and potential difference is: E = ΔV/d. Step 2: Convert distance to metres: d = 5 cm = 0.05 m. Step 3: E = 20 V / 0.05 m = 400 V/m. Option A results from using d = 0.2 m instead of 0.05 m. Options C and D arise from unit conversion errors where the student forgets to convert cm to m correctly.

Charge remains same, potential difference decreases, capacitance increases

Step 1: When the capacitor is disconnected from the battery, the charge Q on the plates is fixed (no path for charge to flow). Step 2: Inserting the dielectric increases capacitance: C = KC₀. Step 3: Since Q = CV and Q is constant, V = Q/C = Q/(KC₀) = V₀/K, which means potential difference decreases by factor K. The dielectric reduces the effective electric field between the plates, lowering V while keeping Q constant. Options B, C, D misunderstand the constraint that charge must remain constant when disconnected from the battery.

108 μC

Step 1: For parallel combination, equivalent capacitance: Cp = C₁ + C₂ = 6 + 3 = 9 μF. Step 2: Total charge: Q = Cp × V = 9 μF × 12 V = 108 μC. Alternatively: Q₁ = 6 × 12 = 72 μC, Q₂ = 3 × 12 = 36 μC, total = 108 μC. Option D (72 μC) is the charge on C₁ alone. Option B (36 μC) is the charge on C₂ alone. Option A results from using the series formula instead of the parallel formula.