Electrostatic Potential and Capacitance — Practice Quiz

C = ε₀A/d

Step 1: For a parallel plate capacitor, the electric field between the plates is E = σ/ε₀ = Q/(ε₀A). Step 2: The potential difference V = Ed = Qd/(ε₀A). Step 3: Capacitance C = Q/V = Q × ε₀A/(Qd) = ε₀A/d. Step 4: This shows C increases with larger area A (more charge stored) and decreases with larger separation d. Option ε₀d/A is the inverse and a common mistake. Option ε₀Ad has wrong dimensions.

2 μF

Step 1: For capacitors in series, 1/C = 1/C₁ + 1/C₂. Step 2: 1/C = 1/6 + 1/3 = 1/6 + 2/6 = 3/6 = 1/2. Step 3: Therefore C = 2 μF. Step 4: Note that in series, equivalent capacitance is always LESS than the smallest individual capacitance. Option 9 μF is incorrect (that's parallel addition). Option 18 μF results from multiplying instead of adding reciprocals.

10 μF

Step 1: For capacitors in parallel, C = C₁ + C₂. Step 2: C = 4 + 6 = 10 μF. Step 3: In parallel, the same voltage is applied across both capacitors but they store different charges. Step 4: The total charge Q = Q₁ + Q₂, which gives C = C₁ + C₂. Option 2.4 μF is the series combination result. Option 24 μF results from multiplying rather than adding.

2.5 × 10⁻⁶ J

Step 1: Use the formula U = ½CV². Step 2: C = 500 pF = 500 × 10⁻¹² F = 5 × 10⁻¹⁰ F, V = 100 V. Step 3: U = ½ × 5 × 10⁻¹⁰ × (100)² = ½ × 5 × 10⁻¹⁰ × 10⁴. Step 4: U = ½ × 5 × 10⁻⁶ = 2.5 × 10⁻⁶ J. Option 5 × 10⁻⁶ J results from forgetting the factor of ½. Always remember U = ½CV².