Two circular plates with diameter 7.0 cm are separated by 1.0 mm. They are charged to 130 V with the battery removed. What is the energy stored in the capacitor?

• A. 3.50×10^-7 J
• B. 2.88×10^-7 J
• C. 1.50×10^-7 J
• D. 4.35×10^-7 J

Answer: (B)

The energy stored in a capacitor is U = (1/2) C V^2. For a parallel-plate capacitor, C = ε0 A / d, where A is the plate area and d is the separation. The diameter is 7.0 cm, so the radius is 3.5 cm = 0.035 m. The area is A = π r^2 = π (0.035)^2 ≈ 3.85 × 10^-3 m^2. The separation is d = 1.0 mm = 1.0 × 10^-3 m, and ε0 = 8.854 × 10^-12 F/m.

So the capacitance is C ≈ (8.854 × 10^-12) × (3.85 × 10^-3) / (1.0 × 10^-3) ≈ 3.41 × 10^-11 F.

With the plates charged to V = 130 V, the energy is U ≈ (1/2) × (3.41 × 10^-11) × (130)^2 ≈ 0.5 × 3.41 × 10^-11 × 1.69 × 10^4 ≈ 2.88 × 10^-7 J.

Answer: 2.88 × 10^-7 J.