A parallel-plate capacitor with a plate area of 20 cm ^2 and a separation of 1.00 mm between the plates is connected to a 12.0 V battery. The plates are subsequently pulled apart to increase the separation to 2.0 mm . Calculate the stored energy in the electric field before and after the process, respectively.

Step-by-step solution

The initial capacitance of the parallel-plate capacitor is given by: C₁ = ₀ A d₁ Substituting the given values A = 20 cm ^2 = 20 10⁻⁴ m ^2 and d₁ = 1.00 mm = 10⁻³ m : C₁ = 8.85 10⁻¹² 20 10⁻⁴ 10⁻³ = 17.7 10⁻¹² F The initial energy stored in the capacitor is: U₁ = 1 2 C₁ V^2 U₁ = 1 2 17.7 10⁻¹² (12.0)^2 = 1274.4 10⁻¹² J 12.7 10⁻¹⁰ J When the plates are pulled apart to a new separation d₂ = 2.0 mm , the new capacitance becomes: C₂ = ₀ A d₂ = C₁ 2 Since the battery remains connected, the potential difference across the