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 . How much energy is absorbed by the battery during this process?

Step-by-step solution

The initial capacitance of the parallel-plate capacitor is given by: C₁ = ₀ A d₁ The final capacitance after the plates are pulled apart is: C₂ = ₀ A d₂ Since the battery remains connected, the potential difference V across the capacitor is constant. The initial and final charges on the capacitor are Q₁ = C₁ V and Q₂ = C₂ V respectively. As the separation increases, the capacitance decreases, causing charge to flow back into the battery. The charge that flows back is: Q = Q₁ - Q₂ = (C₁ - C₂) V The energy absorbed b