Parallel Plate Capacitor: Practice Problem & Solution

Solution Explained:

To solve this problem, we apply the core principles of Parallel Plate Capacitor. Understanding the underlying formula is key to arriving at the correct answer below:

For an isolated capacitor, the charge \(Q\) remains constant. \(Q = C_1 V_1 = C_2 V_2\). Since \(C = \frac{\epsilon_0 A}{d}\), we have \( \frac{\epsilon_0 A}{d_1} V_1 = \frac{\epsilon_0 A}{d_2} V_2 \implies \frac{V_1}{d_1} = \frac{V_2}{d_2} \). Thus, \(V_2 = V_1 \frac{d_2}{d_1} = 60 \text{ V} \times \frac{12 \text{ mm}}{4 \text{ mm}} = 180 \text{ V}\).