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:

Since the capacitor is isolated, its charge \(Q\) remains constant. When the separation \(d\) increases, the capacitance \(C = \frac{\epsilon_0 A}{d}\) decreases. Since \(U = \frac{Q^2}{2C}\), the stored energy increases due to the work done against the electrostatic attraction.