Capacitor With Dielectrics: Practice Problem & Solution

Solution Explained:

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

Once disconnected, charge \(Q\) remains constant. Introducing a slab increases capacitance \(C\). Since potential difference \(V = Q/C\) and stored energy \(U = \frac{Q^2}{2C}\), both potential difference and stored energy decrease.