Capacitors: Practice Problem & Solution

Solution Explained:

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

Initial energy $U_{i} = \frac{Q^{2}}{2C}$. When connected in parallel to an identical capacitor, common potential $V = \frac{Q}{2C}$. Final energy $U_{f} = \frac{1}{2}(2C)V^{2} = \frac{1}{2}(2C)(\frac{Q}{2C})^{2} = \frac{Q^{2}}{4C} = \frac{U_{i}}{2}$. Therefore, the energy decreases by a factor of 2.