Solution:

To solve for the energy stored in the system of three plates (X, Y, Z), let's break the system into simpler components:

1. System Description

• The arrangement forms two capacitors:
  • Capacitor 1: Between plates X and Y.
  • Capacitor 2: Between plates Y and Z.
• Each capacitor has the same plate area $A$ and plate separation $d$.

The capacitance of a parallel plate capacitor is given by:

$$C = \frac{\varepsilon_0 A}{d}$$

Thus, the capacitance of each capacitor is:

$$C_1 = C_2 = \frac{\varepsilon_0 A}{d}$$

2. Effective Capacitance

The two capacitors are in parallel because plate Y is connected to the battery on one side and plate X and Z are on the opposite side. For capacitors in parallel, the effective capacitance is:

$$C_{\text{eq}} = C_1 + C_2$$ $$C_{\text{eq}} = \frac{\varepsilon_0 A}{d} + \frac{\varepsilon_0 A}{d} = \frac{2 \varepsilon_0 A}{d}$$

3. Energy Stored in the System

The energy stored in a capacitor is:

$$U = \frac{1}{2} C_{\text{eq}} V^2$$

Substitute $C_{\text{eq}} = \frac{2 \varepsilon_0 A}{d}$:

$$U = \frac{1}{2} \cdot \frac{2 \varepsilon_0 A}{d} \cdot V^2$$ $$U = \frac{\varepsilon_0 A V^2}{d}$$

Final Answer:

The energy stored in the system is:

$$U = \frac{\varepsilon_0 A V^2}{d}$$