Solution:

To solve this, let us analyze the situation and derive the required expression for the charge:

1. Setup of the System

• The parallel plate capacitor is initially charged, and the battery is disconnected.
• One plate of the capacitor is connected to a spring, and the other is attached to a mass $m$.
• The capacitor plates attract each other due to the opposite charges, generating an electrostatic force.

2. Equilibrium Condition

In equilibrium, the upward force due to the spring's tension balances the downward gravitational force acting on the mass $m$:

$$T = mg$$

This tension $T$ is equal to the electrostatic force $F_{\text{elec}}$ between the plates of the capacitor:

$$F_{\text{elec}} = \frac{Q^2}{2 \epsilon_0 A}$$

3. Force Balance

At equilibrium:

$$F_{\text{elec}} = mg$$

Substitute $F_{\text{elec}} = \frac{Q^2}{2 \epsilon_0 A}$:

$$\frac{Q^2}{2 \epsilon_0 A} = mg$$

4. Solve for $Q$

Rearranging for $Q$:

$$Q^2 = 2 mg \epsilon_0 A$$

Taking the square root:

$$Q = \sqrt{2 mg \epsilon_0 A}$$

Final Answer:

The magnitude of the charge on one of the capacitor plates is:

$$Q = \sqrt{2mgA\epsilon_0}$$