The force of attraction between the plates of a parallel plate capacitor can be derived from the energy stored in the capacitor.

1. Method 1: Using Energy Derivative The potential energy stored in a capacitor with fixed charge $Q$ is given by $U = \frac{Q^2}{2C}$. We know that capacitance $C = \frac{\epsilon_0 A}{x}$, where $x$ is the separation. Thus, $U = \frac{Q^2 x}{2\epsilon_0 A}$. The force $F$ is the rate of change of potential energy with respect to separation distance (magnitude of the gradient of potential energy): $|F| = \frac{dU}{dx} = \frac{d}{dx} \left( \frac{Q^2 x}{2\epsilon_0 A} \right) = \frac{Q^2}{2\epsilon_0 A}$ Substituting $Q = CV$ and $\epsilon_0 A = Cd$ (since $C = \epsilon_0 A / d$): $F = \frac{(CV)^2}{2(Cd)} = \frac{C^2 V^2}{2Cd} = \frac{CV^2}{2d}$

2. Method 2: Using Electric Field The electric field $E$ produced by one charged plate (infinite sheet approximation) is $E = \frac{\sigma}{2\epsilon_0}$, where $\sigma$ is the surface charge density. The force exerted on the other plate carrying charge $Q$ is $F = Q \times E$. $F = Q \left( \frac{\sigma}{2\epsilon_0} \right) = Q \left( \frac{Q/A}{2\epsilon_0} \right) = \frac{Q^2}{2\epsilon_0 A}$ Using $Q = CV$ and $C = \frac{\epsilon_0 A}{d}$ (so $\epsilon_0 A = Cd$): $F = \frac{(CV)^2}{2Cd} = \frac{CV^2}{2d}$