Show that the force on each plate of a parallel plate capacitor has a magnitude equal to `(½) QE`, where Q is the charge on the capacitor, and E is the magnitude of electric field between the plates. Explain the origin of the factor `½`.

Explanation

To show that the force on each plate of a parallel plate capacitor has a magnitude equal to \( \frac{1}{2} QE \), where \( Q \) is the charge on the capacitor and \( E \) is the magnitude of the electric field between the plates, we can follow these steps:

### Step-by-Step Solution

1. **Understanding the Force on the Plates**:
The force \( F \) acting on one of the plates of a parallel plate capacitor can be derived from the energy considerations of the capacitor. When the plates are separated by a small distance \( \Delta x \), work is done against the electric field.

2. **Work Done in Separating the Plates**:
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