The electrostatic force between the metal plates of an isolated parallel plate capacitor $C$ having a charge $

Question

The electrostatic force between the metal plates of an isolated parallel plate capacitor $C$ having a charge $Q$ and area $A$ is:

Accepted Answer

To find the force between the plates, we consider the electric field produced by one plate acting on the charge of the other plate.

1.  **Electric Field of a Single Plate:** The electric field $E$ due to a single large plane sheet of charge with surface charge density $\sigma = Q/A$ is given by $E = \frac{\sigma}{2\epsilon_0}$ [NCERT Class 12, Chapter 1, Application of Gauss's Law].
2.  **Force Calculation:** The force $F$ exerted on the second plate carrying charge $Q$ (placed in the field of the first plate) is $F = Q 	imes E$.
    $$F = Q 	imes \frac{\sigma}{2\epsilon_0} = Q 	imes \frac{Q}{2A\epsilon_0}$$
    $$F = \frac{Q^2}{2\epsilon_0 A}$$
3.  **Conclusion:** The expression for force depends only on the charge $Q$, the area $A$, and the permittivity $\epsilon_0$. It does not contain the separation distance $d$. Therefore, the force is independent of the distance between the plates (provided the distance is small compared to the plate dimensions, maintaining the uniform field approximation).

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