Question

Derive an expression for the capacitance of a parallel plate capacitor.

Answer 1

In a parallel plate, there are two parallel plate with charges $+Q$ and $-Q$ and the area of each plate is $A$.

Now, the electric field in between the plates is $E=\frac{\sigma }{{ϵ}_0}=\frac{Q}{A{ϵ}_0}$ as surface charge density , $\sigma =Q/A$

If $V$ be the potential difference between the plates, the electric field will be $E=V/d$ where $d=$ separation between the plates.

Thus, $\frac{Q}{A{ϵ}_0}=\frac{V}{d}$

or $\frac{Q}{V}=\frac{A{ϵ}_0}{d}$

The capacitance, $C=Q/V=\frac{A{ϵ}_0}{d}$