Question

Derive the expression for the capacitance of a parallel plate capacitore having plate area A and plate separation d.

Answer 1

Let a parallel plate capacitor be applied a voltage V across its terminals due to which a charge Q develops across its ends.

Separation between plates is d and area of plates is A.

From gauss's theorem one can show electric field inside the capaciton plates Q,

$E=\frac{\sigma }{{ϵ}_o}$, where $\sigma =\frac{Q}{A}$ (charge / Area)

$\Rightarrow E=\frac{Q}{A{ϵ}_o}$

Now, voltage across the plates is related by :

$V=Ed$ [In general , $dV=-\overrightarrow{E}.d\overrightarrow{r}$ but we take as $\overrightarrow{E}$ is uniform]

$\Rightarrow V=\frac{Qd}{A{ϵ}_o}\Rightarrow \frac{Q}{V}=\frac{A{ϵ}_o}{d}$

Now, capacitance is defined by $C=\frac{Q}{V}$, thus we get :-

$C=\frac{A{ϵ}_o}{d}$