Question

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

Answer 1

Let $d$ be the separation of two conducting parallel plates , each of area $A$ .

If charge $q$ is given to first plate , then charge $-q$ is induced on inner surface of second plate and charge $+q$ induced on the outer surface , is earthed .

Now , the electric field between the two plates will be ,

$E=V/d$

or $V=Ed$ ...........eq1

If $\sigma$ is he surface density of the plates , then the electric field between two plates is given by ,

$E=\sigma /{\epsilon }_0$ ...............eq2

where ${\epsilon }_0=$ absolute permittivity of the free space.

Putting the value of $E$ from eq2 to eq1 , we get

$V=\frac{\sigma }{{\epsilon }_0}d$ ................eq3

but we have ,

$\sigma =q/A$

hence , $V=\frac{qd}{{\epsilon }_{0}A}$

Now , capacitance of parallel capacitor ,

$C=\frac{q}{qd/{\epsilon }_{0}A}$

or $C=\frac{{\epsilon }_{0}A}{d}$