ELECTRIC POTENTIAL AND CAPACITANCE

Ans.

$A$ is a area of each plate, $d$ is separation between the two parallel plates. $q$ and $–q$ are charges on each plate surface charged density on $\sigma=\frac{q}{A}$.In the inner region between the plates, the electric flelds due to the two charged plates$$E=\frac{\sigma}{2\varepsilon_0}+\frac{\sigma}{2\varepsilon_0}=\frac{\sigma}{\varepsilon_0}=\frac{q}{\varepsilon_0A}$$parallel plates in capacitor behave like equipotential surfaces.
 $\therefore $Potential $V=Ed$ $$V=\frac{q}{\varepsilon_0A}d$$$$\frac{V}{q}=\frac{q}{\varepsilon_0A}$$$$ \frac{1}{C}=\frac{d}{\varepsilon_0A}$$$$ C=\frac{\varepsilon_0A}{d}$$The above equation represents capacitance of a parallel plate capacitor.