Question

Derive an expression for the capacitance of a parallel plate capacitor. Why does the capacitance increase by using dielectric in capacitor?

Answer 1

Consider parallel plate capacitor is place in vacuum and permittivity of vacuum is given by ${ϵ}_o$. Then,

Electric fields between the plates of capacitor is given by,

$E=\frac{\sigma }{{ϵ}_o}=\frac{Q}{A{ϵ}_o}$

potential difference between the plates is given by,

$V=Ed=\frac{Qd}{A{ϵ}_o}$...(1)

But,

Capacitance = $C=\frac{Q}{V}=\frac{{ϵ}_{o}A}{d}$ ... (from eq(1))

If we place dielectric medium in between the plates of capacitor then permittivity of medium increases and

it is denoted by ${ϵ}_m$.Then new capacitance will be,

Capacitance = $C=\frac{Q}{V}=\frac{{ϵ}_{m}A}{d}$ ...(2)...(${ϵ}_m>{ϵ}_o$)

$\therefore$ From equation (2) it is clear that capacitance increases when the dielectric is inserted between the plates of capacitor.