Question

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

Answer 1

Step 1: Given

A capacitor whose capaciatnce is to be calculated

Step 2: Formula used

$E=\frac{\sigma }{2{\in }_0}$ Where $\sigma and{\in }_0$ are charge density and electrical permittivity respectively.

and $C=\frac{Q}{V}$ Q= charge and V= potential

Step 3: Derivation of the expression for capacitance.

Place the two plates parallel to each other separated by a distance d and the cross-sectional area of each plate is A.

The electric field by a single thin plate

$E=\frac{\sigma }{2{\in }_0}where\sigma ischargedensity$

The electric field between the two plates

$E=2\frac{\sigma }{2{\in }_0}=\frac{\sigma }{{\in }_0}$

Also, the potential difference between the two plates is $V=Ed$

⟹$V=\frac{\sigma d}{{\in }_0}=\frac{Qd}{A{\in }_0}$

Capacitance

$C=\frac{Q}{V}=\frac{Q}{Qd/A{\in }_0}=\frac{A{\in }_0}{d}$

Thus we get capacitance of parallel plate capacitor $\frac{{\in }_{0}A}{d}$