Question

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

Answer 1

As shown in the figure the plate area is $A$ and the separation between them is $d$, the assumption is that $d<<\sqrt[2]{2A}$

$\sqrt[2]{2A}$ is supposed to be dimension of the plates.

$Value$ of charge density on either plate is $\sigma =\frac{Q}{A}$

The electric field due to positive plate at point P will be $away$ from this plate i.e towards the negative plate and given

by $E_{+}=\frac{\sigma }{2{ϵ}_0}$

Similarly the electric field due to negative plate at point P will be $towards$ the negative plate and given by $E_{-}=\frac{\sigma }{2{ϵ}_0}$

As both fields are in same $direction$ so net field will be their $sum$

so $E=2\times \frac{\sigma }{2{ϵ}_0}=\frac{\sigma }{{ϵ}_0}$

$\text{as d is very small in comparison to the dimension of plate so the field can be treated as constant through out.}$

So potential difference $\Delta V=E\times d=\frac{Qd}{A{ϵ}_0}$

So the capacitance $C=\frac{Q}{\Delta V}=\frac{A{ϵ}_0}{d}$

If dielectric is there in between the plates then just replace the ${ϵ}_0$ by ${ϵ}_r$ where ${ϵ}_r=k{ϵ}_0$