Question

Derive the expression for the capacitance of a parallel plate capacitor having plate area A and plate separation d.

Answer 1

A parallel plate capacitor consists of two large plane parallel conducting plates separated by a small distance.

Let A be the area of each plate and d be the separation between them. The two plates have charges Q and −Q. Plate 1 has surface charge density, σ = Q/A,
And plate 2 has a surface charge density −σ.

Electric field in different regions:

Outer region I:

`E= sigma/(2epsilon_0)-sigma/(2epsilon_0)=0`

Outer region II:

`E=sigma/(2epsilon_0)-sigma/(2epsilon_0)=0`

In the inner region between plates 1 and 2, the electric fields due to the two charged plates add up.

This gives:

`E=sigma/(2epsilon_0)+sigma/(2epsilon_0)=sigma/epsilon_0=Q/epsilon_0=Q/(epsilon_0A)`

The direction of electric field is from positive to the negative plate. For uniform electric field, potential difference is simply the electric field times the distance between the plates.

 `V=E""d=1/epsilon_0(Qd)/A`

 Capacitance (C) of the parallel plate capacitor, `C=Q/V=(epsilon_0A)/d`