Question

Derive an expression for the capacitance of a parallel plate capacitor in terms of the area of plates and the distance between the plates. Assume there is vacuum between the plates.

Answer 1

Assume there is a vacuum in between the plates with the electric field between the plates E = σ/ ε0. (where the symbols have their usual meaning.) Let the plate A be given a charge +Q. Due to electrostatic induction a charge -Q will appear on the nearer side of B and +Q on its farther side.

Answer 2

Answer:

As shown in the figure the plate area is A and the separation between them is d, the assumption is that d<<2A

​

2A

​ is supposed to be dimension of the plates.

Value of charge density on either plate is σ=AQ​

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+​=2ϵ0​σ​

Similarly the electric field due to negative plate at point P will  be towards  the negative plate and given by E−​=2ϵ0​σ​

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

so E=2×2ϵ0​σ​=ϵ0​σ​

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 ΔV=E×d=Aϵ0​Qd​

So the capacitance C=ΔVQ​=dAϵ0​​

If dielectric is there in between the plates then just replace the ϵ0​ by ϵr​ where  ϵr​=kϵ0​

Explanation: