Question

Obtain the expression for capacitance for a parallel plate capacitor.

Derive an expression for the capacitance of a parallel plate capacitor with air present between the two plates.

Answer 1

Consider a capacitor with two parallel plates each of cross-sectional area A and separated by a distance d. The electric field between two infinite parallel plates is uniform and is given by E = ${\displaystyle \frac{\sigma }{{\epsilon }_0}}$ where σ is the surface charge density on the plates σ = ${\displaystyle \frac{\text{Q}}{\text{A}}}$. If the separation distance d is very much smaller than the size of the plate (d² << A), then the above result is used even for finite-sized
parallel plate capacitor.

Capacitance of a parallel plate capacitor

Capacitance of a parallel plate capacitor

The electric field between the plates is

E = ${\displaystyle \frac{\text{Q}}{\text{A}{\epsilon }_0}}$  ....(1)

Since the electric field is unifonn, the electric potential between the plates having separation d is given by

V = Ed = ${\displaystyle \frac{\text{Qd}}{\text{A}{\epsilon }_0}}$    ....(2)

Therefore the capacitance of the capacitor is given by

C = ${\displaystyle \frac{\text{Q}}{\text{V}}=\frac{\text{Q}}{\left(\frac{\text{Qd}}{\text{A}{\epsilon }_0}\right)}=\frac{{\epsilon }_0\text{A}}{\text{d}}}$   ....(3)

From equation (3), it is evident that capacitance is directly proportional to the area of cross-section and is inversely proportional to the distance between the plates. This can be understood from the following.

1. If the area of cross-section of the capacitor plates is increased, more charges can be distributed for the same potential difference. As a result, the capacitance is increased.
2. If the distance d between the two plates is reduced, the potential difference between the plates (V = Ed) decreases with E constant.

Answer 2

Derivation of the expression for the capacitance.

Let the two plates be kept parallel to each other separated by a distance d and the cross-sectional area of each plate is A. Electric field by a single thin plate E = ${\displaystyle \frac{\sigma }{2{\epsilon }_0}}$

The total electric field between the plates E = ${\displaystyle \frac{\sigma }{{\epsilon }_0}}$ = ${\displaystyle \frac{Q}{A{\epsilon }_0}}$

The potential difference between the plates V = Ed = ${\displaystyle \left[\frac{Q}{A{\epsilon }_0}\right]}$d.

Capacitance C = ${\displaystyle \frac{\text{Q}}{\text{V}}=\frac{A{\epsilon }_0}{d}}$