Obtain an expression for capacitance for a parallel plate capacitor.

Explanation

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 = sigma/epsilon_0` where `sigma` is the surface charge density on the plates `(sigma = Q/A)`. If the separation distance d is very much smaller than the size of the plate `(d^2 ltlt A)`, then the above result is used even for finite-sized parallel plate capacitor .
The electric field between the plates is `E = Q/(Aepsilon_0)`
Since the electric field is uniform, the electric potential between the plates having separation d is given by
`V = Ed = (Qd)/(Aepsilon_0)`
Therefore the capacitance of the capacitor is given by
`C = Q/V = Q/(((Qd)/(Aepsilon_0)))`
` = (epsilon_0A)/d`
From the above equation, it is evident that capacitors is directly proportional to the area of cross section and is inversely proportional to the distance between the plates.