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

Explanation

Consider a parallel plate capacitor consisting of two plates each of surfaces area A , held parallel to each other in air medium such that the separation between the plates is d . Charges on the two plates are `+Q` and `-Q` , respectively .
Then surface density of charge on either plate `|sigma| = (Q)/(A)`
Electric field between the plates of capacitor is uniform every where , having a value `E = E_(1) + E_(2) = (sigma)/(2 in_(0)) + (sigma )/(2 in_(0)) = (sigma)/(in_(0))`
The electric field E is directed from +ve plate to the -ve plate . If V be the potential difference between the plates , then
`V = E d = (sigma)/(in_(0)) d = (Qd)/(in_(0) A)`
`therefore` Capacitance of the parallel plate capacitor
`C = (Q)/(V) = (in_(0) A)/(d)`
If a dielectric medium of dielectric constant .K. is completely filled between the two plates of capacitor , then electric field `E = (sigma)/(K in_(0)) = (Q)/(K A in_(0))`
In that case `V = Ed = (Qd)/(K A in_(0)) implies` Capacitance `C = (Q)/(V) = (K in_(0) A)/(d)`