Obtain the expression for capacitance for capacitance for a parallel plate capacitor .

Explanation

Capacitance of a parallel plate capacitor : 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) ltltA )` 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_(0)A)/(d)`

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 .
(i) 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.
If the distance d between the two plates is reduced the potential difference between the plates (V= Ed) decreases with E constant .