Consider a capacitor of capacitance C, completely uncharged in the begning.
Charhing process of capacitor requires expanditure of energy because while charging a capacitor charge is transferred from plate at lower potential to plate at higher potential.
Now if we start charging capacitor by transporting a charge dQ from negative plate ti the positive plate then work is done against the potential difference across the plate.
If q is the amount of charge on the capacitor at any stage of charging process and φ is the potential difference across the plates of capacitor then magnitude of potential difference is φ=q/C.
Now work dW required to transfer dq is
dW=φdq=qdq/C
To charge the capacitor starting from the uncharged state to some final charge Q work required is
Integrating from 0 to Q
W=(1/C)∫qdq
=(Q2)/2C (14a)
=(CV2)/2
=QV/2
Which is the energy stored in the capacitor and can also be written as
U=(CV2)/2 ---(15)
From equation 14c,we see that the total work done is equal to the average potential V/2 during the charging process ,multiplied by the total charge transferred
If C is measured in Farads ,Q in coulumbs and V in volts the energy stored would in Joules
A parallel plate capacitor of area A and seperation d has capacitance
C=ε0A/d
electric field in the space between the plates is
E=V/d or V=Ed
Putting above values of V and C in equation 14b we find
W=U=(1/2)(ε0A/d)(Ed)2
=(1/2)ε0E2(Ad)
=(1/2)ε0E2.V ---(16)
If u denotes the energy per unit volume or energy density then
u=(1/2)ε0E2 x volume
The result for above equation is generally valid even for electrostatic field that is not constant in space.