Derive an expression for energy density of a parallel plate capacitor.

Explanation

The work done in charging a capacitor is stored in the form of electrical energy.
Let at any instant, a charge q be on the plate of a capacitor. Then potential difference between the plates of the capacitor is given by V = q/C.
If extra charge dq is transferred to the capacitor, then work done to do so is stored as electric potential energy in the capacitor
i.e., `dU =dW =Vdq=q/C dq`
The total increase in potential energy in charging the capacitor from q = 0, to q = Q is the total energy stored in the capacitor
`U= int dU = int_(0)^(Q)q/C dq`
`U =1/C int_(0)^(Q) qdq`
`U=1/(2C) |Q^(2)-0| =1/2Q^(2)/c`
So snergy stored in the capacitor is given by
`U=1/2 Q^(2)/C`
We know, Q =CV
So, `U =1/2 (CV)^(2)/2 =1/2 (C^(2)V^(2))/C`
`therefore U =1/2 CV^(2)`
Also, `C = Q/V`
Put this value of C in eqn. (1)
`U=1/2 QV`
which is required ,expression for the energy stored in a capacitor. Energy is stored in the Capacitor in electrical form.