# Energy of a Charged Conductor or Capacitor

If C is the capacitance of a conductor or capacitor and V is the potential of the conductor (P.D. in case of a capacitor), then the stored electrostatic energy is:

$\displaystyle U = \frac{Q^2}{2 C} = \frac{1}{2}C V^2 = \frac{1}{2}Q V$

$\displaystyle C = \frac{\epsilon_0 A}{d}$ and V = Ed

$\displaystyle U = \frac{Q^2}{2 C} = \frac{1}{2}\frac{\epsilon_0 A}{d}(Ed)^2$

$\displaystyle U = (\frac{1}{2}\epsilon_0 E^2 ) A.d$

Energy density = Energy per unit volume u = U/Volume

$\displaystyle u = \frac{1}{2}\epsilon_0 E^2$

If dielectric is introduced

$\displaystyle u = \frac{1}{2}K\epsilon_0 E^2$

This energy is stored in a capacitor in the electric field between its plates.

##### Force b/w the Plates of a Parallel Plate Capacitor

Consider a parallel plate capacitor with plate area A and charge +Q and -Q. The electric field due to only the positive plates.

$\displaystyle \vec{E} = \frac{\sigma}{2\epsilon_0}\hat{i} = \frac{Q}{2\epsilon_0 A}\hat{i}$

The force on the negatively charged plate in the field of positive charge is

$\displaystyle \vec{F} = (-Q)\vec{E}= -\frac{Q^2}{2\epsilon_0 A}\hat{i}$

Thus the force is attractive and has the magnitude

$\displaystyle F = \frac{Q^2}{2\epsilon_0 A}$