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}$

Also Read :

Capacitance of a Parallel Plate Capacitor
Cylindrical Capacitor & Spherical Capacitor
Series & Parallel Combination of Capacitors
Redistribution of Charge, Common Potential & Loss of Energy
Parallel Plate Capacitor with different Charges
Effect of Dielectrics on Charged Capacitor

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