 # Energy Stored By Capacitor

We know already that a capacitor is used to store energy. In this module we shall discuss how much energy can be stored in a capacitor, the parameters that the energy stored depends upon and their relations.

## How to Calculate the Energy Stored in Capacitor?

Work has to be done to transfer charges onto a conductor, against the force of repulsion from the already existing charges on it. This work done to charges from one plate to the other is stored as potential energy of the electric field of the conductor. Suppose charge is being transferred from plate B to A. At a moment, the charge on the plates is Q’ and –Q’. Then, to transfer a charge of dQ’ from B to A. The work done by an external force will be

$dW=VdQ’=\frac{Q’}{C}dQ'$

Total work done =$\,\int\limits_{0}^{O}{\frac{1}{C}Q’dQ’=\frac{{{Q}^{2}}}{2C}}$

∴ Energy Stored in a Capacitor = $\frac{{{Q}^{2}}}{2C}=\frac{1}{2}C{{V}^{2}}=\frac{1}{2}QV$

### Watch this Video For more Reference ### Energy Density in an Electric Field

Energy stored per unit volume is called energy density. It is given by

U= $\frac{1}{2}{{\varepsilon }_{0}}{{E}^{2}}$

i.e. U = $\frac{1}{2}k\varepsilon {{E}^{2}}$ with a dielectric of dielectric  constant k introduced and E is the net electricfield in dielectric medium.

## Problems on Energy Stored in a Capacitor

Problem 1: A battery of 20 V is connected to 3 capacitors in series as shown in the figure. Two capacitors are of 20μF each and one is of 10μF. Calculate the energy stored in the capacitors in the steady state. Sol:

$\frac{1}{{{C}_{eff}}}=\frac{1}{20}+\frac{1}{20}+\frac{1}{10}=\frac{4}{20}=\frac{1}{5}$

Ceff = 5μF

The energy stored = $\frac{1}{2}C{{V}^{2}}=\frac{1}{2}\times 5\times {{10}^{-6}}\times {{20}^{2}}={{10}^{-3}}J$

Problem 2: A parallel plate capacitor has plates of area 4 m2 separated by a distance of 0.5 mm. The capacitor is connected across a cell of emf 100 volts. Find the capacitance, charge & energy stored in the capacitor if a dielectric slab of dielectric constant k = 3 and thickness 0.5 mm is inserted inside this capacitor after it has been disconnected from the cell.

Sol: When the capacitor is without dielectric

${{C}_{0}}=\frac{{{\varepsilon }_{0}}A}{d}=\frac{8.85\times {{10}^{-12}}\times 4}{0.5\times {{10}^{-3}}}$

${{C}_{0}}=7.08\times {{10}^{-2}}\mu F.$

${{Q}_{0}}={{C}_{0}}{{V}_{0}}$

= ${{(7.08\times {{10}^{-2}}\times 100)}_{\mu C}}=7.08\mu C\,$

${{U}_{0}}=\frac{1}{2}{{C}_{0}}V_{0}^{2}=354\times {{10}^{-6}}J$ as the cell has been disconnected, charge on the capacitor remain constant

C = $\frac{k{{\varepsilon }_{0}}A}{d}=k{{C}_{0}}=0.2124\mu F$

V = $\frac{Q}{C}=\frac{{{Q}_{0}}}{k{{C}_{0}}}\frac{{{V}_{0}}}{k}=\frac{100}{3}volts$

U = $\frac{1}{2}\frac{{{Q}_{0}}}{C}=\frac{1}{2}\frac{Q_{0}^{2}}{k{{C}_{0}}}\frac{{{U}_{0}}}{k}=118\times {{10}^{-6}}J$