Energy Stored in a Capacitor

How much Energy Stored in a Capacitor? This types of question may arise in your mind. In this article we are trying to resolve various queries on Energy Stored in a Capacitor. When a capacitor is charged from it’s original charge less state to the fully charged state the Capacitor absorbs some energy from the source from which it is being charged, and thus the capacitor stores some amount of charge in it.

The energy is stored in the form of electrostatic field in the dielectric medium inside the capacitor, when discharging a capacitor  the electrostatic field in the dielectric medium collapses and the energy stored in the capacitor is released.

When a capacitor is being charged, some energy is lost by the source to move the charge from one plate to another. The force required to move the charges from one plate to another gradually increases as the capacitor gets charged over time, which is overcome by the energy lost by the source, after a point the source can no longer stand the force opposing the  flow of charge through the plates and the capacitor is said to be fully charged or saturated.

Capacitors
Capacitors

Calculation of energy stored in a Capacitor:

Here, Let us calculate the total energy lost by the source or stored in a fully charged capacitor of capacitance “C” with a source of voltage “V”:

Let at any Instance:
The total charge attained by the plates of the capacitor is: q Coulombs
The voltage or potential difference attained by the plates is: v Voltage

So, At any instance the capacitance; C = \dfrac{q}{v}

Or, v = \dfrac{q}{C}

Now, When the charge in the plates is raised by a small amount “dq” coulombs, the work equivalent to v.dq is done to move the charge dq from one plate to another. So the energy stored or added to the capacitor in the interval of moving dq charge is given by dw where:
dw = v.dq = \dfrac{q}{C} . dq

Now, Let the total charge accumulated by the capacitor from the state of totally discharged to the state of totally charged is Q coulombs.
So, The total energy stored by the capacitor from the state of totally discharged to the state of totally charged is the integral of all the instantaneous work done from when the charge accumulated by the capacitor is 0 coulombs up to when the charge accumulated by the capacitor is Q coulombs.
So, The total Work done or Energy stored by the capacitor is:
W = \displaystyle{\int^{Q}_{q=0}} \dfrac{q}{C} . dq

or, W = \dfrac{1}{C} \left[\dfrac{q^2}{2}\right]_{0}^{Q}

or, W = \dfrac{Q^2}{2C}

Also ,
W = \dfrac{Q^2}{2C} = \dfrac{Q}{2} . \dfrac{Q}{C} = \dfrac{1}{2}QV = \dfrac{1}{2}CV^2

Thus, Energy stored in a capacitor of capacitance C which attained a potential difference of V voltage between it’s plates during charging is:
W = \dfrac{Q^2}{2C} = \dfrac{1}{2}QV = \dfrac{1}{2}CV^2