When we charge a capacitor we actually supply certain energy in it. The capacitor stores are energy in its electrostatic field. In other words, we can say the due to this energy capacitor sets up an electrostatic field in its dielectric medium. After that during discharging capacitor release this stored energy by collapsing the electrostatic field. Initially, suppose a capacitor is in uncharged condition. Then we begin charging the capacitor. At that time the capacitor has to do some work to transfer the charge from one plate to another. But when the charging condition increases the capacitor has to work more to transfer the charge against the repulsive force due to already stored charge on the capacitorâ€™s plate. Now, let us find out the requirement of electrical energy in charging a capacitor of capacitor C Farad up to a voltage V volt.

## Derivation of Expression of Energy Stored in a Capacitor

Let us consider at a certain instant during charging, the voltage appears across the capacitor is V. As per the definition of voltage we can say, V is the work to be done to transfer one Coulomb charge from one plate to another of the capacitor. So, the work done for hampering dq Coulomb charge form one plate to another will be vdq. Let us denote this work as dw.

Again we know that charge q=Cv. Therefore dq=Cdv (C is the capacitance of the capacitor and it is constant). Therefore,

After that, we consider that V is the final voltage of the capacitor after full charging. Therefore total work done from initial uncharged condition to final charged condition will be

Again we can write this equation as

Again we can write this equation as

If we measure q in Coulomb and C in Farads then the unit energy stored will be in Jules.

### Non Integral Method of Deriving Expression of Energy Stored in Capacitor

Suppose a capacitor has a constant capacitance C. Again we know that Q=CV, that is the voltage across the capacitor is the product of its capacitance and charge. Since the capacitance of the capacitor (C ) is constant, we can write,

Hence, if we represent this relation graphically we get a straight line passing through the origin. The figure below is showing the graphical representation of this relation. Here X-axis represents the charge Q. Whereas Y-axis represents the voltage across the capacitor. The vertical line AB represents the voltage developed across the capacitor for the charge of Q Coulomb. Therefore the energy required to charge the capacitor for these charge Q and voltage V is nothing but the area bounded by the triangle OAB. Again we know that the area of a triangle is 1/2baseheight.

Therefore, energy stored in the capacitor