# Energy Stored in a Capacitor

The energy stored on a capacitor is expressed in terms of the work performed by the battery. The Voltage shows energy per unit charge, hence the work performed to move a charge element DQ from negative plate to the positive plate equals to V DQ where V is the voltage of the capacitor. The voltage is proportional to a number of charges on the capacitor.

When the capacitor is connected across a battery, the charges enter from the battery and are stored in the plates of capacitors. At the beginning, the capacitor does not hold any potential or charge, i.e, q = 0 C and V = 0 volts.

At the time to switch, the complete battery voltage falls across the capacitor. A +ve charge (q) enters at the capacitor’s positive plate but no work is done in bringing the 1st charge (q) towards the positive plate from the battery.

This occurs because capacitor does not contain any voltage across its plates and the previous voltage exists because of the battery.

## Capacitor Working

First, the charge initiates little voltage across the capacitor’s plate and second +ve charge enters the capacitor’s positive plate but replaced by the 1st charge. Since the battery’s voltage is higher than the capacitor’s voltage, the second charge is stored in positive plate. In this condition, little work has to be performed to store 2nd charge in the capacitor. A similar phenomenon occurs for the 3rd charge. Gradually the charges are stored against pre-stored charges and their small amount of work done hikes up.

The capacitor’s voltage is not constant from the start. It is maximum when the potential of the capacitor is equal to that of a battery. The voltage of capacitor rises and so does the capacitor’s energy as the storage of charges increases.

As the voltage rises the electric field in the capacitor dielectric hikes gradually but from positive to negative plate, i.e. in opposite direction.

$E=-frac{dV}{dX}$

Here, dx indicates the distance between the two plates.

The charge flows from battery to capacitor plate till the capacitor acquires potency same as the battery. Hence we need to determine the capacitor’s energy to the final moment of getting the complete charge.
Suppose a charge is stored in positive plate according to the battery’s voltage and the work done is DW.

If the whole charging time is considered, it can be written as:

Now let’s consider the energy lost by a capacitor during charging time.

Since the battery has a constant voltage, the energy lost always obeys the equation W=V.q. The equation doesn’t apply to the capacitor since it doesn’t carry constant voltage from the point it has started charging.

The collected charge by from the battery is given by:

The charge lost by the battery is given by:

The half energy from the total amount moves to the capacitor and the rest gets off from the battery.

#### Practise This Question

H.C.F of two co prime numbers is