A capacitor stores electric charge and is useful in many electronics circuits. Dielectric is an insulator separates the two parallel plates. Capacitor can store charge in its capacitance (C). The rate of capacitance is directly proportional to the area of plate which is closetogether. It is measure in the farads. Let we see how the energy storedin a capacitor.

## Energy Stored in a Capacitor:

**Energy Stored in a Capacitor:**

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**Image Description – Charge on Capacitor:**

ABattery is connected across the capacitor for charging. The positive ofthe battery attracts the electron from the plate X and negative of battery repels the electron from the plate Y.

**Charging:**

Theelectrons flow round the circuit from X (positive plate) to Y (negativeplate). A current would be sensed by a sensitive ammeter. The voltage builds between X and Y and opposes battery voltage. Charging stops when these two voltages are equal; the electron flow, that the charging current is then zero.

During charging process, electrical energy is transferred from the battery to the capacitor, which then stores the energy.

**Discharging:**

Theelectrons flow round the circuit from Y (negative plate) to X (positiveplate). The charge stored by the capacitor became zero, as does the voltage across it. The stored energy capacitor has been transferred fromcapacitor to the conductor. The delay time taken for a capacitor to fully charge or discharge through a resistor is made use in many electronic circuits.

## Formula - Energy Stoerd in a Capacitor:

**Formula – Energy Stored in a Capacitor:**

Tofind the energy stored in a capacitor, the following formula can be used and the unit of energy stored in a capacitor is joules.

Formula: W = 1/2 C V^{2} or W = 1 / 2 (V (Q))

Where

W =* *energy stored (Joules)

C = capacitance (Farad)

V = potential difference (Voltage)

Q = Charge on capacitor

## Example Problems –Energy Stored in a Capacitor:

**Example 1:**

**Find the energy stored on the capacitor when the value of capacitor is 4 micro farads and the voltage value is 25 V. **

Solution:

Formula: W = 1/2 C V^{2}

W = ½ 4 (10)^{-6}* (25)^{2} = 0.00125 Joules.

**Example 2:**

**Find the energy stored in a capacitor. The charge of the capacitor is 75 coulombs and the value of the voltage drop is 5 V.**

Solution:

Formula: W = 1 / 2 (V (Q))

W = 1/ 2 (5 (75)) = 187.5 Joules.