Capacitance is a property of charge storage exhibited by two conducting surfaces separated by an insulating medium when supplied from a voltage source. The capacitor is a device used to store charge. It consists of two conducting surfaces called electrodes separated from each other by a dielectric medium.

Capacitor stores energy in the form of an electric field (charges). Opposite charges are accumulated on the conducting surfaces such that positive charges are concentrated on the other.

**Definition of capacitance**

Charge per unit voltage a capacitor can store is called capacitance and it is denoted by C.

**Symbol of capacitor**

**Unit of capacitance**

Unit of capacitance is farad and it is denoted by F. One farad can be defined as the amount of capacitance when one coulomb of charge is stored at one volt across the capacitor. If a capacitor is capable of storing more charge per unit voltage, then it is said to have more capacitance.

C= Q/V

**Current through a capacitor**

Current through a capacitor is given by i = C.dv/dt ——————(1)

**Voltage across a capacitor **

Voltage across a capacitor is given by

V = ^{1}/_{c}∫i.dt + v(0) ——————–(2)

From the above equation, it is evident that the voltage across the capacitor is independent of the initial terminal voltage.

Power absorbed by the capacitor is given by

P = vi = vi x

^{dv}/_{dt}The energy stored by the capacitor E = CV

^{2}/2**Key points to be noted about capacitors:**

- From equation (1) it is evident that the current through an inductor varies as a function of voltage. Therefore if the voltage remains constant, the current through the capacitor will become zero. Hence a capacitor acts as an open circuit for DC.
- Even if the current through a capacitor is zero, the capacitor stores a finite amount of energy.
- A pure capacitor never dissipates energy, but stores it the form of the electric field. But in practice, capacitors dissipate a small amount of energy due to its internal resistance.

**Problems based on capacitance.**

Calculate the energy stored in a capacitor of 2µF if it is charged to a voltage of 1000V.

**Solution:**

E = CV

^{2}/2 = 0.5 x 2 x 10^{-6}x 1000^{2}

= 1 joule.