# Capacitance of a Capacitor

The defibrillator is essentially a capacitor whose electrodes are placed on the patient’s chest.

**. Suppose then that plate**

*Q***of this capacitor has electric potential**

*A***and plate**

*V*_{A}**has electric potential**

*B***. The electrical voltage or potential difference between the plates of the capacitor is represented by U. Therefore, we determine the capacity or capacitance of this capacitor using the following equation:**

*V*_{B}Capacity depends:

*– the insulator between the armatures
– the strength and size of each armature, as well as the relative position between them.*

Electric potential energy stored by a capacitor

In order to store energy in a capacitor, it is necessary to do **work** that is transformed into *electrical potential energy* . So, let’s consider the figure above, where we have an electrical circuit with a flat capacitor. If we close the circuit switch ** D** , the capacitor will charge. In this way, its capacitance is given by , which results that the charge of the capacitor is, at each instant, directly proportional to its potential difference.

We can express this result through the ** Q x U** graph shown below. The generator, by charging the capacitor, provided it with electrical potential energy. This energy stored by the capacitor is given, numerically, by the area A represented in the figure below.

*E _{P} = shaded area*

In summary we have: