We already know about electric field and electric potential. We also know that electrostatic field is completely characterized by vector function **E**(**r**). The electric field depicts the force exerted on other electrically charged objects by the electrically charged particle the field is surrounding. Now a question arises why do we need introduction of electric potential when we already have electric field for the description of electric force between charges.

Firstly, the concept of electric potential is very useful not only in physics but as well as in engineering .This is because if we know the potential we can easily calculate the work done by field forces when a charge is displaced from point 1 to point 2 that is

\[{W_{12}} = q({\varphi _1} – {\varphi _2})\]

where \({\varphi _1}\) and \({\varphi _2}\) are the potentials at points 1 and 2. This means that required work is equal to the decrease in the potential energy of charge q when it is displaced from point 1 to 2.

Calculation of the work of field forces with the help of above mentioned formula is not just simple but the only possible method in some cases.

Secondly in some cases of electrostatic field calculation it is often easier to first calculate the potential and then find the gradient of potential \({\varphi}\) to calculate the value of electric field intensity **E**. Also for calculating \({\varphi}\) we only need to evaluate one integral but for calculation of **E **we must take three integrals all for x, y, and z directions since **E** is a vector quantity.

But we must note that for problems with high symmetry we must directly calculate **E **using **Gauss’s Theorem** which is much simpler way to find electric field intensity when charge distribution is symmetrical.

**Related Articles**

Electric charge and electric field questions and answers

How to solve electric force problems

electrostatics in physics