In this article we will understand and derive the expression for potential energy of charges in **external electric field. **so keep reading…

# ELECTRIC POTENTIAL ENERGY

Before going further let’s revise the concept of **electric potential energy**. If we see the term electric potential energy, then we see that this term is self explanatory.

We can define this term as – **It is the energy which is stored in the electrical bodies when it is moved from one point to another point in any external electric field***.* In other words we can say that – **It is the energy acquired by the charged bodies by virtue of change in its position in any external electric field.** [latexpage]

$$

W=Vq

$$

Where V is the electric potential at that point.

There are lots of ways by which we can define **electric potential energy**. A famous definition is – **It is the amount of energy acquired by the charge particle in bringing a positive charge from infinity to a specific point in any electric field without any acceleration.**

## ELECTRIC POTENTIAL ENERGY OF A SINGLE CHARGE IN EXTERNAL ELECTRIC FIELD

Let’s consider a charge particle q placed in external electric field of magnitude E. Here we consider this charge particle very small. Then the potential energy of this charge particle q in this external electric field is equal to the work done in bringing this charge particle from infinity to a specific point in **electric field.**

*Keep in mind that value of electric field and electric potential can vary from point to point because these are dependent on the position vector r. *

The value of **electric potential energy** of charge particle q in external electric field at position vector r from the origin is given as-

\begin{align*}

W&=qV\left(r\right)\\

&=\frac{1}{4\pi\epsilon_0}\frac{q}{r}

\end{align*}

Where $V\left(r\right)$ is electric potential at position r from the origin.

## ELECTRIC POTENTIAL ENERGY OF A SYSTEM OF TWO POINT CHARGE IN EXTERNAL ELECTRIC FIELD

Let’s consider a system of two charges $q_1$ and $q_2$, which is placed at position vector $r_1$ and $r_2$ from the origin in electric field of magnitude E.

Now **what is the potential energy acquired by this system of two charge particles? ** Let the work done in bringing the charge $q_1$ from infinity to position vector $r_1$ is given as $q_1V\left(r_1\right)$ and work done in bringing the charge $q_2$ from infinity to the position vector $r_2$ is given as $q_2V\left(r_2\right)$.

But here we have to note one thing that when are we bringing charge $q_2$ from infinity to the position vector $r_2$ then **we have to do work against the two opposing factors, first is electric field and second is electric field due to charge** $q_1$. So we have to do little more work in bringing charge $q_2$ from infinity to the position $r_2$. So the potential energy acquired by the charge $q_2$ due to charge $q_1$ is given as-

$$

W_2=\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r_{12}}

$$

Where ${r_{12}}$ is the position vector between charge $q_1$ and $q_2$. But the total work done in bringing charge $q_2$ from infinity to position vector $r_2$ is given as-

$$

W_2=q_2V\left(r_2\right)+\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r_{12}}

$$

Now the total work done in bringing both the electric charges from infinity to the present configuration is given as-

$$

U=q_1V\left(r_1\right)+q_2V\left(r_2\right)+\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r_{12}}

$$

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