Explain electric potential energy. Show that the sum of kinetic energy and electric potential energy remains constant.

Consider the field $\overrightarrow{\mathrm{E}}$ due to a charge $\mathrm{Q}$ placed at the origin.

We bring a test charge $q$ from a point $\mathrm{R}$ to a point $\mathrm{P}$ against the repulsive force on it due to the charge $Q$. (This will happen if $Q$ and $q$ are like charges).

Let us take both $Q$ and $q$ as positive.

The test charge $q$ is so small that it does not disturb the original configuration.

In bringing the charge $q$ from $\mathrm{R}$ to $\mathrm{P}$, we apply an external force $\overrightarrow{\mathrm{F}}_{\text {ext }}$ and repulsive electric force on charge $q$ is $\overrightarrow{\mathrm{F}_{\mathrm{E}}}$.

This means there is no net force [Means $\vec{F}_{\text {ext }}=-\vec{F}_{E}$ ] means it is brought with slow constant speed and it has no acceleration.

In this situation, work done by the external force is the negative of the work done by the electric force and gets fully stored in the form of potential energy of the charge $q$.

If the external force is removed on reaching P, the electric force will take the charge away from Q the stored energy at P is used to provide kinetic energy to the charge $q$ in such a way that the sum of the kinetic and potential energies is conserved.

The work done by external forces in moving a charge $q$ from $\mathrm{R}$ to $\mathrm{P}$ is,

$\mathrm{W}_{\mathrm{RP}}=\int_{\mathrm{R}}^{\mathrm{P}} \mathrm{F}_{\mathrm{ext}} \cdot \overrightarrow{d r}$

and work done by electric force

$\mathrm{W}_{\mathrm{RP}}=-\int_{\mathrm{R}}^{\mathrm{P}} \overrightarrow{\mathrm{F}}_{\mathrm{E}} \cdot \overrightarrow{d r}$

This work ${ }^{\mathrm{R}}$ done is gets stored as potential energy of charge $q$,

$\therefore \quad U=\int_{R}^{p} \overrightarrow{F_{\text {ext }}} \cdot \overrightarrow{d r}$