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Obtain the equation of electric potential energy of a system of two electric charges in external electric field.
Solution
Two electric charges $q_{1}$ and $q_{2}$ are brought from infinity distance through distance $r_{1}$ and $r_{2}$ relative to origin in external electric field.
The work done in bringing the charge $q_{1}$ from infinity through $r_{1}$ distance, $\mathrm{W}_{1}=q_{1} \mathrm{~V}\left(\overrightarrow{r_{1}}\right) \quad \ldots$ (1) and work done in bringing $q_{2}$ through distance $r_{2}$. Here work is done not only against the external field $\overrightarrow{\mathrm{E}}$ but also against the field due to $q_{1}$.
$\mathrm{W}_{2}=q_{2} \mathrm{~V}\left(\overrightarrow{r_{2}}\right) \quad \ldots \text { (2) }$
Work done on $q_{2}$ against the field due to $q_{1}$,
$\mathrm{W}_{3}=\frac{k q_{1} q_{2}}{r_{12}} \quad \ldots$ (3)
where $r_{12}$ is the distance between $q_{1}$ and $q_{2}$. The total potential energy of the system = the total work done in assembling the configuration
The total potential or potential energy
$\mathrm{U}=q_{1} \mathrm{~V}\left(\overrightarrow{r_{1}}\right)+q_{2} \mathrm{~V}\left(\overrightarrow{r_{2}}\right)+\frac{k q_{1} q_{2}}{r_{12}}$
