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Derive the formula for the electric potential energy of system of two charges.
Solution
Consider the charges $q_{1}$ and $q_{2}$ initially at infinity. Suppose, first the charge $q_{1}$ is brought from infinity to the point by distance $r_{1}$. There is no external field against which work needs to be done, so work done in bringing $q_{1}$ from infinity by distance $r_{1}$ is,
$\therefore \mathrm{W}_{1}=0$
$\ldots$ $(1)$
This charge $q_{1}$ produces a potential in space is $\mathrm{V}_{1}=\frac{k q_{1}}{r_{1 p}}$
where $r_{1 p}$ is the distance of a point $\mathrm{P}$ in space from the location of $q_{1} .$
Now work done in bringing charge $q_{2}$ from infinity to the given point by distance $r_{2}$ in the electric field of $q_{1}$ is,
$\mathrm{W}_{2}=q_{2} \mathrm{~V}_{1}$
$\therefore \mathrm{W}_{2}=\frac{k q_{1} q_{2}}{r_{12}} \quad \ldots$ (1) where $r_{12}$ is the distance between points $1$ and $2$ at distance $r_{1}$ and $r_{2}$.
Since electrostatic force is conservative this work gets stored in the form of potential energy of the system.
Potential energy of system of two charges $q_{1}$ and $q_{2}$ is,
$\mathrm{U}=\mathrm{W}_{1}+\mathrm{W}_{2}$
$\therefore \mathrm{U}=\frac{k q_{1} q_{2}}{r_{12}}$
If $q_{2}$ was brought first to its present location and $q_{1}$ brought later the potential energy would be according to equation $(2)$.
Generally, the potential energy expression is unaltered whatever way the charges are brought to the specified locations because the path independence of work for electrostatic force.
$\therefore U=k\left[0+\frac{q_{1} q_{2}}{r_{12}}+\frac{q_{1} q_{3}}{r_{13}}+\frac{q_{2} q_{3}}{r_{23}}\right]$
$\therefore U=k\left[\frac{q_{1} q_{2}}{r_{12}}+\frac{q_{1} q_{3}}{r_{13}}+\frac{q_{2} q_{3}}{r_{23}}\right]$
Equation $(2)$ is true for any sign of $q_{1}$ and $q_{2}$.
If $q_{1} q_{2}>0$ potential energy will be positive. And if $q_{1} q_{2}<0$ potential energy will be negative.
