Deduce an expression for the potential energy of a system of two point charges q1 and q2 located at position r1 and r2 respectively in an external field (E) 

Answers (1)

Work done to bring q1 from infinity to r1 = q1v(r1)

Work done to bring q2 from infinity to r= q2v(r2)

Also, work done on q2 to move it against the field due to \mathrm{q}_{1}=\frac{k q_{1} q_{2}}{r_{12}}
Where,
\mathrm{k}=\frac{1}{4 \pi \epsilon_{o}}
r_{12}= distance between q1 and q2
. . potential energy of the system =\mathrm{q}_{1} \mathrm{v}\left(\mathrm{r}_{1}\right)+\mathrm{q}_{2} \mathrm{v}\left(\mathrm{r}_{2}\right)+\frac{k q_{1} q_{2}}{r_{12}}

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