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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) 

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

Work done to bring q2 from infinity to r2 = 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}}

Posted by

Satyajeet Kumar

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