A charge of $10\, e.s.u.$ is placed at a distance of $2\, cm$ from a charge of $40\, e.s.u.$ and $4\, cm$ from another charge of $20\, e.s.u.$ The potential energy of the charge $10\, e.s.u.$ is (in $ergs$)
A$87.5$
B$112.5$
C$150$
D$250$
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D$250$
d (d) Energy $ = \frac{{10 \times 40}}{2} + \frac{{10 \times 20}}{4} = 250\,\,erg$
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