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 $ergs$)

Question

Dr. Abubacker · Accepted Answer

(d) Energy $= \frac{10 	imes 40}{2} + \frac{10 	imes 20}{4} = 250\ erg$

Q. 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 $ergs$)

Explanation

Reviewed by Dr. Abubacker

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