Gujarati
2. Electric Potential and Capacitance
easy

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

$250$

Solution

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

Standard 12
Physics

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