Question

A charge 10 esu is placed at a distance of 2 cm from a charge 40 esu and 4 cm from another charge of 20 esu. The potential energy 10 esu in ergs is

• A. 87.5
• B. 112.5
• C. 150
• D. 250

Answer 1

The correct option is D 250

Potential energy of system of $2$ charges

$U=\frac{k{Q}_{1}Q2}{r}(S.I)U=\frac{Q_1{Q}_2}{r}(C.G.S)$ where $H=\frac{1}{4\pi {ϵ}_0}$

Potential Energy of $10e.s.u$ charge

$U=\frac{10\times 40}{2}+\frac{10\times 20}{4}=250erg$