Question

Two equal charges of magnitude $Q$ each are placed at a distance d apart. Their electrostatic energy is $E.A$ third charge $-Q/2$ is brought midway between these two charges. The electrostatic energy of the system is now

• A. -2 E
• B. -E
• C. 0
• D. E

Answer 1

Answer: (B)

Electrostatic energy of two equal charges of magnitude $Q$ placed $d$ distance apart is
$E=\frac{k{q}_1{q}_2}{r_{12}}=\frac{k{Q}^2}{d}...(i)$
Now, when a third charge $(-\frac{Q}{2})$ is placed
at mid-point of these charges,
then electrostatic energy of system is
$E^{\prime }=\frac{k{q}_1{q}_2}{r_{12}}+\frac{k{q}_2{q}_3}{r_{23}}+\frac{k{q}_1{q}_3}{r_{13}}$
$=\frac{k{Q}^2}{d}-\frac{k{Q}^2/2}{d/2}-\frac{k{Q}^2/2}{d/2}$
$=-\frac{k{Q}^2}{d}=-E$
[from Eq. (i)]