Question

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

• A. -2E
• B. -E
• C. 0
• D. E

Answer 1

The correct option is B

-E

${\mu }_i=\frac{K{Q}^2}{d}=E$

${\mu }_f=\frac{K{Q}^2}{d}+\frac{K{(-Q)}^2}{d}+\frac{K{(-Q)}^2}{d}$

$=-\frac{K{Q}^2}{d}=-E$