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 betway 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 $-E$

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

Final energy is equal to

$E+2K\times \frac{Q(-Q/2}{d/2}=E-2E=-E$