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
$E = \frac{k q_1q_2}{r_{12}} = \frac{kQ^2}{d} \,...(i)$
Now, when a third charge $(-\frac{Q}{2})$ is placed
at mid-point of these charges,
then electrostatic energy of system is
$E' = \frac{kq_1q_2}{r_{12}} + \frac{kq_2q_3}{r_{23}} + \frac{kq_1q_3}{r_{13}}$
$=\frac{kQ^2}{d} - \frac{kQ^2/2}{d/2} - \frac{kQ^2/2}{d/2}$
$ = -\frac{kQ^2}{d} = -E$
[from Eq. (i)]