Question

Two equal charge of magnitude Q each are placed at a distance d apart. Their electrostatic energy is E. A third charge -Q/2 is bought 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 $-E$

Given two charges of magnitude $Q$ are at a distance $d$ apart. Their electrostatic energy is $E$. Now, a third charge $\frac{-Q}{2}$ is brought midway between these two charges.

We have to find the electrostatic energy of the system now.

Given $E$ is the electrostatic energy of two charges of magnitude $Q$ apart a distance $d$, so, $E=\frac{1}{4\pi {\epsilon }_0}\frac{Q^2}{d}$

Now, when a third charge $\frac{-Q}{2}$ is brought midway then,

distance of each $Q$ charge from $\frac{-Q}{2}$ is $\frac{d}{2}$

So, now electrostatic energy is $E^{\prime }=\frac{1}{4\pi {\epsilon }_0}[\frac{Q(-Q/2)}{d/2}+\frac{Q\left(Q\right)}{d}+\frac{Q(-Q/2)}{d/2}]$

$E^{\prime }=\frac{1}{4\pi {\epsilon }_0}[\frac{-Q^2}{d}+\frac{Q^2}{d}-\frac{Q^2}{d}]$

$=-\frac{1}{4\pi {\epsilon }_0}\frac{Q^2}{d}=-E$