Question

Charges $+q$ and $-q$ are placed at points $A$ and $B$ respectively which are a distance $2L$ apart, $C$ is the midpoint between $A$ and $B$ . The work done in moving a charge $+Q$ along the semicircle $CRD$ is

• A. $\frac{qQ}{4\pi {\epsilon }_{0}L}$
• B. $\frac{qQ}{2\pi {\epsilon }_{0}L}$
• C. $\frac{qQ}{6\pi {\epsilon }_{0}L}$
• D. $-\frac{qQ}{6\pi {\epsilon }_{0}L}$

Answer 1

Answer: (D)

Work done is equal to change in potential energy. In $I^{st}$ case, when charge $+Q$ is situated at $C$.

Electric potential energy of system
$U_1=\frac{1}{4\pi {\epsilon }_0}\frac{\left(q\right)\left(-q\right)}{2L}+\frac{1}{4\pi {\epsilon }_0}\frac{\left(-q\right)Q}{L}+\frac{1}{4\pi {\epsilon }_0}\cdot \frac{qQ}{L}$
In $I{I}^{nd}$ case, when charge $+Q$ is moved from $C$ to $D$.

Electric potential energy of system in that case
$U_2=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{\left(q\right)\left(-q\right)}{2L}+\frac{1}{4\pi {\epsilon }_0}\cdot \frac{qQ}{3L}+\frac{1}{4\pi {\epsilon }_0}\frac{\left(-q\right)\left(Q\right)}{L}$
$\therefore$ Work done $=\Delta U=U_2-U_1$
$=\left(-\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{2L}+\frac{1}{4\pi {\epsilon }_0}\frac{qQ}{3L}-\frac{1}{4\pi {\epsilon }_0}\frac{qQ}{L}\right)$
$-\left(-\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{2L}-\frac{1}{4\pi {\epsilon }_0}\cdot \frac{qQ}{L}+\frac{1}{4\pi \epsilon }\cdot \frac{qQ}{L}\right)$
$=\frac{qQ}{4\pi {\epsilon }_0}\cdot \left[\frac{1}{3L}-\frac{1}{L}\right]=\frac{qQ}{4\pi {\epsilon }_0}\frac{\left(1-3\right)}{3L}$
$=\frac{-2qQ}{12\pi {\epsilon }_{0}L}=-\frac{qQ}{6\pi {\epsilon }_{0}L}$