Question

A charge $+q$ is placed at the origin $O$ of $X-Y$ axes as shown in the figure. The work done in taking a charge $Q$ from $A$ to $B$ along the straight line $AB$ is

• A. $\frac{qQ}{4\pi {\epsilon }_0}\left(\frac{a-b}{ab}\right)$
• B. $\frac{qQ}{4\pi {\epsilon }_0}\left(\frac{b-a}{ab}\right)$
• C. $\frac{qQ}{4\pi {\epsilon }_0}\left(\frac{b}{a^2}-\frac{1}{b}\right)$
• D. $\frac{qQ}{4\pi {\epsilon }_0}\left(\frac{a}{b^2}-\frac{1}{b}\right)$

Answer 1

Answer: (A)

We know that

$W=q\Delta V$
and $V=\frac{Q}{4\pi {\epsilon }_{0}r}$
Hence, $W=q\left[\frac{Q}{4\pi {\epsilon }_{0}b}-\frac{Q}{4\pi {\epsilon }_{0}a}\right]$
$=\frac{Qq}{4\pi {\epsilon }_0}\left[\frac{a-b}{ab}\right]$