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. $\frac{qQ}{4 \pi \varepsilon _{0}}\left(\frac{a -b}{ab}\right)$
• B. $\frac{qQ}{4 \pi \varepsilon _{0}}\left(\frac{b -a}{ab}\right)$
• C. $\frac{qQ}{4 \pi \varepsilon_{0} }\left(\frac{b}{a^{2}}-\frac{1}{b}\right)$
• D. $\frac{qQ}{4 \pi \varepsilon_{0} }\left(\frac{a}{b^{2}}-\frac{1}{b}\right)$

Answer 1

Answer: (A)

We know that

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