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 $A B$ is

• Correct answer
  A

  $\frac{q Q}{4 \pi \varepsilon_{0}}\left(\frac{a-b}{a b}\right)$
• B

  $\frac{q Q}{4 \pi \varepsilon_{0}}\left(\frac{b-a}{a b}\right)$
• C

  $\frac{q Q}{4 \pi \varepsilon_{0}}\left(\frac{b}{a^{2}}-\frac{1}{b}\right)$
• D

  $\frac{q Q}{4 \pi \varepsilon_{0}}\left(\frac{a}{b^{2}}-\frac{1}{b}\right)$

Solution

We know that and $\quad V=\frac{Q}{4 \pi \varepsilon_{0} r}$ $\begin{aligned} \text { Hence, } \quad W &=q\left[\frac{Q}{4 \pi \varepsilon_{0} b}-\frac{Q}{4 \pi \varepsilon_{0} a}\right] \\ &=\frac{Q q}{4 \pi \varepsilon_{0}}\left[\frac{a-b}{a b}\right] \end{aligned}$

Hence, the correct answer is option: 'A'.