question_answer

A charge +q is placed at the origin 0 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)   \[\frac{qQ}{4\pi {{\varepsilon }_{0}}}\left( \frac{a}{{{b}^{2}}}-\frac{1}{b} \right)\]

Correct Answer: A

Solution :

 We know that \[W=q\Delta V\]and \[V=\frac{Q}{4\pi {{\varepsilon }_{0}}r}\] Hence,   \[W=q\left[ \frac{Q}{4\pi {{\varepsilon }_{0}}b}-\frac{Q}{4\pi {{\varepsilon }_{0}}a} \right]\]  \[=\frac{Qq}{4\pi {{\varepsilon }_{0}}}\left[ \frac{a-b}{ab} \right]\]