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. $\frac{qQ}{4\pi {\epsilon }_0}(\frac{b}{a^2}-\frac{1}{b})$
• D. $\frac{qQ}{4\pi {\epsilon }_0}(\frac{a}{b^2}-\frac{1}{b})$

Answer 1

The correct option is A $\frac{qQ}{4\pi {\epsilon }_0}\left(\frac{a-b}{ab}\right)$

Charge $=+q$

Work done in taking charge $=Q$ from $A$ to $B$ will be equal to change in potential energy between that point.

Work done $=W=U_f-U_i=\frac{qQ}{4\pi {ϵ}_0}(\frac{1}{b}-\frac{1}{a})=\frac{qQ}{4\pi {ϵ}_0}\left(\frac{a-b}{ab}\right)$