Question

As per the diagram a point charge +q is placed at the origin O. work done in taking another point charge -Q from the point A [ coordinates (0, a)] to another point B [coordinates (a, 0)] along the straight line AB is:

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

Answer 1

The correct option is A zero

Work done is equal to zero because the potential of $A$ and $B$ are the same $=\frac{1}{4\pi {\epsilon }_0}\frac{q}{a}$
No work is done if a particle does not change its potential energy.
ie initial potential energy $=$ final potential energy.