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 (1a, 0)] along the straight path AB is

• A. zero
• B. $\left(\frac{qQ}{4\pi {\epsilon }_0}\frac{1}{a^2}\right)\cdot \sqrt{2}a$
• C. $\left(\frac{-qQ}{4\pi {\epsilon }_0}\frac{1}{a^2}\right)\cdot \sqrt{2}a$
• D. $\left(\frac{qQ}{4\pi {\epsilon }_0}\frac{1}{a^2}\right)\cdot \frac{a}{\sqrt{2}}$

Answer 1

Answer: (A)

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.
i.e. initial potential energy = final potential energy.