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. $\left( \frac{qQ}{4\pi\varepsilon_{0}}\frac{1}{a^{2}} \right)\sqrt{2}a$
• C. $\left( \frac{- qQ}{4\pi\varepsilon_{0}}\frac{1}{a^{2}} \right)\sqrt{2}a$
• D. $\left( \frac{qQ}{4\pi\varepsilon_{0}}\frac{1}{a^{2}} \right)\frac{a}{\sqrt{2}}$

Answer 1

Answer: (A)

Zero

Answer 2

: Work done is equal to zero because the potential of

A and B are the same $= \frac{1}{4\pi\varepsilon_{0}}\frac{q}{a}$ no work is done if a particle does not change its potential energy.

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