Q. As per this diagram a point charge $+ q$ is placed at the origin $O.$ Work done in taking another point charge $- Q$ from the point $A$ [co-ordinates $(0,a)$] to another point $B$[co-ordinates $(a,0)$] along the straight path $AB$ is
A) Zero
B) $\left( \frac{- qQ}{4\pi\varepsilon_{0}}\frac{1}{a^{2}} \right)\sqrt{2}a$
C) $\left( \frac{qQ}{4\pi\varepsilon_{0}}\frac{1}{a^{2}} \right)\frac{a}{\sqrt{2}}$
D) $\left( \frac{qQ}{4\pi\varepsilon_{0}}\frac{1}{a^{2}} \right)\sqrt{2}a$
Explanation
(a) Since $A$ and $B$ are at equal potential so potential difference between $A$ and $B$ is zero. Hence $W = Q.\mathrm{\Delta}V = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.