Q. There are two equipotential surfaces as shown in figure. The distance between them is $r.$ The charge of $- q$ coulomb taken from the surface $A\ to\ B,$ the resultant work done will be
A) $W = \frac{1}{4\pi\varepsilon_{0}}\frac{q}{r}$
B) $W = \frac{1}{4\pi\varepsilon_{0}}\frac{q}{r^{2}}$
C) $W = - \frac{1}{4\pi\varepsilon_{0}}\frac{q}{r^{2}}$
D) $W =$ zero
Explanation
(d) The work done is given by $= q\left( V_{2} - V_{1} \right) = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.