MCQ
There are two equipotential surface as shown in figure. The distance between them is r. The charge of –q coulomb is 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$W=-\frac{1}{4 \pi \varepsilon_0} \frac{q}{r^2}$
- ✓$W = Zero$
