Question

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. $W=\frac{1}{4 \pi \varepsilon_{o}} \frac{q}{r}$
• 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

Answer 1

Answer: (D)

The work done is given by $=q\left(V_{2}-V_{1}\right)=0$