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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$W = zero$
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