The figure shows two parallel equipotential surfaces $A$ and $B$ kept a small distance $r$ apart from each other. $A$ point charge of $q$ coulomb is taken from the surface $A$ to $B$. The amount of net work done will be

  • A
    $ - \frac{1}{{4\pi {\varepsilon _0}}}\frac{q}{r}$
  • B
    $ \frac{1}{{4\pi {\varepsilon _0}}}\frac{q}{r^2}$
  • C
    $- \frac{1}{{4\pi {\varepsilon _0}}}\frac{q}{r^2}$
  • D
    Zero

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Reason: Two equipotential surfaces are parallel to each other.

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