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Two parallel surfaces, A and B having the same potential, are separated by a distance r as shown in the figure. The work done in moving a charge -q from A to B is ( $ε_{o}$ - permittivity of free space)

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$\frac{1}{4 π ε_{0}} \cdot \frac{q}{r}$

$\frac{1}{4 π ε_{0}} \cdot \frac{- q}{r}$

$\frac{1}{4 π ε_{0}} \cdot \frac{q}{r^{2}}$

zero

answer is D.

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Detailed Solution

$w o r k d o n e t o m o v e t h e c h a r g e W = q (V_{B} - V_{A}) A s t h e y a r e e q u i p o t e n t i a l s u r f a c e V_{B} - V_{A} = 0 w o r k d o n e = 0$

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