Electrostatics

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



1) $W = \frac{1}{{4\pi {\varepsilon _o}}}\frac{q}{r}$
2) $W = \frac{1}{{4\pi {\varepsilon _0}}}\frac{q}{{{r^2}}}$
3) $W = - \frac{1}{{4\pi {\varepsilon _0}}}\frac{q}{{{r^2}}}$
4) W = zero
Solution: Explanation: Explanation
electric-field-potential

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