There are two equipotential surfaces as shown | vedclass

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Question

(D) By definition,an equipotential surface is a surface where the electric potential is the same at every point.Let the potential of surface $A$ be $V

Vedclass · Accepted Answer

$W = 0$

Vedclass · Answer

(D) By definition,an equipotential surface is a surface where the electric potential is the same at every point.Let the potential of surface $A$ be $V_A$ and the potential of surface $B$ be $V_B$.Since both $A$ and $B$ are equipotential surfaces,the potential difference between any two points on the same surface is zero.However,the question implies moving a charge between two distinct equipotential surfaces. If the surfaces are at different potentials,work is done. But in standard physics problems of this type,if the surfaces are defined as part of a system where no potential difference is specified or if they are meant to be at the same potential,the work done is zero.Given the options,the correct answer is $W = 0$,which implies $V_A = V_B$ or that the work done in moving a charge along an equipotential surface is zero. Since the charge is moved from one equipotential surface to another,and the options suggest zero,we conclude $V_A = V_B$.

There are two equipotential surfaces as shown in the figure. The distance between them is $r$. If a charge of $-q \text{ C}$ is moved from surface $A$ to surface $B$,the resultant work done will be:

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