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(D) By definition,an equipotential surface is a surface where the electric potential is the same at every point.Since $A$ and $B$ are two distinct equ

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Zero

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(D) By definition,an equipotential surface is a surface where the electric potential is the same at every point.Since $A$ and $B$ are two distinct equipotential surfaces,the potential at any point on surface $A$ is constant (let it be $V_A$) and the potential at any point on surface $B$ is constant (let it be $V_B$).However,the question implies moving a charge between two equipotential surfaces. If the surfaces are at different potentials,work is done. But if the question implies moving a charge along an equipotential surface,work is zero.Re-evaluating the standard physics problem context: If $A$ and $B$ are equipotential surfaces,the work done in moving a charge $q$ from $A$ to $B$ is $W = q(V_B - V_A)$.If the question implies moving the charge *along* the surface or if the surfaces are at the same potential,the work is zero. Given the options,the standard interpretation for this specific textbook problem is that the work done in moving a charge between two points on the same equipotential surface is zero. If moving between two different surfaces,it depends on the potential difference. Assuming the question asks for work done *along* an equipotential surface or that the surfaces are at the same potential,the answer is Zero.

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

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