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A charge \[q_0\] is moved from point A to point B on the same equipotential surface. If the potentials at A and B are \[V_A\] and \[V_B\] respectively, the work done is:

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Question

A charge \[q_0\] is moved from point A to point B on the same equipotential surface. If the potentials at A and B are \[V_A\] and \[V_B\] respectively, the work done is:

Options

  • \[W = q_0(V_A - V_B) \neq 0\]

  • \[W = q_0(V_A + V_B)\]

  • \[W = \frac{q_0}{(V_A - V_B)}\]

  • \[W = q_0(V_A - V_B) = 0\]

MCQ
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Solution

The work done in moving a charge \[q_0\] from A to B is \[W = q_0(V_A - V_B)\]. Since both points lie on the same equipotential surface, \[V_A = V_B\], which forces the work done to be exactly zero regardless of the path taken along the surface.

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