Question:
Two equipotential surfaces A and B are separated by a distance x. The work done in moving a charge -q from A to B is (Assume $\varepsilon_0$ = permittivity of free space)
Two equipotential surfaces A and B are separated by a distance x. The work done in moving a charge -q from A to B is (Assume $\varepsilon_0$ = permittivity of free space)
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No work is done in moving a charge along or between equipotential surfaces.
Updated On: May 12, 2025
- $\dfrac{1}{4\pi\varepsilon_0} \cdot \dfrac{q}{r}$
- $\dfrac{1}{4\pi\varepsilon_0} \cdot \dfrac{-q}{r}$
- $\dfrac{1}{4\pi\varepsilon_0} \cdot \dfrac{q}{r^2}$
- zero
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The Correct Option is D
Solution and Explanation
By definition, the potential difference between two points on an equipotential surface is zero.
Since $W = q \cdot \Delta V$, and $\Delta V = 0$ between points A and B on equipotential surfaces,
$\Rightarrow W = -q \cdot 0 = 0$
Since $W = q \cdot \Delta V$, and $\Delta V = 0$ between points A and B on equipotential surfaces,
$\Rightarrow W = -q \cdot 0 = 0$
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