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Correct answer: B. The electric field lines
An equipotential surface is defined as a surface where every point has the same electric potential ($V = \text{constant}$).
1. **Work and Potential Difference**: The work done ($W$) in moving a charge $q$ from one point to another on a surface is given by $W = q(V_2 - V_1)$.
2. **Zero Work**: Since $V_1 = V_2$ for any two points on an equipotential surface, the potential difference $\Delta V = 0$. Therefore, the work done in moving a charge along an equipotential surface is zero ($W = 0$).
3. **Force and Displacement**: Work is also defined by the dot product of the electric force ($\vec{F} = q\vec{E}$) and the displacement vector ($d\vec{s}$):
$$W = \int q\vec{E} \cdot d\vec{s} = 0$$
4. **Perpendicularity**: For the dot product $\vec{E} \cdot d\vec{s}$ to be zero for any displacement $d\vec{s}$ along the surface (where $\vec{E} \neq 0$), the electric field vector $\vec{E}$ must be perpendicular to the displacement vector.
Conclusion: Since electric field lines represent the direction of the electric field vector at any point, the electric field lines must be perpendicular to the equipotential surface.
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