Question

In the derivation proving \[\vec{E}\] is perpendicular to an equipotential surface, a charge \[q_0\] undergoes a small displacement \[d\vec{l}\] along the surface, with \[\vec{E}\] making an angle \[\theta\] with the displacement. Equating the work expressions leads to which conclusion?

विकल्प

• \[\tan\theta = 0 \implies \theta = 0^\circ\]

• \[\cos\theta = 0 \implies \theta = 90^\circ\]

• \[\cos\theta = 1 \implies \theta = 0^\circ\]

• \[\sin\theta = 1 \implies \theta = 90^\circ\]

Answer 1

The work done is \[W = q_0 E \cdot dl \cdot \cos\theta\]. Since A and B lie on the same equipotential surface, \[W = q_0(V_A - V_B) = 0\]. Equating gives \[q_0 E \cdot dl \cdot \cos\theta = 0\]. Since \[q_0 \neq 0\], \[E \neq 0\], and \[dl \neq 0\], we must have \[\cos\theta = 0\], so \[\theta = 90^\circ\] — proving the field is perpendicular (normal) to the equipotential surface.