Question

Why electric field $\vec{E}$ at a point on an equipotential surface must be perpendicular to the surface at that point ?}

Hint:

If the field were not flawlessly perpendicular, it would inherently possess a non-zero horizontal component physically running along the surface. This renegade component would actively exert force and do real work on charges, violently violating the defining $dW=0$ condition.

Answer 1

Concept:
• By foundational definition, an equipotential surface is a continuous 3D geometric boundary across which the absolute electric potential is physically exactly constant at every single point.
• Because there exists absolutely zero potential difference directly between any two arbitrary points safely lying on this surface ($dV = 0$), the rigorous physical work required to move a tiny test charge anywhere along it is mathematically zero.

Step 1: Establish the mathematical relation for work done
The physical work $dW$ continuously done in moving a tiny test charge $q$ over a microscopically small displacement $d\vec{l}$ entirely along any equipotential surface is rigidly given by the standard dot product equation:
\[ dW = \vec{F} \cdot d\vec{l} = q(\vec{E} \cdot d\vec{l}) \]
\[ dW = q E dl \cos\theta \]
where $\theta$ actively represents the explicit angle strictly existing between the electric field vector $\vec{E}$ and the surface displacement vector $d\vec{l}$.

Step 2: Apply the equipotential condition
As logically stated, the absolute potential difference $dV$ across any equipotential surface is definitively zero.
Since work done is also deeply defined as $dW = -q dV$, it robustly follows that:
\[ dW = -q(0) = 0 \]

Step 3: Deduce the required angle
Equating our two independent work expressions yields:
\[ q E dl \cos\theta = 0 \]
Since the test charge $q$ is non-zero, the local electric field $E$ is non-zero, and the displacement $dl$ is certainly non-zero, the purely mathematical burden of establishing this zero strictly falls entirely upon the cosine term:
\[ \cos\theta = 0 \]
The only viable physical angle that beautifully satisfies this stringent trigonometric condition is:
\[ \theta = 90^\circ \]

Step 4: Conclusion
This elegantly proves mathematically that the electric field vector $\vec{E}$ must absolutely, without exception, be oriented strictly perpendicular (normal) to the equipotential surface at every single point upon it.