Question

Explain the following statement giving reason :
An equipotential surface through a point is normal to the electric field at that point.

Hint:

Always remember that if an electric field had a parallel component along a surface, it would actively exert a force on charges, doing physical work to move them. Since no work can be done on an equipotential surface, the parallel component must strictly be zero!

Answer 1

Concept:
• An equipotential surface is defined as a region in space where the electric potential $V$ is completely uniform and constant at every single point.
• The relationship between the electric field vector $\vec{E}$ and the change in electric potential $dV$ is given by the fundamental calculus relation $dV = -\vec{E} \cdot d\vec{l}$.
• Work done in moving a test charge along an equipotential surface must be exactly zero.

Step 1: Define the potential difference along the surface
By the very definition of an equipotential surface, the potential difference between any two infinitesimally close points on this specific surface is exactly zero.
Mathematically, this means $dV = 0$.

Step 2: Apply the electric field and potential relation
The change in potential $dV$ for a tiny displacement $d\vec{l}$ along the surface is expressed using the dot product:
\[ dV = -\vec{E} \cdot d\vec{l} \]
Substitute $dV = 0$ strictly into this equation:
\[ 0 = -\vec{E} \cdot d\vec{l} \]

Step 3: Expand the dot product mathematically
Expanding the vector dot product gives:
\[ E \cdot dl \cdot \cos\theta = 0 \]
Where $\theta$ represents the exact angle between the electric field vector $\vec{E}$ and the displacement vector $d\vec{l}$ lying on the surface.
Since neither the electric field magnitude $E$ is zero nor the displacement $dl$ is zero (for a non-trivial movement), the only mathematical possibility remaining is:
\[ \cos\theta = 0 \]

Step 4: Conclusion
Solving for $\theta$ yields:
\[ \theta = 90^\circ \]
This rigorously proves that the electric field must always be perfectly perpendicular (normal) to the equipotential surface at any given point to ensure no work is done moving charges along it.