Q12CQ - College Physics (Urone)

Question

Explain in your own words why equipotential lines and surfaces must be perpendicular to electric field lines.

Step-by-Step Solution

Answer

As the work done in moving a charge on an equipotential surface is zero but the electric field and charge is non zero, the electric field must be perpendicular to the direction of motion of charge.

Step 1: Definition of Equipotential surface

On the equipotential surface, the potential has the same value at every point. The equipotential lines are 2-D manifestations of the surfaces, on which potential is the same all along the line.

Step 2: Explanation of why equipotential lines and surfaces must be perpendicular to electric field lines

The work required  (W)  to move a charge along an equipotential line is zero; 

$\begin{array}{rcl}\mathbf{W}& \mathbf{=}& \mathbf{-}\mathbf{q}\mathbf{\times }\mathbf{dV}\\ & \mathbf{=}& \mathbf{-}\mathbf{q}\mathbf{\times }\mathbf{\left(}\mathbf{0}\mathbf{\right)}\\ & \mathbf{=}& \mathbf{0}\end{array}$

where q is electric charge and dv is potential difference.

 

This work is related to the electric field as: 

 $\begin{array}{rcl}\mathbf{W}& \mathbf{=}& {\mathbf{F}}_{\mathbf{e}}\mathbf{\times }\mathbf{d}\mathbf{\times }\mathbf{cos\theta }\\ & \mathbf{=}& \mathbf{qE}\mathbf{\times }\mathbf{d}\mathbf{\times }\mathbf{cos\theta }\\ & & \end{array}$

where F_e = qE is the force on charge q due to electric field E .

Since neither charge q nor electric field strength E  is zero, $\cos \left(\mathrm{\theta }\right)$ must be zero.

Hence, $\mathrm{\theta }$  must be $90°$ .

 

This means that the equipotential lines and surfaces are perpendicular to the electrical field lines.