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Found in: Page 694

### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

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

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.

See the step by step solution

## 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;

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

where q is electric charge and dv is potential difference.

This work is related to the electric field as:

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

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

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

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

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