Suggested languages for you:

Americas

Europe

Q12CQ

Expert-verifiedFound in: Page 694

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.

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.

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, $\mathrm{cos}\left(\mathrm{\theta}\right)$ must be zero.

Hence, $\mathrm{\theta}$ must be $90\xb0$ .

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

94% of StudySmarter users get better grades.

Sign up for free