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Found in: Page 696

### College Physics (Urone)

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

# Show that units of V/m and N/m for electric field strength are indeed equivalent.

The units V/m and N/m are equivalent.

See the step by step solution

## Step 1: Principle

In a homogeneous electric field of magnitude E , the potential difference between two locations separated by a distance d is

${\mathbf{\Delta V}}{\mathbf{=}}{\mathbf{Ed}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{\left(1\right)}}$

The electric force Fe at any point due to a test charge q and electric field $\stackrel{\to }{\mathrm{E}}$ is given by,

$\stackrel{\mathbf{\to }}{\mathbf{E}}{\mathbf{=}}\frac{{\stackrel{\mathbf{\to }}{\mathbf{F}}}_{\mathbf{e}}}{\mathbf{q}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{\left(2\right)}}$

## Step 2: Explanation

The electric field strength is, according to Equation (1), equal to:

${\mathbf{E}}{\mathbf{=}}\frac{{\mathbf{F}}_{\mathbf{e}}}{\mathbf{q}}$

Where Fe is measured in Newton and q is in Coulombs, therefore, the electric field strength is measured in Newton/Coulomb or N/C.

From equation (2), electric field strength is:

${\mathbf{E}}{\mathbf{=}}\frac{\mathbf{\Delta V}}{\mathbf{d}}$

Where $∆\mathrm{V}$ is measured in volts and d is measured in meters, the electric field strength is measured in volt/meter or V/m .

Hence, V/m is equivalent to N/m.