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

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

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

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

$\overrightarrow{\mathbf{E}}\mathbf{=}\frac{{\overrightarrow{\mathbf{F}}}_{\mathbf{e}}}{\mathbf{q}}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{(2)}$

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

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

Where F_e 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.