Deriving the equation for electric field strength between parallel plates

By definition electric field strength is force per unit positive charge;

$E = \frac{F}{q}$     (1)

Work done is the product of force and distance, $W = Fd$. This can be rearranged for $F$;

$F = \frac{W}{d}$     (2)

Additionally, potential difference is the work done per unit charge, $V = \frac{W}{q}$. This can be rearranged for $q$;

$q = \frac{W}{V}$     (3)

Substituting equation (2) and (3) into (1) gives:

$E = \frac{\frac{W}{d}}{\frac{W}{V}}$

Work done cancels to give:

$E = \frac{\frac{1}{d}}{\frac{1}{V}}$

This then leaves:

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

Showing that $Vm^{-1}$ is equivalent to $NC^{-1}$

The same process as above can be used but by using units instead of quantities:

So, by definition electric field strength is force per unit positive charge and so:

$E \rightarrow NC^{-1}$     (1)

Work done is the product of force and distance, $W \rightarrow Nm$. This can be rearranged for $N$;

$N \rightarrow Jm^{-1}$     (2)

Additionally, potential difference is the work done per unit charge, $V \rightarrow JC^{-1}$. This can be rearranged for $C$;

$C \rightarrow JV^{-1}$     (3)

Substituting (2) and (3) into (1) gives:

$NC^{-1} \rightarrow Jm^{-1} \times (JV^{-1})^{-1}$

Work done cancels to give:

$NC^{-1} \rightarrow m^{-1} \times (V^{-1})^{-1}$

$NC^{-1} \rightarrow m^{-1} \times V$

This then leaves:

$NC^{-1} \rightarrow Vm^{-1}$