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}