Formula: Plate Capacitor Electric field (E field) Voltage Distance
$$\class{purple}{E} ~=~ \frac{U}{d}$$
$$\class{purple}{E} ~=~ \frac{U}{d}$$
$$U ~=~ \class{purple}{E}\,d$$
$$d ~=~ \frac{U}{\class{purple}{E}}$$
Electric field (E field)
$$ \class{purple}{\boldsymbol E} $$ Unit $$ \frac{\mathrm{V}}{\mathrm{m}} = \frac{\mathrm{N}}{\mathrm{C}} = \frac{\mathrm{kg} \, \mathrm{m}}{\mathrm{A} \, \mathrm{s}^3} $$
Electric field tells how large the electric force on a test charge would be if it were placed between the capacitor plates (electrodes). In a plate capacitor whose electrodes are larger compared to the distance between the electrodes, the electric field between the electrodes is homogeneous. This means that it does not matter where the test charge is placed between the electrodes, the charge will always experience the same force.
Voltage
$$ U $$ Unit $$ \mathrm{V} = \frac{ \mathrm J }{ \mathrm C } = \frac{ \mathrm{kg} \, \mathrm{m}^2 }{ \mathrm{A} \, \mathrm{s}^3 } $$
Voltage tells how large the potential difference between the two electrodes is, that is how large the difference in potential energies per charge is. Thus, the voltage indicates how much energy a test charge \(q\) gains or loses as it travels from one electrode to the opposite electrode: \(q \, U \).
Distance
$$ d $$ Unit $$ \mathrm{m} $$
Distance between the two electrodes. The greater the distance, the smaller the electric field \(E\) (with constant voltage \(U\)).