Formula: Plate Capacitor Attraction Force Electric charge Area Relative permittivity Vacuum Permittivity
$$F ~=~ \frac{Q^2}{2\varepsilon_0 \, \varepsilon_{\text r} A}$$
$$F ~=~ \frac{Q^2}{2\varepsilon_0 \, \varepsilon_{\text r} A}$$
$$Q ~=~ \sqrt{ 2\varepsilon_0 \, \varepsilon_{\text r} \, A \, F }$$
$$A ~=~ \frac{ Q }{ 2\varepsilon_0 \, \varepsilon_{\text r}\, F }$$
$$\varepsilon_{\text r} ~=~ \frac{Q^2}{2 \varepsilon_0 \, F \, A}$$
Attraction Force
$$ F $$ Unit $$ \mathrm{N} $$
Force with which the two plates (electrodes) attract each other.
Electric charge
$$ Q $$ Unit $$ \mathrm{C} = \mathrm{As} $$
Amount of charge on one of the plates. Here it is assumed that both plates contain the same amount of charge, but are charged with different sign (+ and -).
Area
$$ A $$ Unit $$ \mathrm{m}^2 $$
Area of one side of the capacitor plate.
Relative permittivity
$$ \varepsilon_{\text r} $$ Unit $$ - $$
Relative permittivity is a dimensionless number describing the dielectric (e.g. air, water, glass) between the two capacitor plates. In vacuum, the relative permittivity has the value \( \varepsilon_{\text r} = 1 \).
Vacuum Permittivity
$$ \varepsilon_0 $$ Unit $$ \frac{\mathrm{As}}{\mathrm{Vm}} $$
The vacuum permittivity is a physical constant that appears in equations involving electromagnetic fields. It has the following experimentally determined value:
$$ \varepsilon_0 ~\approx~ 8.854 \, 187 \, 8128 ~\cdot~ 10^{-12} \, \frac{\mathrm{As}}{\mathrm{Vm}} $$