Formula: Electric (Electrostatic) Potential of a Point Charge Distance Electric charge Relative permittivity
$$V_{\text e} ~=~ \frac{1}{4\pi \varepsilon_0 \, \varepsilon_{\text r}} \, \frac{ \class{red}{Q} }{ r }$$
$$V_{\text e} ~=~ \frac{1}{4\pi \varepsilon_0 \, \varepsilon_{\text r}} \, \frac{ \class{red}{Q} }{ r }$$
$$r ~=~ \frac{1}{4\pi \varepsilon_0 \, \varepsilon_{\text r}} \, \frac{ \class{red}{Q} }{ V_{\text e} }$$
$$\class{red}{Q} ~=~ 4\pi \varepsilon_0 \, \varepsilon_{\text r} \, r \, V_{\text e}$$
Electric Potential
$$ V_{\text e} $$ Unit $$ \frac{\mathrm{J}}{\mathrm{C}} $$
Electric potential at a distance from a charge \( \class{red}{Q} \) which generates this potential. The potential indicates the potential energy that a test charge gains or must release when placed from an infinite distance to a distance \( r \) from the source charge.
Distance
$$ r $$ Unit $$ \mathrm{m} $$
Distance of a test charge from the source charge \( \class{red}{Q} \).
Source charge
$$ \class{red}{Q} $$ Unit $$ \mathrm{C} = \mathrm{As} $$
Electric charge generating potential \( V_{\text e} \).
Relative permittivity
$$ \varepsilon_{\text r} $$ Unit $$ - $$
Relative permittivity describes the medium in which the two charges are located. If the two charges are in vacuum then \( \varepsilon_{\text r} = 1 \). And, for example, if they are in lukewarm water then \( \varepsilon_{\text r} = 80 \). The greater the relative permittivity of the medium, the more this medium weakens the potential.
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:
$$ \varepsilon_0 ~\approx~ 8.854 \, 187 \, 8128 ~\cdot~ 10^{-12} \, \frac{\mathrm{As}}{\mathrm{Vm}} $$