Answer to Question #120968 in Electric Circuits for Faithh

Question

Two bodies of equal charges each of 1C are separated in air by a distance of 1km. find the magnitude of the electric force between them.

Answer 1

According to the Coulomb's law (https://en.wikipedia.org/wiki/Coulomb%27s_law) the force can be calculated as

$F = k\dfrac{q_1q_2}{\varepsilon r^2},$ where $q_1$ and $q_2$ are the charges, $r$ is the distance between charges, $k \approx 9\cdot10^9\,\mathrm{N\,m^2\,C^{-2}}$ and $\varepsilon$ is the permittivity, for air $\varepsilon = 1.000594 \approx 1.$

Therefore,

$F = 9\cdot10^9\,\mathrm{N\,m^2\,C^{-2}}\cdot\dfrac{1\,\mathrm{C}\cdot1\,\mathrm{C}}{(1000\,\mathrm{m})^2} = 9\cdot10^3\,\mathrm{N}.$