If two equal charges each of\({\rm{1 C}}\)each are separated in air by a distance of\({\rm{1 km}}\), what is the magnitude of the force acting between them? You will see that even at a distance as large as\({\rm{1 km}}\), the repulsive force is substantial because\({\rm{1 C}}\)is a very significant amount of charge.

The electrostatic force between two point charges is directly proportional to the product of the magnitude of charges.

\(F \propto {q_1}{q_2}\) ….. (1)

The electrostatic force is inversely proportional to the distance of separation between them.

\(F \propto \frac{1}{{{r^2}}}\) ….. (2)

From equation (1) and (2), you have

\(F \propto \frac{{{q_1}{q_2}}}{{{r^2}}}\)

\(F = \frac{{K{q_1}{q_2}}}{{{r^2}}}\) ….. (3)

The magnitude of the force between two charges can be calculated using equation (3).

\(F = \frac{{K{q_1}{q_2}}}{{{r^2}}}\)

Here, \(F\) is the electrostatic force, \(K\) is the electrostatic force constant, \({q_1}\)is the charge on the first body, \({q_2}\) is the charge on the second body, and \(r\) is the separation between the charges.

Consider the known data as below.

The separation between the charges, \(r = 1.0{\rm{ }}km\)

The electrostatic force constant, \(k=9\times 10^{9} N.m^{2}/C^{2}\)

The charge on the first body, \({q_1} = 1.0{\rm{ }}C\)

The charge on the second body, \({q_2} = 1.0{\rm{ }}C\)

Substituting all known values into equation (3).

\(\begin{aligned} F &= \frac{{\left( {9 \times {{10}^9}{\rm{ }}N \times {m^2}/{C^2}} \right) \times \left( {1{\rm{ }}C} \right) \times \left( {1{\rm{ }}C} \right)}}{{{{\left( {1.0{\rm{ }}km} \right)}^2}}}\\ &= \frac{{\left( {9 \times {{10}^9}{\rm{ }}N \times {m^2}/{C^2}} \right) \times \left( {1{\rm{ }}C} \right) \times \left( {1{\rm{ }}C} \right)}}{{{{\left( {\left( {1.0{\rm{ }}km} \right) \times \left( {\frac{{1000{\rm{ }}m}}{{1{\rm{ }}km}}} \right)} \right)}^2}}}\\ &= 9 \times {10^3}{\rm{ }}N\end{aligned}\)

Hence, the magnitude of force acting between two equal charges is \(9 \times {10^3}{\rm{ }}N\).