Option 4 : 20 mN

__Concept__:

- Coulomb’s law: When two charge particles of charges q1 and q2 are separated by a distance r from each other then the electrostatic force between them is directly proportional to multiplication of charges of two particles and inversely proportional to square of distance between them.

Force (F) ∝ q1 × q2

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

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

Where K is a constant = 9 × 109 Nm2/C2

__Explanation__:

**Case 1:**

Given that charges are: Q_{1} and Q_{2}

Distance = d

Force (F_{1}) = 20 mN

\(F \propto \frac{{{Q_1}{Q_2}}}{{{d^2}}}\)

**Case 2:**

The charges of both particles are doubled and distance between them is also doubled.

Given that charges are: 2Q_{1} and 2Q_{2}

Distance = 2d

\(F \propto \frac{{2{Q_1} \times 2{Q_2}}}{{{{\left( {2d} \right)}^2}}} = \frac{{{Q_1}{Q_2}}}{{{d^2}}}\)

Therefore, the force remains same and it is equal to 20 mN