The electrostatic force on a small sphere of charge $0.4 \; \mu C$ due to another small sphere of charge $-0.8 \; \mu C$ in air is $0.2 \; N$.
$(a)$ What is the distance between the two spheres?
$(b)$ What is the force on the second sphere due to the first?

(A) Electrostatic force on the first sphere,$F = 0.2 \; N$.
Charge on the first sphere,$q_{1} = 0.4 \; \mu C = 0.4 \times 10^{-6} \; C$.
Charge on the second sphere,$q_{2} = -0.8 \; \mu C = -0.8 \times 10^{-6} \; C$.
Electrostatic force between the spheres is given by Coulomb's Law: $F = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{|q_{1} q_{2}|}{r^{2}}$.
Here,$\frac{1}{4 \pi \varepsilon_{0}} = 9 \times 10^{9} \; N \cdot m^{2} \cdot C^{-2}$.
Rearranging for $r^{2}$,we get $r^{2} = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{|q_{1} q_{2}|}{F}$.
$r^{2} = \frac{9 \times 10^{9} \times (0.4 \times 10^{-6}) \times (0.8 \times 10^{-6})}{0.2} = \frac{2.88 \times 10^{-3}}{0.2} = 144 \times 10^{-4} \; m^{2}$.
$r = \sqrt{144 \times 10^{-4}} = 12 \times 10^{-2} = 0.12 \; m$.
The distance between the two spheres is $0.12 \; m$.
$(b)$ According to Newton's third law of motion,the force exerted by the first sphere on the second is equal in magnitude and opposite in direction to the force exerted by the second on the first. Therefore,the force on the second sphere due to the first is $0.2 \; N$.