$(a)$ Two insulated charged copper spheres $A$ and $B$ have their centres separated by a distance of $50 \; cm$. What is the mutual force of electrostatic repulsion if the charge on each is $6.5 \times 10^{-7} \; C$? The radii of $A$ and $B$ are negligible compared to the distance of separation.
$(b)$ What is the force of repulsion if each sphere is charged double the above amount,and the distance between them is halved?

(A) Charge on sphere $A, q_{A} = 6.5 \times 10^{-7} \; C$
Charge on sphere $B, q_{B} = 6.5 \times 10^{-7} \; C$
Distance between the spheres,$r = 50 \; cm = 0.5 \; m$
Force of repulsion between the two spheres is given by Coulomb's Law:
$F = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q_{A} q_{B}}{r^{2}}$
Where,$\frac{1}{4 \pi \varepsilon_{0}} = 9 \times 10^{9} \; N \cdot m^{2} \cdot C^{-2}$
$F = \frac{9 \times 10^{9} \times (6.5 \times 10^{-7})^{2}}{(0.5)^{2}} = \frac{9 \times 10^{9} \times 42.25 \times 10^{-14}}{0.25} = 1.521 \times 10^{-2} \; N$
$(b)$ New charge on each sphere,$q'_{A} = q'_{B} = 2 \times 6.5 \times 10^{-7} = 1.3 \times 10^{-6} \; C$
New distance between the spheres,$r' = \frac{0.5}{2} = 0.25 \; m$
New force of repulsion,$F' = \frac{9 \times 10^{9} \times (1.3 \times 10^{-6})^{2}}{(0.25)^{2}} = \frac{9 \times 10^{9} \times 1.69 \times 10^{-12}}{0.0625} = 0.24336 \; N \approx 0.243 \; N$