(a) Two insulated charged copper spheres A and B have their centers separated by a distance of 50 cm.

Solution:

(a) Charge on sphere $A, q_{A}=$ 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,

$F=\frac{q_{\mathrm{A}} q_{\mathrm{B}}}{4 \pi \in_{0} r^{2}}$

Where,

$\epsilon_{0}=$ Free space permittivity

$\therefore F=\frac{9 \times 10^{9} \times\left(6.5 \times 10^{-7}\right)^{2}}{(0.5)^{2}}$

$=1.52 \times 10^{-2} \mathrm{~N}$

Therefore, the force between the two spheres is $1.52 \times 10^{-2} \mathrm{~N}$.

(b) After doubling the charge, charge on sphere $A, q_{A}=$ Charge on sphere $B, q_{B}=2 \times 6.5 \times 10^{-7} C=1.3 \times 10^{-6} C$ The distance between the spheres is halved.

$\therefore r=\frac{0.5}{2}=0.25 \mathrm{~m}$

Force of repulsion between the two spheres,

$F=\frac{q_{\AA} q_{\mathrm{B}}}{4 \pi \epsilon_{0} r^{2}}$

$=\frac{9 \times 10^{9} \times 1.3 \times 10^{-6} \times 1.3 \times 10^{-6}}{(0.25)^{2}}$

$=16 \times 1.52 \times 10^{-2}$

= 0.243 N

Therefore, the force between the two spheres is 0.243 N.