Answer
(a) $8.5\times 10^{-7}C$
(b) $2.5 \times 10^{-6}C;-8.4\times 10^{-7}C$
Work Step by Step
(a) We know that
$q=\sqrt{\frac{F_2d^2}{K}}$
We plug in the known values to obtain:
$q=\sqrt{\frac{(0.032N)(0.45m)^2}{8.99\times 10^9Nm^2/C^2}}$
$q=8.5\times 10^{-7}C$
(b) We know that
$q_1=\frac{d}{\sqrt K}(\sqrt{F_2}+\sqrt{F_1+F_2})$
We plug in the known values to obtain:
$q_1=\frac{0.45m}{\sqrt{8.99\times 10^9Nm^2/C^2}}(\sqrt{0.032N}+\sqrt{0.032N+0.095N})$
$\implies q_1=2.5\times 10^{-6}C$
The charge on the other sphere can be calculated as
$q_2=\frac{2d}{\sqrt{K}}\sqrt{F_2}-\frac{d}{\sqrt{K}}(\sqrt{F_2}+\sqrt{F_1+F_2})$
We plug in the known values to obtain:
$q_2=\frac{0.45m}{\sqrt{8.99\times 10^9Nm^2/C^2}}(\sqrt{0.032N}-\sqrt{0.095N+0.032N})$
$\implies q_2=-8.4\times 10^{-7}C$
