#### Answer

(a) The force on object A is 0.45 N
(b) $q_A = 1.0\times 10^{-6}$
$q_B = 5.0\times 10^{-7}~C$

#### Work Step by Step

(a) Since both objects experience the same magnitude of electric force, the force on object A is also 0.45 N
(b) Let $q$ be the charge on object B. Then the charge on object A is $2q$. We can find the charge $q$.
$F = \frac{k~q_A~q_B}{r^2}$
$F = \frac{k~(2q)~q}{r^2}$
$2q^2 = \frac{F~r^2}{k}$
$q^2 = \frac{F~r^2}{2k}$
$q = \sqrt{\frac{F}{2k}}~r$
$q = \sqrt{\frac{0.45~N}{(2)(9.0\times 10^9~N~m^2/C^2)}}~(0.10~m)$
$q = 5.0\times 10^{-7}~C$
Since $q_A = 2q$, then $q_A = 1.0\times 10^{-6}$ and $q_B = 5.0\times 10^{-7}~C$.