#### Answer

(a) $F = 0.90~N$
(b) $a = 0.90~m/s^2$

#### Work Step by Step

(a) We can find the magnitude of the electric force on each mass:
$F = \frac{1}{4\pi ~\epsilon_0}\frac{q_1q_2}{r^2}$
$F = \frac{1}{(4\pi)(8.85\times 10^{-12}~C^2/N~m^2)}\frac{(10\times 10^{-6}~C)(10\times 10^{-6}~C)}{(1.0~m)^2}$
$F = 0.90~N$
(b) We can find the acceleration.
$a = \frac{F}{m}$
$a = \frac{0.90~N}{1.0~kg}$
$a = 0.90~m/s^2$