Answer
The negative charge on a sphere is $~~-1.00\times 10^{-6}~C$
Work Step by Step
After the wire is removed, both spheres have a positive charge of $q$
We can find $q$:
$F = \frac{q^2}{4\pi~\epsilon_0~r^2} = 0.0360~N$
$q^2 = (4\pi~\epsilon_0~r^2)(0.0360~N)$
$q = \sqrt{(4\pi~\epsilon_0~r^2)(0.0360~N)}$
$q = \sqrt{(4\pi)~(8.854\times 10^{-12}~F/m)~(0.500~m)^2(0.0360~N)}$
$q = 1.00\times 10^{-6}~C$
Let the initial charges on the spheres be $q_1$ and $q_2$
Note that $q_1+q_2 = 2.00\times 10^{-6}~C$
We can find $q_1$:
$F = \frac{q_1~q_2}{4\pi~\epsilon_0~r^2} = -0.108~N$
$\frac{q_1~(2.00\times 10^{-6}~C-q_1)}{4\pi~\epsilon_0~r^2} = -0.108~N$
$-q_1^2+q_1~(2.00\times 10^{-6}~C) = (-0.108~N)(4\pi~\epsilon_0~r^2)$
$q_1^2-q_1~(2.00\times 10^{-6}~C) -(0.108~N)(4\pi)~(8.854\times 10^{-12}~F/m)~(0.500~m)^2 = 0$
$q_1^2-q_1~(2.00\times 10^{-6}~C) -(3.00\times 10^{-12}~C^2) = 0$
$(q_1-3.00\times 10^{-6}~C)(q_1+1.00\times 10^{-6}~C)= 0$
$q_1 = 3.00\times 10^{-6}~C, -1.00\times 10^{-6}~C$
This result tells us the charge on the positively charged sphere and the negatively charged sphere.
The negative charge on a sphere is $~~-1.00\times 10^{-6}~C$
