Answer
$\theta = 4.4^{\circ}$
Work Step by Step
The vertical component of the tension in each thread is equal in magnitude to the weight of one charge:
$F_T~cos~\theta = mg$
$F_T = \frac{mg}{cos~\theta}$
Since $\theta$ is a small angle, then $sin~\theta \approx \theta$ and $tan~\theta \approx \theta$
We can find the angle $\theta$:
$F_T~sin~\theta = \frac{kq^2}{r^2}$
$(\frac{mg}{cos~\theta})~(sin~\theta) = \frac{kq^2}{(2~L~sin~\theta)^2}$
$tan~\theta~sin^2~\theta = \frac{kq^2}{4L^2~mg}$
$\theta^3 = \frac{kq^2}{4L^2~mg}$
$\theta = (\frac{kq^2}{4L^2~mg})^{1/3}$
$\theta = [\frac{(9.0\times 10^9~N~m^2/C^2)(100\times 10^{-9}~C)^2}{(4)(1.0~m)^2~(0.0050~kg)(9.8~m/s^2)}]^{1/3}$
$\theta = 0.077~rad$
$\theta = (0.077~rad)(\frac{180^{\circ}}{\pi~rad})$
$\theta = 4.4^{\circ}$
