Answer
$\theta = 14.3^{\circ}$
Work Step by Step
The vertical component of the tension in the string is equal in magnitude to the weight of the ball:
$F_T~cos~\theta = mg$
$F_T = \frac{mg}{cos~\theta}$
We can find the angle $\theta$:
$F_T~sin~\theta = q~E$
$\frac{mg~sin~\theta}{cos~\theta} = q~E$
$mg~tan~\theta = q~E$
$tan~\theta = \frac{q~E}{mg}$
$tan~\theta = \frac{(25\times 10^{-9}~C)(2\times 10^5~N/C)}{(0.0020~kg)(9.8~m/s^2)}$
$tan~\theta = 0.2551$
$\theta = tan^{-1}~(0.2551)$
$\theta = 14.3^{\circ}$
