## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

$q = +1.78\times 10^{-7}~C$
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 charge $q$: $q~E = F_T~sin~\theta$ $q~E = \frac{mg~sin~\theta}{cos~\theta}$ $q = \frac{mg~tan~\theta}{E}$ $q = \frac{(0.0050~kg)(9.8~m/s^2)~tan~20^{\circ}}{10^5~N/C}$ $q = +1.78\times 10^{-7}~C$