Problem 19.39

19.39.
This problem is computationally no different from the ones done back in mechanics. We just have to sum the forces in the two directions. In the $\hat{y}$ direction,
$\sum F_{y} = -mg + T\cos 15{}^{\circ}= 0$
Solving for the tension in the string, we find,
$T = 2.03 \times 10^{-2} N$
Now, we turn out attention to the $\hat{x}$ direction.
$\sum F_{x} = qE + T\sin 15{}^{\circ}= 0$
Solving for q, we find that
$q = \frac{T \sin 15{}^{\circ}}{E}$
Plugging in the numbers, we find that
$q = 5.25 \mu C$.