Problem 19.41

19.41. a)
Once again, we are interested in summing the components to find out an unknown quantity. This time, we will need to combine the two equations. Starting out, we find that
$\sum F_{x} = qE_{x} - T\sin \theta ~=~ 0$
$\sum F_{y} = qE_{y} + T\cos \theta - mg ~=~ 0$
Solving the $\hat{x}$ equation for T, we find
$T = \frac{qE_{x}}{\sin \theta}$
Substituting that into the $\hat{y}$ equation, we find that
$q = \frac{mg}{E_{x}\cot \theta + E_{y}}$
$q = \frac{(.001)(9.8)}{(3 \cot 37{}^{\circ}+ 5) \times 10^{5}}$
$q = 1.09 \times 10^{-8} C$

b)
From part a), we found that
$T = \frac{qE_{x}}{\sin \theta}$.
So, plugging in the numbers, we find that
$T = \frac{q E_{x}}{\sin 37{}^{\circ}}$
$T = 5.43 \times 10^{-3} N$