## College Physics (4th Edition)

$Q = 2.0\times 10^{-16}~C$
We can find the angle between the vertical and each thread: $tan~(\theta/2) = \frac{0.01~m}{0.98~m}$ $\frac{\theta}{2} = tan^{-1}(\frac{0.01~cm}{0.98~m})$ $\frac{\theta}{2} = 0.585^{\circ}$ The vertical component of the tension in each thread is equal in magnitude to a ball's weight: $T ~cos \frac{\theta}{2}= mg$ $T= \frac{mg}{cos \frac{\theta}{2}}$ The horizontal component of the tension in each thread is equal in magnitude to the electric force due to the two spheres. We can find the charge $Q$: $\frac{kQ^2}{d^2} = T~sin\frac{\theta}{2}$ $\frac{kQ^2}{d^2} = \left(\frac{mg}{cos \frac{\theta}{2}}\right)~sin\frac{\theta}{2}$ $\frac{kQ^2}{d^2} = mg~tan\frac{\theta}{2}$ $Q^2 = \frac{d^2~mg~tan\frac{\theta}{2}}{k}$ $Q = \sqrt{\frac{d^2~mg~tan\frac{\theta}{2}}{k}}$ $Q = \sqrt{\frac{(0.020~m)^2~(9.0\times 10^{-18}~kg)(9.80~m/s^2)~tan~0.585^{\circ}}{9.0\times 10^9~N~m^2/C^2}}$ $Q = 2.0\times 10^{-16}~C$