Chapter 16 - Electric Charge and Electric Field - Problems - Page 469: 16

$Q=4.68\times10^{-6}C$

vertical direction $F_{Ty}=mg=(0.021kg)(9.8\frac{m}{s^2})=0.2058N$ $F_{Tx}=k\frac{Q_1Q_2}{r^2}=(9.0\times10^9\frac{Nm^2}{C^2})\frac{(0.5Q)^2}{(0.78cos(26^o))^2m}$ $=4.58\times10^9Q^2$ $tan(26.0^o)=\frac{F_{Tx}}{F_{Ty}}$ $F_{Tx}=0.4877F_{Ty}$ $4.58\times10^9Q^2=(0.4877)(0.2058)$ $Q=\sqrt{\frac{0.1004}{4.58\times10^9}}=4.68\times10^{-6}C$