Two small identical balls, each of mass 2.0 g, hang from the same point on silk threads 1.0 m long…

Step 1. The angle. $\sin\theta = \dfrac{d/2}{L} = \dfrac{0.050\ \text{m}}{1.0\ \text{m}} = 0.050$, so $\theta = 2.87^\circ$ and $\tan\theta = 0.0501$.

Step 2. The balance. Vertically, $T\cos\theta = mg$. Horizontally, $T\sin\theta = F$. Divide: $\tan\theta = \dfrac{F}{mg}$. So $F = mg\tan\theta = 2.0 \times 10^{-3}\ \text{kg} \times 9.8\ \text{m/s}^2 \times 0.0501 = 9.8 \times 10^{-4}\ \text{N}$.

Step 3. The charge. $F = \dfrac{kq^2}{d^2}$. We want $q$, so $q = d\sqrt{\dfrac{F}{k}} = 0.10\ \text{m} \times \sqrt{\dfrac{9.8 \times 10^{-4}\ \text{N}}{9.0 \times 10^9\ \text{N m}^2/\text{C}^2}} = 3.3 \times 10^{-8}\ \text{C} = 33\ \text{nC}$.