`\sin \alpha = \frac{r/2}{L} = \frac{0,15 / 2}{0,6} = \frac{0,075}{0,6} = 0,125`

`\cos \alpha = \sqrt{1 - \sin^2 \alpha} = \sqrt{1 - 0,125^2} \approx 0,992156`

`\tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{0,125}{0,992156} \approx 0,126`

$F_e = P \cdot \tan \alpha$

`k \frac{q^2}{r^2} = mg \cdot \tan \alpha`

`q^2 = \frac{mg \cdot \tan \alpha \cdot r^2}{k}`

`q^2 = \frac{(2,0 \times 10^{-3}) \cdot 9,8 \cdot 0,126 \cdot (0,15)^2}{9 \times 10^9}`

`q^2 \approx \frac{5,5566 \times 10^{-5}}{9 \times 10^9} \approx 6,174 \times 10^{-15}`

`q \approx \sqrt{6,174 \times 10^{-15}} \approx 7,857 \times 10^{-8} \text{ C}`

Vậy đáp số cần tìm là 7,86