Đáp án:
$\left| q \right| = 1,{67.10^{ - 7}}C$
Giải thích các bước giải:
Hai quả cầu cân bằng
$\begin{array}{l}
\overrightarrow P + \overrightarrow F + \overrightarrow T = \overrightarrow 0 \\
\Rightarrow \overrightarrow P + \overrightarrow F = - \overrightarrow T \\
\tan \alpha = \frac{F}{P} = \frac{{\frac{r}{2}}}{{\sqrt {{l^2} - {{\left( {\frac{r}{2}} \right)}^2}} }}\\
\Rightarrow \frac{{k\frac{{{q^2}}}{{{r^2}}}}}{{mg}} = \frac{{\frac{r}{2}}}{{\sqrt {{l^2} - {{\left( {\frac{r}{2}} \right)}^2}} }} \Rightarrow \frac{{{{9.10}^9}.\frac{{{q^2}}}{{0,{{05}^2}}}}}{{0,2.10}} = \frac{{\frac{{0,05}}{2}}}{{\sqrt {0,{5^2} - {{\left( {\frac{{0,05}}{2}} \right)}^2}} }}\\
\Rightarrow \left| q \right| = 1,{67.10^{ - 7}}C
\end{array}$