Đáp án:

$q = {8.10^{ - 8}}C$ 

Giải thích các bước giải:

Ta có:

$\begin{array}{l}
\tan \dfrac{{{\alpha _1}}}{2} = \dfrac{{{F_d}}}{P}\\
 \Rightarrow {F_d} = P.\tan \dfrac{{{\alpha _1}}}{2}\\
 \Leftrightarrow \dfrac{{k{q_1}{q_2}}}{{{r^2}}} = P.\tan \dfrac{{{\alpha _1}}}{2}\\
 \Leftrightarrow \dfrac{{k{q_1}{q_2}}}{{{l^2}{{\sin }^2}\dfrac{{{a_1}}}{2}}} = P.\tan \dfrac{{{\alpha _1}}}{2}\\
 \Leftrightarrow {q_1}{q_2} = \dfrac{{P.{l^2}\tan \dfrac{{{\alpha _1}}}{2}.{{\sin }^2}\dfrac{{{a_1}}}{2}}}{k}
\end{array}$

Tương tự khi truyền cho q2 điện tích q ta có:

${q_1}\left( {{q_2} + q} \right) = \dfrac{{P.{l^2}\tan \dfrac{{{\alpha _2}}}{2}.{{\sin }^2}\dfrac{{{a_2}}}{2}}}{k}$

Từ đó suy ra:

$\begin{array}{l}
\dfrac{{{q_2} + q}}{{{q_2}}} = \dfrac{{\tan \dfrac{{{\alpha _2}}}{2}.{{\sin }^2}\dfrac{{{a_2}}}{2}}}{{\tan \dfrac{{{\alpha _1}}}{2}.{{\sin }^2}\dfrac{{{a_1}}}{2}}} = \dfrac{{\tan {{60}^o}{{\sin }^2}{{60}^o}}}{{\tan {{30}^o}.{{\sin }^2}{{30}^o}}} = 9\\
 \Rightarrow q = 8{q_2} = {8.10^{ - 8}}C
\end{array}$