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