Bài 6:
Ta có:
\(\begin{array}{l}
F = k\dfrac{{\left| {{q_1}.{q_2}} \right|}}{{\varepsilon .{r^2}}}\\
\Rightarrow 0,{5.10^{ - 3}} = {9.10^9}.\dfrac{{\left| {{q_1}.{q_2}} \right|}}{{3.0,{{12}^2}}}\\
\Rightarrow \left| {{q_1}.{q_2}} \right| = 2,{4.10^{ - 15}}\\
\Rightarrow {q_1}.{q_2} = - 2,{4.10^{ - 15}}\\
\Rightarrow {q_1}\left( { - 5,{{10}^{ - 8}} - {q_1}} \right) = - 2,{4.10^{ - 15}}\\
\Rightarrow \left\{ \begin{array}{l}
{q_1} = {3.10^{ - 8}}C\\
{q_2} = - {8.10^{ - 8}}C
\end{array} \right.
\end{array}\)
Bài 7:
Ta có:
\(\begin{array}{l}
F = k\dfrac{{\left| {{q_1}.{q_2}} \right|}}{{\varepsilon .{r^2}}}\\
\Rightarrow 1,8 = {9.10^9}.\dfrac{{\left| {{q_1}.{q_2}} \right|}}{{{1^2}}}\\
\Rightarrow \left| {{q_1}.{q_2}} \right| = {2.10^{ - 10}}\\
\Rightarrow {q_1}.{q_2} = - {2.10^{ - 10}}
\end{array}\)
Mà:
\(\begin{array}{l}
\dfrac{{{q_1} + {q_2}}}{2} = {10^{ - 7}} \Rightarrow {q_1} + {q_2} = {2.10^{ - 7}}\\
\Rightarrow {q_1}\left( {{{2.10}^{ - 7}} - {q_1}} \right) = - {2.10^{ - 10}}\\
\Rightarrow \left\{ \begin{array}{l}
{q_1} = 1,{42.10^{ - 5}}C\\
{q_2} = - 1,{4.10^{ - 5}}C
\end{array} \right.
\end{array}\)
