Câu 1:
Lực điện q2, q3 tác dụng lên q1 là:
\(\begin{array}{l}
{F_3} = k\dfrac{{{q_1}{q_3}}}{{A{C^2}}} = 150000{q_3}\\
{F_2} = k\dfrac{{{q_2}{q_1}}}{{A{B^2}}} = 16,2N
\end{array}\)
Ta có:
\({F_2} + {F_3} = 14,2 \Rightarrow {q_3} = - \dfrac{1}{{75000}}C\)
Câu 2:
Khi chưa nối:
\(\left| {{q_1}{q_2}} \right| = \dfrac{{{F_1}r}}{k} = {6.10^{ - 12}}\)
Sau khi nối:
\(\begin{array}{l}
{\left( {\dfrac{{{q_1} + {q_2}}}{2}} \right)^2} = \dfrac{{{F_2}r}}{k} = {2.10^{ - 12}}\\
\Rightarrow \left[ \begin{array}{l}
{q_1} + {q_2} = {2,83.10^{ - 6}}\\
{q_1} + {q_2} = - {2,83.10^{ - 6}}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
{q_1} = {4,24.10^{ - 6}}\\
{q_2} = - {1,41.10^{ - 6}}
\end{array} \right.\\
\left\{ \begin{array}{l}
{q_1} = - {4,24.10^{ - 6}}\\
{q_2} = + {1,41.10^{ - 6}}
\end{array} \right.
\end{array} \right.\\
{F_2} + {F_3} = 14,2 \Rightarrow {q_3} = -
\end{array}\)