Đáp án:
\(\begin{array}{l}
{q_1} = \dfrac{8}{3}{.10^{ - 7}}C\\
{q_2} = - \dfrac{{20}}{3}{.10^{ - 8}}C
\end{array}\)
Giải thích các bước giải:
-Trước khi tiếp xúc:
\(\begin{array}{l}
{F_1} = k\dfrac{{|{q_1}{q_2}|}}{{{r^2}}}\\
\Rightarrow |{q_1}{q_2}| = \dfrac{{{F_1}{r^2}}}{k} = \dfrac{{1,{{6.10}^{ - 2}}.0,{1^2}}}{{{{9.10}^9}}} = \dfrac{{16}}{9}{.10^{ - 14}}\\
{q_1}{q_2} = - \dfrac{{16}}{9}{.10^{ - 14}}(1)
\end{array}\)
-Sau khi tiếp xúc:
\(\begin{array}{l}
{F_2} = k\dfrac{{|{q_1}'.{q_2}'|}}{{{r^2}}}\\
\Rightarrow |{q_1}'.{q_2}'| = \dfrac{{{F_2}{r^2}}}{k} = \dfrac{{{{9.10}^{ - 3}}.0,{1^2}}}{{{{9.10}^9}}} = {10^{ - 14}}
\end{array}\)
Ta có:
\(\begin{array}{l}
{q_1}' = {q_2}' = \dfrac{{{q_1} + {q_2}}}{2}\\
\Rightarrow {q_1}'.{q_2}' = \dfrac{{{{({q_1} + {q_2})}^2}}}{4}\\
\Rightarrow {10^{ - 14}} = \dfrac{{{{({q_1} + {q_2})}^2}}}{4}\\
\Rightarrow {q_1} + {q_2} = {2.10^{ - 7}}(2)\\
(1),(2) \Rightarrow \left\{ \begin{array}{l}
{q_1} = \dfrac{8}{3}{.10^{ - 7}}C\\
{q_2} = - \dfrac{{20}}{3}{.10^{ - 8}}C
\end{array} \right.
\end{array}\)
