Đáp án:
$\left\{ \begin{align}
& {{q}_{1}}=\pm 6,{{14.10}^{-9}}C \\
& {{q}_{2}}=\mp 4,{{8.10}^{-10}}C \\
\end{align} \right.$
Giải thích các bước giải:
Lực hút: $F=2,{{7.10}^{-6}}N$
=> 2 điện tích trái dấu
$\begin{align}
& F=2,{{7.10}^{-6}}N=k.\dfrac{\left| {{q}_{1}}.{{q}_{2}} \right|}{{{r}^{2}}} \\
& \Rightarrow \left| {{q}_{1}}{{q}_{2}} \right|=\dfrac{2,{{7.10}^{-6}}.0,{{1}^{2}}}{{{9.10}^{9}}}={{3.10}^{-18}} \\
& \Rightarrow {{q}_{1}}.{{q}_{2}}=-{{3.10}^{-18}}{{C}^{2}}(1) \\
\end{align}$
sau khi cho chúng tiếp xúc nhau :
$\begin{align}
& q{{'}_{1}}=q{{'}_{2}}=\frac{{{q}_{1}}+{{q}_{2}}}{2} \\
& F'=k.\dfrac{\left| q{{'}_{1}}.q{{'}_{2}} \right|}{0,{{1}^{2}}}=7,{{2.10}^{-6}} \\
& \Leftrightarrow \dfrac{{{q}_{1}}+{{q}_{2}}}{2}=\pm \sqrt{\dfrac{0,{{1}^{2}}.7,{{2.10}^{-6}}}{{{9.10}^{9}}}}=\pm 2\sqrt{2}{{.10}^{-9}} \\
& \Rightarrow {{q}_{1}}+{{q}_{2}}=\pm 4\sqrt{2}{{.10}^{-9}}C(2) \\
\end{align}$
Từ (1) và (2) ta có:
$\left\{ \begin{align}
& {{q}_{1}}.{{q}_{2}}=-{{3.10}^{-18}} \\
& {{q}_{1}}+{{q}_{2}}=\pm 4\sqrt{2}{{.10}^{-9}}C \\
\end{align} \right.$
$\Rightarrow \left\{ \begin{align}
& {{q}_{1}}=\pm 6,{{14.10}^{-9}}C \\
& {{q}_{2}}=\mp 4,{{8.10}^{-10}}C \\
\end{align} \right.$
