Đáp án:
$\begin{align}
& a)\varepsilon =1,8 \\
& b)r=10cm \\
\end{align}$
Giải thích các bước giải:
${{q}_{1}}=1,{{3.10}^{-9}}C;{{q}_{2}}=6,{{5.10}^{-9}}C;F$
a) Điện tích mỗi quả sau khi tiếp xúc:
$\begin{align}
& q{{'}_{1}}=q{{'}_{2}}=\dfrac{{{q}_{1}}+{{q}_{2}}}{2} \\
& =\dfrac{1,{{3.10}^{-9}}+6,{{5.10}^{-9}}}{2} \\
& =3,{{9.10}^{-9}}C \\
\end{align}$
ta có: $\begin{align}
& F=k.\dfrac{\left| {{q}_{1}}.{{q}_{2}} \right|}{{{r}^{2}}} \\
& F'=k.\dfrac{\left| q{{'}_{1}}.q{{'}_{2}} \right|}{\varepsilon .{{r}^{2}}} \\
\end{align}$
mà:
$\begin{align}
& F=F' \\
& \Leftrightarrow k.\dfrac{\left| {{q}_{1}}.{{q}_{2}} \right|}{{{r}^{2}}}=k.\dfrac{\left| q{{'}_{1}}.q{{'}_{2}} \right|}{\varepsilon .{{r}^{2}}} \\
& \Leftrightarrow 1,{{3.10}^{-9}}.6,{{5.10}^{-9}}=\dfrac{3,{{9.10}^{-9}}.3,{{9.10}^{-9}}}{\varepsilon } \\
& \Rightarrow \varepsilon =1,8 \\
\end{align}$
b) lực tương tác:
$\begin{align}
& F=k.\dfrac{\left| {{q}_{1}}.{{q}_{2}} \right|}{{{r}^{2}}} \\
& \Leftrightarrow 4,{{5.10}^{-6}}={{9.10}^{9}}.\dfrac{1,{{3.10}^{-9}}.3,{{9.10}^{-9}}}{{{r}^{2}}} \\
& \Rightarrow r=0,1m=10cm \\
\end{align}$
