Đáp án:
Lực điện
\(\begin{array}{l}F = k\dfrac{{{q^2}}}{{{r^2}}}\\ \Rightarrow \left| q \right| = \sqrt {\dfrac{{F.{r^2}}}{k}} = \sqrt {\dfrac{{1,{{6.10}^{ - 4}}.0,{{02}^2}}}{{{{9.10}^9}}}} = \dfrac{8}{3}{.10^{ - 9}}C\end{array}\)
Ta có \(F \sim \dfrac{1}{{{r^2}}} \Rightarrow \dfrac{{{F_2}}}{{{F_1}}} = \dfrac{{r_1^2}}{{r_2^2}} \Leftrightarrow \dfrac{{2,5,{{10}^{ - 4}}}}{{1,{{6.10}^{ - 4}}}} = \dfrac{{0,{{02}^2}}}{{r_2^2}} \Rightarrow {r_2} = 0,016m = 1,6cm\)
