Đáp án:
\(F=0,015N\)
Giải thích các bước giải:
\(\begin{array}{*{35}{l}}
\text{ }\!\!~\!\!\text{ } & {{F}_{13}}=k.\dfrac{\left| {{q}_{1}}{{q}_{2}} \right|}{A{{C}^{2}}}={{9.10}^{9}}.\dfrac{{{2.10}^{-8}}{{.5.10}^{-8}}}{0,{{03}^{2}}}=0,01N \\
\text{ }\!\!~\!\!\text{ } & {{F}_{23}}=k.\dfrac{\left| {{q}_{2}}{{q}_{3}} \right|}{BC}={{9.10}^{9}}.\dfrac{{{4.10}^{-8}}{{.5.10}^{-8}}}{0,{{04}^{2}}}=0,01125N \\
\text{ }\!\!~\!\!\text{ } & {} \\
\end{array}\)
ta có:
\(F=\sqrt{F_{13}^{2}+F_{23}^{2}}=0,015N\)