a, $C$ là trung điểm của $AB$
$F=|F_{13}-F_{23}|=|k.\dfrac{q_{1}.q_{3}}{AC²}-k.\dfrac{q_{2}.q_{3}}{BC²}|$
$=|9.10^{9}.\dfrac{10^{-7}.4.10^{-7}}{0,05²}-9.10^{9}.\dfrac{5.10^{-7}.4.10^{-7}}{0,05²}|$
$=0,576N$
c, $ΔABC$ vuông tại $C$
$F=\sqrt{F_{13}²+F_{23}²}$
$=\sqrt{(k.\dfrac{q_{1}.q_{3}}{AC²})²+(k.\dfrac{q_{2}.q_{3}}{BC²})²}$
$=\sqrt{(9.10^{9}.\dfrac{10^{-7}.4.10^{-7}}{0,06²})²+(9.10^{9}.\dfrac{5.10^{-7}.4.10^{-7}}{0,08²})²}$
$=0,1025N$
d, $ΔABC$ đều
$F=\sqrt{F_{13}²+F_{23}²+2.F_{13}.F_{23}.cos(60)}$
$=\sqrt{(k.\dfrac{q_{1}.q_{3}}{AB²})²+(k.\dfrac{q_{2}.q_{3}}{AB²})²+2.k.\dfrac{q_{1}.q_{3}}{AB²}.k.\dfrac{q_{2}.q_{3}}{AB²}.cos(60)}$
$=\sqrt{(9.10^{9}.\dfrac{10^{-7}.4.10^{-7}}{0,1²})²+(9.10^{9}.\dfrac{5.10^{-7}.4.10^{-7}}{0,1²})²+2.9.10^{9}.\dfrac{10^{-7}.4.10^{-7}}{0,1²}.9.10^{9}.\dfrac{5.10^{-7}.4.10^{-7}}{0,1²}}$
$≈0,2N$