$AB=AC+BC\Rightarrow A.B.C$ thẳng hàng và $C$ nằm giữa $AB$
Ta có: $F_{BA}=F_{AB}=k\frac{q_1q_2}{AB^2}=9.10^9\frac{10^{-7}.5.10^{-8}}{0,05^2}=0,018 (N)$ $F_{AC}=F_{CA}=k\frac{q_1q_3}{AC^2}=0,0225(N)$
$F_{CB}=F_{BC}=k\frac{q_2q_3}{BC^2}=0,18 (N)$
Vậy $F_A=F_{CA}+F_{BA}=0,0405 (N)$
$F_B=F_{CB}+F_{AB}=0,198 (N)$
$F_C=F_{BC}-F_{AC}=0,1575 (N)$