$|\vec{F_{13}}|=k.\dfrac{q_{1}²}{AC²}$
$|\vec{F_{23}}|=k.\dfrac{q_{1}²}{BC²}$
a, $AC=4cm=0,04m$; $BC=2cm=0,02m$
$⇒\vec{F_{13}}↑↑\vec{F_{23}}$
$⇒|\vec{F_{3}}|=|k.\dfrac{q_{1}²}{AC²}+k.\dfrac{q_{1}²}{BC²}|=|9.10^{9}.\dfrac{(8.10^{-8})²}{0,04²}+9.10^{9}.\dfrac{8.10^{-8}}{0,02²}|=0,18N$
b, $AC=4cm=0,04m$; $BC=10cm=0,1m$
$⇒\vec{F_{13}}↑↓\vec{F_{23}}$
$⇒|\vec{F_{3}}|=|k.\dfrac{q_{1}²}{AC²}-k.\dfrac{q_{1}²}{BC²}|=|9.10^{9}.\dfrac{(8.10^{-8})²}{0,04²}-9.10^{9}.\dfrac{8.10^{-8}}{0,1²}|=0,03024N$
c, $AC=BC=5cm=0,05m$
$ΔABC$ là $Δ$ cân
$\vec{F_{13}}=\vec{F_{23}}$
Gọi $\alpha$ là góc ngoài tại đỉnh $C$
$⇒\alpha=2.\widehat{CAB}$
$⇒cos(\dfrac{\alpha}{2})=cos(\widehat{CAB})=\dfrac{3}{5}$
$⇒|\vec{F}|=2.F_{13}.cos(\dfrac{\alpha}{2})=2.k.\dfrac{q_{1}²}{AC²}.\dfrac{3}{5}=2.9.10^{9}.\dfrac{(8.10^{-8})²}{0,05²}.\dfrac{3}{5}=0,027648N$
