Giải thích các bước giải:

`q_{1}=80(nC)=8.10^{-8}(C)`

`q_{2}=120(nC)=12.10^{-8}(C)`

`q_{3}=40(nC)=4.10^{-8}(C)`

c) 

`F_{13}=k.\frac{|q_{1}q_{3}|}{r_{13}^{2}}``=9.10^{9}.\frac{|8.10^{-8}.4.10^{-8}|}{0,12^{2}}``=2.10^{-3}(N)`

`F_{23}=k.\frac{|q_{2}q_{3}|}{r_{23}^{2}}``=9.10^{9}.\frac{|12.10^{-8}.4.10^{-8}|}{0,16^{2}}``=1,6875.10^{-3}(N)`

Vì `\vec{F_{13}}`$\bot$`\vec{F_{23}}` nên:

`F=\sqrt{F_{13}^{2}+F_{23}^{2}}``=\sqrt{(2.10^{-3})^{2}+(1,6875.10^{-3})^{2}}``≈2,62.10^{-3}(N)`

d)

`r=20(cm)=0,2(m)`

`F_{13}=k.\frac{|q_{1}q_{3}|}{r^{2}}``=9.10^{9}.\frac{|8.10^{-8}.4.10^{-8}|}{0,2^{2}}``=7,2.10^{-4}(N)`

`F_{23}=k.\frac{|q_{2}q_{3}|}{r^{2}}``=9.10^{9}.\frac{|12.10^{-8}.4.10^{-8}|}{0,2^{2}}``=1,08.10^{-3}(N)`

Lực điện tổng hợp tác dụng lên điện tích `q_{3}` là:

`F=\sqrt{F_{13}^{2}}+F_{23}^{2}+2F_{13}.F_{23}.cos\alpha``=\sqrt{(7,2.10^{-4})^{2}+(1,08.10^{-3})^{2}+2.7,2.10^{-4}.1,08.10^{-3}.cos60°}``=1,57.10^{-3}(N)`

`q_1=80nC=8.10^{-8}C; q_2=120nC=12.10^{-8}C; q_3=40nC=4.10^{-8}C`

c) $E_{13}=k\dfrac{|q_1|}{r_{13}^2}=9.10^{9}.\dfrac{8.10^{-8}}{0,12^2}=50000 \ (V/m)$

$E_{23}=k\dfrac{|q_2|}{r_{23}^2}=9.10^{9}.\dfrac{12.10^{-8}}{0,16^2}=42187,5 \ (V/m)$

$E_3=\sqrt{E^{2}_{13}+E^{2}_{23}}=\sqrt{50000^{2}+42187,5^{2}}=65420 \ (V/m)$

`F_3=|q_3|E_3=4.10^{-8}.65420=2,6168.10^{-3} \ (N)`

d) $E_{13}=k\dfrac{|q_1|}{r_{13}^2}=9.10^{9}.\dfrac{8.10^{-8}}{0,2^2}=1800 \ (V/m)$

$E_{23}=k\dfrac{|q_2|}{r_{23}^2}=9.10^{9}.\dfrac{12.10^{-8}}{0,2^2}=2700 \ (V/m)$

$E_3=\sqrt{E_{13}^2+E_{23}^2+2E_{13}E_{23}\cos60^o}=3923 \ (V/m)$

`F_3=|q_3|E_3=4.10^{-8}.3923=1,5692.10^{-4} \ (N)`