Đáp án: `F≈7,655.10^{-4}N`
Giải:
`BC=\sqrt{AB^2+AC^2}=\sqrt{4^2+3^2}=5 \ (cm)`
Ta có:
`F_{13}=k\frac{|q_1q_3|}{AC^2}=9.10^9.\frac{|4.10^{-8}.2.10^{-9}|}{0,03^2}=8.10^{-4} \ (N)`
`F_{23}=k\frac{|q_2q_3|}{BC^2}=9.10^9.\frac{|-12,5.10^{-8}.2.10^{-9}|}{0,05^2}=9.10^{-4} \ (N)`
`cos\alpha=cos(\vec{F_{13}};\vec{F_{23}})=-cosACB=-\frac{AC}{BC}=-\frac{3}{5}`
Lực điện tác dụng lên q3:
`\vec{F}=\vec{F_{13}}+\vec{F_{23}}`
→ `F=\sqrt{F_{13}^2+F_{23}^2+2F_1F_2cos\alpha}`
`F=\sqrt{(8.10^{-4})^2+(9.10^{-4})^2-2.8.10^{-4}.9.10^{-4}.\frac{3}{5}}≈7,655.10^{-4} \ (N)`
