a, N là trung điểm \(NA=NB=0,1\left(m\right)\)
\(F_{13}=k\dfrac{\left|q_1q_3\right|}{NA^2}=3,6\left(N\right)\)
\(F_{23}=k\dfrac{\left|q_2q_3\right|}{NB^2}=7,2\left(N\right)\)
\(\overrightarrow{F}=\overrightarrow{F_{13}}+\overrightarrow{F_{23}}\Rightarrow F=F_{13}+F_{23}=10,8\left(N\right)\)
b, MA=0,1 MB=0,3
\(F_{13}=k\dfrac{\left|q_1q_3\right|}{MA^2}=3,6\left(N\right)\)
\(F_{23}=k\dfrac{\left|q_2q_3\right|}{MB^2}=0,8\left(N\right)\)
\(\Rightarrow F=\left|F_{23}-F_{13}\right|=2,8\left(N\right)\)
c, ta thấy \(12^2+16^2=20^2\)
góc \(\alpha\) giữa hai q3 vs 1 2 là 90 độ
\(F_{13}=k\dfrac{\left|q_1q_3\right|}{PA^2}=2,5\left(N\right)\)
\(F_{23}=k\dfrac{\left|q_2q_3\right|}{PB^2}=5\left(N\right)\)
\(\Rightarrow F=\sqrt{F_{13}^2+F_{23}^2}=\dfrac{5\sqrt{5}}{2}\left(N\right)\)
