cạnh bằng 0,03m
\(q_1=18.10^{-8}C\)
\(q_3=-36.10^{-8}C\)
a)\(\overrightarrow{F}=\overrightarrow{F_{13}}+\overrightarrow{F_{23}}\)
\(\Leftrightarrow F=\sqrt{F^2_{13}+F_{23}^2+2.F_1.F_2.cos\left(\frac{45}{2}\right)}\) (4)
F13=\(\frac{k.\left|q_1.q_3\right|}{AC^2}\) (1)
\(F_{23}=\frac{k.\left|q_2.q_3\right|}{BC^2}\) (2)
AC=\(\frac{BC}{cos45^0}=\frac{3}{50\sqrt{2}}m\) (3)
từ (1),(2), (3), (4)
\(\Rightarrow F\approx\)0,95541N
b)
\(F_{34}=\frac{k.\left|q_3.q_4\right|}{DC^2}\)
\(F_{14}=\frac{k.\left|q_1.q_4\right|}{AD^2}\)
\(F_{24}=\frac{k.\left|q_2.q_4\right|}{BD^2}\)
BD=AC
\(\overrightarrow{F'}=\overrightarrow{F_{14}}+\overrightarrow{F_{34}}+\overrightarrow{F_{24}}\)
cho: \(\overrightarrow{F_{34}}+\overrightarrow{F_{14}}=\overrightarrow{F''}\)
\(\Leftrightarrow F''=\sqrt{F_{34}^2+F_{14}^2}\)\(\approx\)\(0,16099N\)
ta có F'=\(\sqrt{F''^2+F_{24}^2}\)\(\approx\)0,176356N
