Thus, the field from one wire is given by: \[\begin{aligned} E&= \frac{2k\lambda}{R}\sin(\frac{\pi}{6})\\ E &= \frac{k\lambda}{R}\end{aligned}\] Given that the charge \(Q\) is evenly distributed along...Thus, the field from one wire is given by: \[\begin{aligned} E&= \frac{2k\lambda}{R}\sin(\frac{\pi}{6})\\ E &= \frac{k\lambda}{R}\end{aligned}\] Given that the charge \(Q\) is evenly distributed along the rod of length \(L\), we can rewrite the charge density as \(\frac{Q}{L}\), which gives: \[\begin{aligned} E &= \frac{k Q}{RL} = \frac{k Q}{\frac{L\sqrt{3}}{6}L} = \frac{6k Q}{\sqrt{3}L^2}\end{aligned}\] This is the magnitude of the electric field for each side of the triangle.