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    • https://phys.libretexts.org/Courses/Berea_College/Introductory_Physics%3A_Berea_College/16%3A_Electric_Charges_and_Fields/16.07%3A_Sample_problems_and_solutions
      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.

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