Thus, the field from one wire is given by: E=2kλRsin(π6)E=kλR Given that the charge Q is evenly distributed along...Thus, the field from one wire is given by: E=2kλRsin(π6)E=kλR Given that the charge Q is evenly distributed along the rod of length L, we can rewrite the charge density as QL, which gives: E=kQRL=kQL√36L=6kQ√3L2 This is the magnitude of the electric field for each side of the triangle.