Two electrons are located at two of the vertices of an equilateral triangle,one at each vertex.The length of one side of the triangle is 1.00 nm.What is the magnitude of the net electric field at the third vertex?

Question

Quizplus · Accepted Answer

The net electric field at the third vertex can be found by vector addition of the electric fields due to each of the electrons individually. The electric field due to one electron at a distance of 1.00 nm is given by:
E = kq/r^2 = (9 x 10^9 N*m^2/C^2)(1.6 x 10^-19 C)/(1.00 x 10^-9 m)^2 = 2.304 N/C
Using the law of cosines, the magnitude of the net electric field at the third vertex can be found:
E^2 = (2.304 N/C)^2 + (2.304 N/C)^2 - 2(2.304 N/C)(2.304 N/C)cos(60°)
E = 2.49 x 10^10 N/C, which is closest to choice C.

Two Electrons Are Located at Two of the Vertices of an Equilateral