I'm having trouble with this question. Where am I going wrong? Three charges are at the corners of an equilateral trangle. The bottom left corner has a charge of 2.00e-6 C, bottom right -4.00e-6 C, and top with 7.00e-6 C and separated by 0.500m. Calculate the electric field at the position of the 2.00e-6 C charge due to the 7.00e-6 C and -4.00e-6 C charges. ___________________________ E1 = [(K_e)(q1)(q2)/(r^2)] =[-(8.99e9)(7.00e-6)(2.00e-6) / (.500m)^2] * [-cos(60)i - sin(60)j] = 0.25172 i + 0.8660 j ___________________________ E2 = [(K_e)(q1)(q2)/(r^2)] = [(8.99e9)(2.00e-6)(-4.00e-6) / (.500m)^2] * [cos(0)i] = -0.28768 i ___________________________ (0.25172 i - 0.28768 i) = -0.03596 i So, -0.03596 i and 0.8660 j ___________________________ I know this isn't right, where did I go wrong? Any help is greatly appreciated Thanks.