Hello everyone, i'm having problems finding The resultant intensity at point b, i know its the vector sum of E1 and E2. q1 = -6x10^-9; q2 = 6x10^9 they are 12cm apaprt. Determine the electrict feild at A and B, i already figured out A. This is the free body diagram. ^ E2 ..\ ...\ ...(B) ....| ....| ....v E1 (q1)----------------------------------(q2) The angle that E2 forms is a 37 degree angle. E1 is 9cm from q1. Point B is 15 cm from q2 diagnoally. Okay I figued the feild intensity at B due to q1 is directed downward and is equal to E1 = kq1/r^2 = -[(9x10^9)(6x10^-9)]/(9x10^-2m)^2; E1 = -.667x10^4 N/C; To find E2 E2 = [(9x10^9)(6x10^-9)]/(15x10^-2m)^2; E2 = .240x10^4N/C 37 degree N of W. Now this is where I get lost... I need to find the X and Y component of E2 which seems easy enough....but i f it up anyways. So i figure, you have a triangle that looks like this: ^ |E2y | 37 degrees <------ E2x E2y = E2*sin37; E2x = E2*cos37; E2x = -.192x10^4 N/C E2y = .144x10^4 N/C They did the following: Ex = -E2x = -(.240x10^4 N/C)*cos37; Ex = -.144x10^4 N/C; //is this a misprint? Thanks.