Two test charges are located in the x–y plane. If q1 = -2.75 nC and is located at x = 0.00 m, y = 0.800 m and the second test charge has magnitude of q2 = 3.20 nC and is located at x = 1.00 m, y = 0.400 m,
calculate the x and y components, Ex and Ey, of the electric field, , in component form at the origin, (0,0). The Coulomb Force constant is 1/(4*pi*ε) = 8.99 × 10^9 N·m^2/C^2
q1 = -2.75 nC; x = 0.00 m, y = 0.800 m
q2 = 3.20 nC; x = 1.00 m, y = 0.400 m
The Attempt at a Solution
So what I did was take the q1 and q2 values (given in nC) and convert them to C.
Next I took found the distance from the origin of the two points, solved using the E=kq/r^2 for each point where k=8.99e+9.
I found the components of the E1 (has only j component) and E2 (by solving the angle from the origin using tan^-1(y-distance/x-distance) and multiplying that by the E2 value)
then I added the two vectors together but still do not get the right answer.
q1 = -2.75e-9 C
q2 = 3.20e-9 C
r1 = .8
r2 = sqrt(1^2 + .4^2) = 1.077
theta = tan^-1(.4/1) = 21.8 deg
I solved the equation
so E1 = -38.6 j (N/C)
and |E2| = 24.8 N/C => separate into components
E2 = 23.03 i + 9.21 j (N/C)
thus Ex = 23.03 N/C
Ey = -15.6 N/C