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## Homework Statement

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

given:

q1 = -2.75 nC; x = 0.00 m, y = 0.800 m

q2 = 3.20 nC; x = 1.00 m, y = 0.400 m

## Homework Equations

E=kq/r^2

## 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.

I Calculated:

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