Find the vector components of the electric force on q1 and the vector's magnitude and direction.

Using the same triangle, find the vector components of the electric force on q1 and the vector's magnitude and direction. (Use the charges given in the Practice It section.)

A) Fx = N
B) Fy = N
C) magnitude=
D) direction= ° counter-clockwise from the +x-axis

Practice it section it is referring to: "Use the worked example above to help you solve this problem. Consider three point charges at the corners of a triangle, as shown in the figure, where q1 = 6.44 10^ -9 C, q2 = 1.59 10^ -9 C, and q3 = 5.15 10^ -8 C."

Image of charges/orientation http://www.webassign.net/sercp8/15-figure-08.gif

Thanks so much!

Look at the picture :

Fa = F12 = kq1*q2/r12^2 = 9 x 10^9 * 6.44 10^ -9 * 1.59 10^ -9/3^2
= 1.02 x 10^-8 N

Fb = F13 = kq1*q3/r13^2 = 9 x 10^9 * 6.44 10^ -9 * 5.15 10^ -8/3^2
= 1.19 x 10^-7N

a) Fx = Fax + Fbx
= 1.02 x 10^-8 cos 90 + 1.19 x 10^-7 cos 37
= 0 + 9.5 x 10^-8 = 9.5 x 10^-8 N

b) Fy = Fay + Fby
= 1.02 x 10^-8 sin 90 + 1.19 x 10^-7 sin 37
= 8.18 x 10^-8 N

c) R = √(Fx^2 + Fy^2)
= √(9.5 x 10^-8^2 + 8.18 x 10^-8^2)
= 1.25 x 10^-7 N

d) tan θ = Fy/Fx
θ = arc tan (8.18 x 10^-8/ 9.5 x 10^-8)
= 40.73 degrees

of course, you're right
I was wrong to see the image, where the sign of q2 is negative it should positive.
thx

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