Electric Force in the y-z plane

[StephenDoty]

Two positive charges q0 at the origin and q1 at (0,d1,0) and one negative charge -q2 at (0,d2,0). A positively charged particle, q3 at (0,d2,d2) is added. What is the net force on particle 0 due to particle 3? (See Picture CPartD)

F on 0 due to 3=kq0q3/r^2 where r would be the distance from q0 at the origin to q3 at (0,d2,d2). Thus, r= $$\sqrt{0^2 + d2^2 + d2^2}$$. Thus F on 0 due to 3 = kq0q3/(d2^2 + d2^2) = kq0q3/2d2^2.

And for the vector components: F= -Fcos($$\theta$$) - Fsin($$\theta$$) since q3 repels q0 the F vector would be in the negative y direction and the negative z direction
and since the y magnitude and the z magnitude are equal theta would equal 45 degrees. Making sin(45) and cos 45 = $$\sqrt{2}$$/2

So F= -kq0q3/2d^2 *$$\sqrt{2}$$ /2 $$\hat{y}$$- kq0q3/2d^2 * $$\sqrt{2}$$/2 $$\hat{z}$$

F= -$$\sqrt{2}$$kq0q3/4d2^2 $$\hat{y}$$ - $$\sqrt{2}$$kq0q3/4d2^2 $$\hat{z}$$

Did I do this right? Is this what I type in Mastering physics?

Thank you for your help.
Stephen

 

[rl.bhat]

Yes. You are right.