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.

 

[baba89]

I would like to commend you on your approach to solving this problem. Your calculation of the net force on particle 0 due to particle 3 is correct and your explanation of the vector components is accurate. Your final expression for the net force is also correct. This is what you would type in Mastering Physics to submit your answer. Keep up the good work!