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StephenDoty
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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
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