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Homework Help: Electric Force in the y-z plane

  1. Aug 30, 2008 #1
    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= [tex]\sqrt{0^2 + d2^2 + d2^2}[/tex]. Thus F on 0 due to 3 = kq0q3/(d2^2 + d2^2) = kq0q3/2d2^2.

    And for the vector components: F= -Fcos([tex]\theta[/tex]) - Fsin([tex]\theta[/tex]) 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 = [tex]\sqrt{2}[/tex]/2

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

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

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

    Thank you for your help.
    Stephen

    attachment.php?attachmentid=15211&d=1219955156.jpg
    1. The problem statement, all variables and given/known data



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  2. jcsd
  3. Aug 30, 2008 #2

    rl.bhat

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    Yes. You are right.
     
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