1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    [​IMG]
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 30, 2008 #2

    rl.bhat

    User Avatar
    Homework Helper

    Yes. You are right.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?