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!

Find charge due to two particles

  1. Aug 30, 2008 #1
    Suppose that the magnitude of the charge on the yellow sphere is determined to be 2q and at (d2cos(theta), -d2cos(theta). Calculate the charge q_red on the red sphere at (d1,0) with a positive blue sphere,q, at the origin and whose net force,F, is in the direction of -y. The yellow sphere is negative and the red sphere is positive. ( See picture bellow)
    Express your answer in terms of q, d1,d2, and theta.

    Well the Fx on blue due to the yellow sphere= k(2q)(q)/(d2cos(theta))^2
    and the Fx on blue due to the red sphere = - k(q_red)(q)/(d1)^2

    Now what? I am unsure as to what to do now. Any help would be appreciated.

    Thank you.

  2. jcsd
  3. Aug 30, 2008 #2
    Would you set them equal to each other?
  4. Aug 30, 2008 #3
    I would? because of the fact that there is no net force along the x axis. Hence all x components of the force must cancel out to equal 0. Do you not get the right answer by doing that?
  5. Aug 31, 2008 #4
    I set them equal to each other and solved for q_red and got:

    q_red= 2qd1^2/(d2^2)(cos(theta)^2)

    Did I set up the original equations correctly? And did I find q_red correctly in terms of q, d1,d2, and theta?

    Thank you. I really appreciate the help.

  6. Aug 31, 2008 #5
    I have to turn this in soon so any help would be appreciated.

    thank you.

  7. Aug 31, 2008 #6
    I don't see anything wrong with it. Does anyone else see anything wrong with it?
  8. Sep 1, 2008 #7
    Does everyone agree with nova-ex? Are my equations correct?
  9. Sep 1, 2008 #8
    Not to be a bother but I have to turn this in tomorrow. So if you can please tell me whether or not I set the problem up correctly and got the correct formula for q_red: q_red= 2qd1^2/[(d2cos(theta))^2]?

    Thanks I would really appreciate it.

  10. Sep 2, 2008 #9
    Any opinions????
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?