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Charges are Great Need more assistance!

  1. Feb 2, 2008 #1
    [SOLVED] Charges are Great!! Need more assistance!

    1. The problem statement, all variables and given/known data

    In the figure particle 1 of +q and particle 2 of charge +4q are held at a separation of L=.09 m If particle 3 of charge q_3 is to be placed such that the three particles remain in place when released, what must be (a) the ratio q_3/q and the (b) x and (c) y components?

    Okay, I solved b and c easy enough. There is no y component and x=.03.

    I am having trouble with a. Any hints? I know it must be a similar procedure, that is, I need to use the fact that there is NO NET force....but I can't seem to spot the correct relationship such that I am left with the desired variables (they don't cancel).

  2. jcsd
  3. Feb 2, 2008 #2
    Okay, I looked it up on Cramster (which I absolutely hate doing) and in their solution they assume that [itex]F_{31}=F_{21}[/itex] , but I cannot reason that out.

    Any ideas?
  4. Feb 3, 2008 #3
    Good Morning! Maybe I'll have some luck, now let's see here. . .:smile:
  5. Feb 3, 2008 #4
    For your answer of b you used the fact that the net force on q3 must be 0. now use the fact that the net force on q1 and q2 must be 0.
  6. Feb 3, 2008 #5
    Okay, I am kind of with you. So since the net force on all three particles is 0, then. . . crap I lost it.

    Can someone give ne a little more of a nudge?
  7. Feb 3, 2008 #6
    Anyone? I know it may be obvious to some, but not to me.
  8. Feb 3, 2008 #7
    What are the expressions for the total force on each charge? For each one just add the forces from each of the other two.
  9. Feb 3, 2008 #8
    You mean like [itex]F_{1,net}=F_{12}+F_{13}[/itex]?
  10. Feb 3, 2008 #9
    Sure, why not? Now, what are [tex]F_{12}[/tex] and [tex]F_{13}[/tex]?
  11. Feb 3, 2008 #10



    Okay, I have values for L and X. What about the [itex]F_{1,net}[/itex] would that just be 0? It must be since the condition is that they stay in place, thus F_net-0.

    Does this sound reasonable?
    Last edited: Feb 4, 2008
  12. Feb 4, 2008 #11
    I cannot get this to come out right!!! I am trying to solve for the ratio [itex]\frac{q_3}{q}[/itex]


    [tex]\Rightarrow F_{1,net}=\frac{k|q||4q|}{L^2}+\frac{k|q||q_3|}{(L-x)^2}[/tex]

    I thought I could say that [itex]F_{1,net}=0[/itex] And then solve the above for


    But this is not working?
  13. Feb 4, 2008 #12
    Got IT! r_2 not equal to L-x. . . duh. It's just x!

    Thanks everyone!
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