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Point charges and static equilibrium

  1. Jan 12, 2008 #1
    1. The problem statement, all variables and given/known data

    Two point charges q and 4q are distance L apart and free to move. A third charge is placed so that the entire three-charge system is in static equilibrium.

    2. Relevant equations

    [tex]\Sigma[/tex]F=0 This is the equation for static equilibrium
    [tex]\vec{E}[/tex]=[tex]q/4\pi\epsilon_{0}r^2[/tex]

    3. The attempt at a solution

    If the three point charges are in static equilibrium, the net electric field is zero. So, [tex]\vec{E}[/tex]=0, which means that the individual electric fields of the point charges must add up to zero. Thus, the third charge must be -5q, as q + 4q = 5q.

    Is any of this right?
     
  2. jcsd
  3. Jan 12, 2008 #2

    Shooting Star

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    No. There is a r^2 in the denominator. What happened to that? You have to take distance into consideration.
     
  4. Jan 12, 2008 #3

    mda

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    Don't forget that the force on each of the three particles must be zero. Due to symmetries, and the fact that you need to find two quantities (charge and distance) you actually only need to consider any two particles, so that you get two simultaneous equations.
     
  5. Jan 12, 2008 #4
    Sorry, forgot to include the actual question. The question asks for the magnitude and sign of the third charge. I don't really understand the wording of this question. That is, should I assign another variable for the distance from the third charge? As well, is the magnitude of the third charge related to q, I don't think it is, because it does not ask for inclusion of the q variable in the answer...but I don't know how to calculate it otherwise. Thank you!
     
  6. Jan 12, 2008 #5

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    Yes, take the dist of the 3rd charge from the 1st as x and then the dist q3-q2 will be L-x, if q3 is in between them.
     
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