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Electrostatic Force

  1. Jan 16, 2009 #1
    1. The problem statement, all variables and given/known data

    Four particles form a square. The charges are [tex]q_1 = q_4 = Q[/tex] and [tex]q_2 = q_3 = q[/tex].

    (a) What is [tex]\frac Qq[/tex] if the net electrostatic force on particles 1 and 4 is zero?

    (b) Is there any value of q that makes the net electrostatic force on each of the four particles 0? Explain.

    The particles are arranged as such:

    1 2
    3 4

    All separated by a distance a, the axes are drawn as standard up = +y and right = +x.

    2. Relevant equations

    [tex]F_E = k\frac{q_1 q_2}{r^2}[/tex]

    3. The attempt at a solution

    (The way I solve for Force, I say that the force from 2 to 1 as [tex]F_21[/tex])

    First I find the net force on particle 1:

    [tex]\sum F_1 = F_{21} + F_{31} + F_{41}[/tex]

    [tex]= k\frac{qQ}{a^2} + k\frac{qQ}{a^2} + k\frac{Q^2}{(a\sqrt{2})^2}[/tex]

    [tex]= 2k\frac{qQ}{a^2} + k\frac{Q^2}{2a^2}[/tex]

    [tex]= 2k\frac{2qQ + Q^2}{2a^2}[/tex]

    [tex] = k\frac{2qQ + Q^2}{a^2}[/tex]

    I determined that the net force on 4 must be the same, so the net force on Q is then:

    [tex]\sum F_Q = 2k\frac{2qQ + Q^2}{a^2} = 0[/tex]

    [tex]= \frac{2k}{a^2}(2qQ + Q^2) = 0[/tex]

    I determined that [tex](2qQ + Q^2)[/tex] must be 0, so:

    [tex]2qQ + Q^2 = 0[/tex]

    [tex]Q^2 = -2qQ[/tex]

    [tex]Q = -2q[/tex]

    [tex]\frac Qq = -2.0[/tex]

    Is that right? I don't even know how to do part b and I know I forgot something in part a that is important... Which is exactly why I'm here asking for help.
  2. jcsd
  3. Jan 16, 2009 #2


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    Welcome to PF.

    That looks like the right result.

    By symmetry wouldn't the only way the they could all be 0 net force would be if they were all equal charge magnitudes?
  4. Jan 16, 2009 #3


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    In the above calculation, the fourth step is wrong.
    It should be k[4qQ +Q^2]/2a^2
  5. Jan 16, 2009 #4
    Thanks for the help guys. And I can't believe I missed that little error.

    So, my [tex]\frac Qq[/tex] is actually -4.0 and not -2.0.

    Hey LowlyPion, would you be willing to tell me why (b) is the answer it is. I can kinda see it but I can't make it make sense to me. And thanks for the welcome.
  6. Jan 16, 2009 #5


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    Other than the trivial case ([itex]Q=q=0[/itex]), I don't imagine there is a way that the force on all 4 particles could be zero.

    If you were told that instead of the force on particles 1 and 4 being zero, the force on 2 and 3 was zero you could apply the same method as in part a and obtain (due to symmetry) [tex]\frac{q}{Q}=-4[/tex] right?

    If the force on all four is zero,. then both cases would have to apply simultaneously. Is it possible for [tex]\frac{q}{Q}=-4[/tex] and [tex]\frac{Q}{q}=-4[/tex] to both be true?
  7. Jan 16, 2009 #6
    Ah, that makes perfect sense.

    Thank you very much for the help. I appreciate it.
  8. Jan 16, 2009 #7


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