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Point Charges in a Square: Could the Force on Each Be Zero?

  1. Aug 31, 2008 #1
    1. The problem statement, all variables and given/known data
    A charge Q is fixed at each of two opposite corners of a square, while a charge q is fixed at the other two corners. (a) if the resultant electrical force on Q is zero, how are Q and q related? (b) could q be choosen to make the resultant electrical force on every charge zero? explain.
    F1 = force between Q's
    F2 = force between one Q and q's
    a = side of the square

    2. Relevant equations
    F = (1/(4*pi*E))(q1*q2/r^2)

    3. The attempt at a solution

    (a) F1 = (1/(4*pi*E))(Q^2/2a^2)
    F2 = (1/(4*pi*E))((Qq*sqrt(2))/(a^2))
    I set them equal to each other and got: Q = -2q*sqrt(2)
    The answers weren't in the back of the book, so I'm not sure if that's correct or not

    The part I'm having trouble with is part (b). I don't think there can be a value for q that would make the forces on all the charges zero, but I don't know how to explain that.

  2. jcsd
  3. Aug 31, 2008 #2
    whats the distance between the Q's?

    and where did the sqr(2) come from in your F2 equation?
  4. Aug 31, 2008 #3
    the distance of the side of the square is a, so the diagonal is a*sqrt(2)
  5. Aug 31, 2008 #4
    I misread it as (2a)^2.

    still dont understand the F2 equation.

    you cant just set them equal to each other. they are vectors.
  6. Aug 31, 2008 #5
    i saw the equation: (1/(4*pi*E))((Qq)/(a^2)) as the force on one of the sides of the square between Q and q. and i know there's another force pointing perpendicular on the adjacent side of the square. the addition of those two vectors would give me the hypotenuse of a 45-45-90 triangle. therefore, i saw that the magnitude of the hypotenuse is that force times sqrt(2).
    would that be right?
  7. Aug 31, 2008 #6
    sounds good.
  8. Aug 31, 2008 #7
    that's good.
    do you know about part (b) ?
  9. Aug 31, 2008 #8
    you know what q must be for the force on Q to be zero. what must the charge on Q be to make the force on q be zero?
    Last edited: Aug 31, 2008
  10. Sep 1, 2008 #9
    Since it's a square, the charge for q would be the same formula as it was for Q, just the Q and q are switched:
    q = -2Qsqrt(2)
  11. Sep 1, 2008 #10
    yes. and so? what is your conclusion?
  12. Sep 1, 2008 #11
    so would that mean that the value of q would be -Q ?
    i know q =/= Q because then the charges would all repel each other.
    if i plug one equation into the other, like: q = -2(-2q*sqrt(2))sqrt(2), i just get that q=8q. which is why i think there can't be a value for q. though, i don't know the explanation.
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