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

  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.

    Thanks!
     
  2. jcsd
  3. whats the distance between the Q's?

    and where did the sqr(2) come from in your F2 equation?
     
  4. the distance of the side of the square is a, so the diagonal is a*sqrt(2)
     
  5. 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. 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. sounds good.
     
  8. that's good.
    do you know about part (b) ?
     
  9. 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. 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. yes. and so? what is your conclusion?
     
  12. 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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