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Homework Help: Simple coloumb's law problem

  1. Feb 2, 2008 #1
    1. The problem statement, all variables and given/known data
    4 charges are arranged in the corners of a square of lengths L as follows:

    (-Q) --------- (+q)
    (+4Q) ------- (-Q)

    What is the magnitude of the force on the (+q)?

    3. The attempt at a solution
    My answer doesn't match up with my book's.

    x-component of the force is, where K is [itex]\frac{1}{4 \pi \epsilon_0}[/itex]:

    [tex]F_x = K q(\frac{- Q}{L^2} + \frac{4Q}{2 L^2}) = K q \frac{Q}{L^2}[/tex]

    The x and y components of the force are equal by symmetry:

    [tex]\vec{F} = K q \frac{Q}{L^2} (\hat{x} + \hat{y})[/tex]

    Therefore the magnitude is:

    [tex]F = \sqrt{2} K q \frac{Q}{L^2}[/tex]

    However, my textbook says the answer is:

    [tex]F = (2 - \sqrt{2}) K q \frac{Q}{L^2}[/tex]

    I have no idea where they got the "2" term from. I probably just made a careless mistake somewhere, but I can't see it.
    Last edited: Feb 2, 2008
  2. jcsd
  3. Feb 2, 2008 #2
    Bump -- anyone?
  4. Feb 2, 2008 #3
    so, in F_x,when computing F(4Q)

    you must 1 compute L, and then compute projection in x, i guess that wasn't made...
    i made [tex]Kq \frac{4Q}{\frac{L}{cos(45º)}^{2}}*cos(45º)[/tex]


    P.s, Sorry, but i give up, 10m to try making the equation and i still couldn't do it...
    Last edited: Feb 2, 2008
  5. Feb 2, 2008 #4
    Ah, that's exactly it! I always forget to do that. Thanks.
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