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Coulomb law - 3 point charges

  1. Apr 3, 2013 #1
    1. The problem statement, all variables and given/known data
    2 questions regarding the answer I have been given for this problem. Attachments are the problem & relevant worked answer I disagree with.

    Three charges are arranged in the xy-plane as shown in attachment. A charge Q is at the point A with (x, y) coordinates (–2a, 4a), a charge –Q is at the point B which is the origin (0, 0), and a charge 2Q is at point C with coordinates (3a, 4a).

    Find an expression for the force FC on the charge at C due to the other two charges and hence determine the magnitude |FC| of the force and the unit vector that specifies the direction of FC

    2. Relevant equations
    Coulomb's law
    F = q1 q2 / 4 ∏ ε0 r123 rhat12

    F = ke q1 q2 / r123 * rhat

    where ke = 1 / 4 ∏ ε0

    3. The attempt at a solution
    FCA = 1/4∏ε0 qC qA / rCA3 * rCA

    FCA = ke 2Q Q / (5a)3 * (5a ex)
    OK, but then goes to

    FCA = ke Q2 / a2 * 2/53 * (5a ex)

    probably simple algebra but can't see how (5a)3 goes to a2 &53

    Next query is regarding
    FCB = ke * 2Q * -Q / (5a)3 * (3aex + 4aey)

    How can the denominator be (5a)3 too, surely this is √20
    Please assure me (or otherwise) that it's not (5a)3

    Thank you

    Attached Files:

    Last edited: Apr 3, 2013
  2. jcsd
  3. Apr 3, 2013 #2
    There's an "a" in the numerator which leaves a2 in the denominator.

    Remember that rCA is the vector from C to A and is not the position vector of A
  4. Apr 3, 2013 #3
    Thank you.
    Yes, easy points to remedy.
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