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Homework Help: Force on a charge, 4 charges

  1. Sep 15, 2008 #1
    1. The problem statement, all variables and given/known data
    A charge of 5.00 mC is placed at each corner of a square 0.100 m on a side.

    Determine the magnitude of the force on each charge.

    2. Relevant equations
    F= (kq1q2)/r2

    k= 9*10^9

    3. The attempt at a solution

    I drew 4 charges in a square .1m apart from each other, then used the formula above.

    Q1 Q2

    Q3 Q4

    I did the calculation for Q2 (figuring they all should be the same):

    F(2 due to 1) = [(9*10^9)(5*10^-6)^2]/(.01) = 22.5 N
    F(2 due to 4) = [(9*10^9)(5*10^-6)^2]/(.01) = 22.5 N
    F(2 due to 3) = [(9*10^9)(5*10^-6)^2]/(.02) = 11.25 N


    Ftotal = ([tex]\sqrt{(22.5^2+22.5^2}[/tex]) + 11.25 = 43.07 N

    Did I do this wrong? Or is the distance from each charge not .1 m?
  2. jcsd
  3. Sep 15, 2008 #2


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    Gold Member

    You seem to have forgotten to square the distances between charges. Also, think again about what the distance between Q2 and Q3 is. The length of the diagonal of a square is not just double the length of each side.
  4. Sep 15, 2008 #3
    I did square the dist between charges, .1^2 = .01

    You're right Q2 to Q3 is not the double of the length of the sides, I used Pythagorean theorem and squared both sides and took the sqrt of that;

    .1^2 + .1^2 = .02

    sqrt .02 = .1414

    but the dist. needed to be squared for the formula so .1414^2 = .02

    lmk if I have mistaken what you posted, and thanks for the reply
  5. Sep 15, 2008 #4


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    Gold Member

    Sorry i read the length of the sides as 0.01m instead of 0.1m; my mistake :smile:
  6. Sep 15, 2008 #5
    No worries :). Any more suggestions?
  7. Sep 16, 2008 #6
    Figured it out..... mC does not mean micro it means MILLA!!!! 10^-3 instead of 10^-6
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