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How much energy is need to place four positive charges

  1. Jul 24, 2010 #1
    1. The problem statement, all variables and given/known data
    How much energy is need to place four positive charges, each of magnitude +5.0 mC, at the vertices of a square of side 2.5 cm?

    2. Relevant equations
    E = kQ/r^2

    3. The attempt at a solution
    I'm really not sure if what I did is close to correct, but I got an answer.

    1) I calculated the Force experienced by 1 charge by another (arbitrarily picked):
    F = k Q^2/d^2
    = 3.6 x 10^8

    Therefore, if I picked the bottom left charge of the square, it is experiencing a repulsive force in the -y and -x directions each of 3.6 x 10^8 N.

    2) The third charge, the upper right, was also taken into account after finding the distance from the bottom left to the top right:

    F = k q^2/ r^2 = 1.8 x 10^6 N

    Therefore, the top right charge is repelling the bottom right charge by 1.8 x 10^6 N at a 45 degree angle south of west.

    3) I added up the vectors to find a resultant vector of the force acting on the bottom left charge: (tr = top right)

    a) F(tr)x = 1.8 x 10^6 (cos 45) = 1.27 x 10^6 N (in the negative x direction)
    b) F(tr)y = 1.8 x 10^6 (sin 45) = 1.27 x 10^6 N (in the negative y direction)

    The resultant vector consists of:
    F(x) = (1.27 x 10^6) + (3.6 x 10^8) = 3.61 x 10^8 (negative x direction)
    F(y) = 3.61 x 10^8 (negative y direction)

    Resultant force = sqrt [(3.61 x 10^8)^2 + (3.61 x 10^8)^2]
    = 5.1 x 10^8 N (at 45 degrees south of west)

    I'm not sure where to go from here.

    I'm wondering if I should have calculated work from each of the seperate charges acting on the arbitrarily picked bottom left charge and added them up?:

    a) W = Fd = (3.6 x 10^8 N) x (0.025 m) = 9.0 x 10^6 J
    b) W = Fd = (3.6 x 10^8 N) x (0.025 m) = 9.0 x 10^6 J
    c) W= Fd = (1.8 x 10^6 N) x (0.035m) = 6.3 x 10^4 J

    Total Energy = 1.81 x 10^7 J

    Then the TOTAL ENERGY for all 4 charges should be 1.81 x 10^7 multiplied by 4? (Because there are 4 charges):

    = 7.23 x 10^7 J[/B]

  2. jcsd
  3. Jul 24, 2010 #2
    I believe this formula [tex]W=k\frac{q_1q_2}{r}[/tex] must be given somewhere in your textbook. Check it :smile:
  4. Jul 24, 2010 #3

    Doc Al

    User Avatar

    Staff: Mentor

    Careful! Work = Force X distance only when the force is constant, which is not the case here. Instead of calculating the force, calculate the change in electric potential energy. (Look up the formula for electric potential energy.)
  5. Jul 24, 2010 #4
    My notes say Force = kQ1Q2/r^2 .... I don't see how I can get work from that? Or is that the wrong formula?

    DocAl: Why isn't force constant here? Is each charge not experiencing a constant charge from each of the other three since they are repelling eachother?
    Last edited: Jul 24, 2010
  6. Jul 24, 2010 #5

    Doc Al

    User Avatar

    Staff: Mentor

    That's the correct formula to find the force, but not the one you need for this problem. hikaru1221 gave the formula you need. Read more about it in your textbook and here: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elepe.html" [Broken]

    The force depends on the distance between the charges, which is changing as the charges are moved into position. Imagine that the charges were initially very far apart from each other (so their force was minimal) and you brought them to their final positions. The forces between them would change as they are brought together.

    You can certainly derive the formula for work from the force equation, but you'd need to use calculus since the force isn't constant. Check your text.
    Last edited by a moderator: May 4, 2017
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