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Calculating work done by E field/voltage

  1. Oct 19, 2009 #1
    1. The problem statement, all variables and given/known data
    Four point particles with charges .6 microC, 2.2 microC, -3.6microC, and 4.8 microC are placed at the corners of a square of side 10cm. what is the external work needed to bring a charge of -5 microC from infinity to the center of the square? (Assume the speed of the -5 microC charge is kept constant)


    2. Relevant equations
    V=kq/r(from infinity)
    w = Q * v
    Vtotal = V1 + v2...Vn


    3. The attempt at a solution
    9x10^9 * (-5 x 10^-12) / squareroot(.1^2 + .1^2) * (.6 + 2.2 - 3.6 + 4.8) = -1.28 J
    answer in book: -2.55J
    my answer is nearly a perfect factor of 2 away. What's going on here?
     
  2. jcsd
  3. Oct 19, 2009 #2
    I'm not too sure what you did in your calculations but..
    First step is to do a diagram with the 4 charges each at a corner of the square. Then you have to find the potential energy at the center of the square (which is acquired by adding the V=kq/r for each charge). Now that you have the potential energy at the center of the square you can just apply w=qv to get the correct answer.
     
  4. Oct 19, 2009 #3
    that's what i did, and there is no need for a diagram due to the simplicity of a cube


    let me explain my calculations: w = k Qq/r

    i factored the k out, i factored the r out, and the Q(the charge brought to the middle) is also factored out. I also factored out a x10^-6 (the -12 comes from that), so all that is left is the sum of the charges on the corner.
     
  5. Oct 19, 2009 #4
    your method is fine, you must have done a calculation error somewhere. just redo it on a piece of paper.
     
  6. Oct 19, 2009 #5
    I have! Could you do it yourself to confirm my answer? Erroneous book answers are not unheard of.
     
  7. Oct 19, 2009 #6
    yeah, I had just done it and I got -2.55J as the answer right before I posted my reply. here:

    W=(kQq)/r
    get it into W=(k)*(Q)/(r)*(sum of qs) form
    W=(9*10^9)*(-5microC)/(0.0707m) *(.6microC + 2.2microC - 3.6microC + 4.8microC)
    W=-636396*(4microC)
    W=-2.55J
     
  8. Oct 19, 2009 #7
    I'm kind of sleepy so i couldnt pinpoint the problem earlier but I'm pretty sure you just used the wrong value of r, thats why I recommended to draw a diagram. its the middle of a square with 10 cm sides so if you do pythagoras you have to use .05m and not .1m like you did
     
  9. Oct 19, 2009 #8
    omg... of course. a factor of two was no coincidence! I forgot to divide my length by two!
     
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