1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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:

    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)
  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!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook