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Electric fields between disks

  1. Sep 5, 2008 #1
    Two 10cm-diameter charged disks face each other, 18 cm apart. Both disks are charged to - 30 nC. What is the electric field strength a point on the axis 5.0 cm from one disk between them?

    [tex]\eta[/tex]/2[tex]\epsilon[/tex]0 * [1-1/[tex]\sqrt{1+ R^2/z^2}[/tex]]
    with [tex]\eta[/tex] = -4.0 X 10^(-6) C/m^2
    R=.05m z=.05cm for the left one and z=.13m for the right one

    As I read this question, it is asking for the net electric field at a point .05m from one disk and .13m from the other. And since both disks are negative,the point will be attracted to both disks, making E1 and E2 opposite.

    E1= -66190.6 N/C
    E2= -15063.2 N/C
    Both are negative meaning both are pointing towards their respective ring.
    So to find Enet do I add E1 and E2? Making Enet= -81253.8 N/C = 81253.8 N/C toward the left ring??

    Any help would be appreciated. Thanks.
    Stephen
     
  2. jcsd
  3. Sep 6, 2008 #2
    Yes, electric fields obey the principle of superposition, which means that you must sum the vectors of all electric fields in order to get the total. Just note that in your case the vectors are facing opposite directions, so you should subtract.
     
  4. Sep 6, 2008 #3
    -66190.6 N/C - -15063.2 N/C

    or
    66190.6N/C - 15063.2 N/C??????
     
  5. Sep 6, 2008 #4

    Doc Al

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    Good.

    The way to describe these fields is (I will assume that your arithmetic is OK and that E1 is the field from the left disk):
    E1 = 66190.6 N/C towards the left disk
    E2 = 15063.2 N/C towards the right disk

    If you call left negative and right positive, that means:
    E1 = -66190.6 N/C (since it points to the left)
    E2 = +15063.2 N/C (since it points to the right)

    Now that you're using a consistent sign convention, to find the total field just add them up:
    Enet = E1 + E2.

    As for your final answer, I would just give the magnitude of the field strength as the sign is arbitrary.
     
  6. Sep 6, 2008 #5
    so E=51127.4 N/C???
     
  7. Sep 6, 2008 #6

    Doc Al

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    When I checked your calculations I got a bit of a different answer. Don't round off your value for [itex]\eta[/itex] to 2 sig figs and expect accurate answers to 6 figs! Redo those calculations--don't round off until the last step.
     
  8. Sep 6, 2008 #7
    redid it and got 48823.1.
    Is this what you got?
     
  9. Sep 7, 2008 #8

    Doc Al

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    That looks much better.
     
  10. Sep 7, 2008 #9
    I rounded 48823.1 to two sig figs or 4.9*10^4
    but when I put it into mastering physics it said it was wrong. Any suggestions????

    Thanks for the help.
    Stephen
     
  11. Sep 8, 2008 #10
    What might I have done wrong? Any ideas?
     
  12. Sep 8, 2008 #11

    Doc Al

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    How many significant figures does Mastering Physics expect?

    Edit: I found a webpage with tips for using Mastering Physics. One of the things it said is this:
    Use three significant figures for all calculations! Your answer may be incorrect if you used only 2 significant figures during steps of the calculation or 2 significant figures for your answer.​
    This might be the problem.

    Here's the webpage: http://phystec4.cosam.calpoly.edu/141/masteringphysics.pdf
     
    Last edited: Sep 8, 2008
  13. Sep 8, 2008 #12
    2 sig figs
     
  14. Sep 8, 2008 #13

    Doc Al

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    See the comment I added to my last post. (Of course, it may not apply to your situation.)
     
  15. Apr 21, 2009 #14
    I'm just wondering where the -30 nC comes into play? I see that StephenDoty has eta listed as -4.0 * 10^(-6). Where did that come from? Thanks.
     
  16. Apr 22, 2009 #15

    Doc Al

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    eta ≡ charge/area; since charge and dimensions are given, one can calculate eta. The value of -4.0 * 10^(-6) is incorrect.
     
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