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Surface Charge Density (Electric Fields)

  1. Jul 19, 2015 #1
    A disk with a uniform positive surface charge density lies in the x-y plane, centered on the origin. The disk contains 2.5 x 10-6 C/m2 of charge, and is 7.5 cm in radius. What is the electric field at z = 15 cm?

    I have used the formula:

    http://edugen.wileyplus.com/edugen/courses/crs7165/halliday9781118230725/c22/math/math121.gif
    (if this picture doesn't appear it is E=o/2e0 x (1-(z/sqrt(z^2+r^2)))

    Using 2.5x10-6C/M2 as my surface charge density, I get a final result of 1.5x10^4 N/C

    However it isn't one of the answer options, although it is exactly half of one of the options, so I'd lean towards that if I was guessing.

    Here are the possible options:

    30 N/C

    300 N/C

    3000 N/C

    3.0 x 10^4 N/C

    3.0 x 10^7 N/C
     
  2. jcsd
  3. Jul 19, 2015 #2
    Your equation is good. (assuming o/(2e0) is σ/(2ε0)
    Is this supposed to say 7.5 cm in diameter? Or is the radius acually 7.5 cm? did you convert cm to m?
     
  4. Jul 19, 2015 #3
    I copied the question exactly as it is written. I did convert to m.
     
  5. Jul 19, 2015 #4

    gneill

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    Staff: Mentor

    The question does not state a thickness for the disk, and it's unlikely to be comprised of conductive material if its uniformly charged in that shape. I guess you have to assume that it's much thinner than it is wide. How many (flat) sides does a disk have? Does that give you any ideas?
     
  6. Jul 19, 2015 #5
    Oh, so given a disk has two sides, then I would just x2 my answer. What would I do if I was given a thickness?
    Does the equation only tell you the electric field of one particular side of an object normally?
     
  7. Jul 19, 2015 #6

    gneill

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    Staff: Mentor

    You would have to take into account the actual distance of each surface from the location of interest.
    You should check the derivation of the formula. Applying a formula without knowing what it actually represents can be risky.

    I can tell you that the formula ##E = \frac{δ}{2 ε_o}## yields the field strength on either side of a uniform sheet of charge.
     
  8. Jul 19, 2015 #7
    Thanks
     
  9. Jul 19, 2015 #8
    The disk is of finite radius. It is not an infinite sheet.

    I have done the derivation of that formula many times. It's for one side of a uniform conducting disk of surface charge density σ.

    Edit: I get the same answer. Notice if you doubled your answer you would have 3.0x10^4. I don't see why you would double it though unless there is other given information.
     
    Last edited: Jul 19, 2015
  10. Jul 19, 2015 #9
    Your equation is correct in the limit that r→∞
     
  11. Jul 19, 2015 #10

    gneill

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    Yes, hence the (1 - ...) term that accompanies the ##\frac{δ}{2 ε_o}##.

    The point was to recognize the portion of the formula that corresponds to an infinite sheet has the "2" in the denominator for a reason.
     
  12. Jul 19, 2015 #11
    Ok, misunderstanding. I thought you were mentioning a formula to be used in this instance.
    P.s nice Tardis, I have one on my desk in front of me.
     
  13. Jul 19, 2015 #12

    gneill

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    Staff: Mentor

    Okay, so assuming that the disk is actually infinitely thin (like a disk punched out of a sheet of charge), then Jimbob999's answer would be correct, and the supplied list of possible answers does not include it.

    This can happen sometimes due to a typo in the problem parameters, or when an author tweaks a problem to make it "new" and fails to update the answer key. Sometimes the problem can be a problem :smile:
     
  14. Jul 19, 2015 #13
    So it wouldn't be twice my answer as there are two sides to the disk?
     
  15. Jul 19, 2015 #14

    gneill

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    Staff: Mentor

    Not if the disk is only a single layer of charge. Come to think of it, I suppose they would have called it a cylinder if it had vertical thickness. That's me not thinking o:)
     
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