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The electric field on the axis of a charged disk

  1. Nov 10, 2014 #1
    1. The problem statement, all variables and given/known data
    A disk of radius 10 [cm] is charged with 5E-9 [Coulon]. where on it's axis is the greatest electric field.

    2. Relevant equations
    The field on the axis of a charged disk:
    ##E=\frac{1}{4\pi\varepsilon_0}\frac{q\cdot \cos\alpha}{s^2}##

    3. The attempt at a solution
    I express [itex]\cos\alpha[/itex] in terms of the radius R and the diagonal distance s:
    ##\cos\alpha=\frac{\sqrt{s^2-R^2}}{s}##
    I insert that into part of the equation of the field:
    ##\frac{\cos\alpha}{s^2}=\frac{\sqrt{s^2-R^2}}{s^3}##
    The derivative of this expression is (maybe i have a mistake here):
    ##\left( \frac{\sqrt{s^2-R^2}}{s^3}\right)'=\frac{3R^2-2s^2}{s^4\sqrt{s^2-R^2}}##
    I equal the numerator to 0 and get s=1.22R=12.2[cm]
    The horizontal distance on the axis that corresponds to this diagonal is:
    ##\cos\alpha=\frac{1.22R}{R}\rightarrow\alpha=34.95^0##
    To this angle the horizontal distance is 14.3[cm] while it should be around 8[cm]
     

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  2. jcsd
  3. Nov 10, 2014 #2

    rude man

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    I don't think that's right!
    Think of the disc as an infinite number of infinitely small annuli ranging in radius from 0 to R
    where R = radius of disc.
    You probably have the formula for the E field along the axis for a ring (thin annulus) somewhere.

    Extra thought: what is the E field very close to a charged plane? I.e. for you that would imply s = R.
    Hint: don't try setting a derivative to zero to find the max E field.
     
  4. Nov 10, 2014 #3

    rude man

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    Extra credit: what if there is a hole in the center of the disc, of radius R2? What then would be the point of maximum E field along the axis?
    (Warning: don't try this at home without good math software!)
     
  5. Nov 11, 2014 #4
    The question is about a thin annulus
     
  6. Nov 11, 2014 #5

    haruspex

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    So there are two radii specified? Or is it so thin it's effectively a ring, not a disc?
     
  7. Nov 11, 2014 #6
    yes, a ring, i didn't know how to say in english a flat disk. and what is an annulus? also a ring?
     
  8. Nov 11, 2014 #7

    rude man

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    @haruspex, the OP's problem clearly states that it's a disc, not an annulus (ring). Even a thin disc is not the same thing as an annulus. A disc has area = pi R^2 while an annulus has an area = pi(R2^2 - R1^2).

    I added the "extra credit" as a new, far more challenging problem, mathematically speaking. There is only one radius in the OP's problem.

    My "extra credit" was a NEW problem, and in fact a tease since the math, which WOULD involve taking a derivative and setting it to zero, would probably blow your mind if not your computer!
    [/quote]

    No it's not. An annulus is a ring and this problem is a solid disc.
     
  9. Nov 11, 2014 #8

    rude man

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    A ring and an annulus are somewhat the same thing, although typically a ring is said to have zero dimensions so charge is in terms of linear (coulombs/meter) rather than surface density (coulombs/sq. meter). An annulus can have an arbitrarily large width so it can be a disc with a small hole in it and still qualify as an annulus.

    A disc is a solid circle of one radius only, and with charge density typically defined in terms of surface charge density.
     
  10. Nov 11, 2014 #9
    I am almost sure the question is about a ring, look at the drawing, the book explained about that ring and in the explanation gave also only one radius R.
     
  11. Nov 11, 2014 #10

    haruspex

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    OK. In the OP, your differentiation looks wrong.
    You will find the algebra much easier if you take an angle as the independent variable instead of a distance.
     
  12. Nov 12, 2014 #11

    rude man

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    I agree, the drawing looks like a ring. But a ring is not a disc, and the wording says disc.

    I will let others take over here ...
     
  13. Nov 13, 2014 #12
    I express s in terms of [itex]\alpha[/itex]:
    ##\frac{\cos\alpha}{s^2}=\frac{\cos^2\alpha}{R^2}##
    ##\left( \frac{\cos^2\alpha}{R^2}\right)'=\frac{1}{R^2}(\cos^2\alpha)'=\frac{-2}{R^2}\cos\alpha\cdot\sin\alpha##
    And if i equal this to 0 then sin and cos may become 0 and i get 00 and 900
     
  14. Nov 13, 2014 #13
    I made a mistake.
    ##\frac{\cos\alpha}{s^2}=\frac{\cos\alpha\cdot\sin^2\alpha}{R^2}##
    ##\left(\cos\alpha\cdot\sin^2\alpha\right)'=2\sin\alpha\cos^2\alpha-\sin\alpha\cdot\sin^2\alpha##
    ##\tan^2\alpha=2\rightarrow\alpha=54.7^0\rightarrow x=7.1[cm]##
    Better but still not true, it should be between 8 and 10
     
  15. Nov 13, 2014 #14

    haruspex

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    I agree, but no need to calculate the angle. The distance you want to calculate is ##R/\tan\alpha = 5\sqrt 2cm##
    How do you know?
     
  16. Nov 13, 2014 #15
    I was told to calculate the field E at distances 0, 5, 8, 10 and 15 [cm] from the origin.
    ##E=\frac{1}{4\pi\varepsilon_0}\frac{q\cdot \cos\alpha}{s^2}##
    I calculate only:
    ##\frac{q\cdot \cos\alpha}{s^2}##
    At 0 [cm] it's 0.001
    At 5 [cm] it's 0.00356
    At 8 [cm] it's 0.00381
    At 10 [cm] it's 0.0035
    At 15 [cm] it's 0.00256
    The peak is either between 5 and 8 or 8 and 10, but i guess closer to 8. i will draw a graph and attach
     

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  17. Nov 13, 2014 #16
    r=R/sqrt(2)
     
  18. Nov 14, 2014 #17

    haruspex

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    Exactly. If you want to find out which, by the same method, you need to calculate some more points. Why not just trust the caclulus method?
     
  19. Nov 14, 2014 #18
    what is the calculus method? do you mean just to calculate more points or make a graph?
     
  20. Nov 14, 2014 #19

    haruspex

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    The differentiation method in post #12.
     
  21. Nov 14, 2014 #20

    Zondrina

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    The question looks like a ring of radius ##r## and radial width ##dr##. In this case the electric field of the "ring" would be given by:

    $$\vec E(x) = \frac{kqx}{(x^2 + R^2)^{3/2}}$$

    Where you would take the derivative...
     
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