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Potential of thin ring of charge

  1. May 11, 2017 #1
    1. The problem statement, all variables and given/known data

    A thin disc, radius 3 cm, has a circular hole of radius 1 cm in the middle. There is a surface charge of -4 esu/##cm^2## on the disc.

    a) what is the potential at the centre of the hole ?

    b) An electron starting from rest at centre of the hole moves out along the axis. What is the ultimate velocity it gains ?

    2. Relevant equations


    3. The attempt at a solution

    For (a) :
    path4206.png
    ##R = \sqrt{y^2 + s^2}## and ## dq = 2\pi s\sigma ds##

    I did not drew ##\theta## but it is the angle between radius vector ##r## and ##y## axis.

    Then,
    ##\displaystyle \phi(0,y,0) = \int^3_1 {dq \over R} = \int^3_1 {2\pi \sigma s ds \over \sqrt{y^2 + s^2}} =2\pi \sigma (\sqrt{y^2 + 9} - \sqrt{y^2 + 1})##

    For centre, ##\phi = 2\pi\sigma##

    For (b):-

    Since ##\cos \theta = \dfrac{y}{\sqrt{y^2 + s^2}}##

    ##\displaystyle E = \int^3_1 {dq \over R^2} \cos \theta = \int^3_1 {2\pi \sigma s y \over (y^2 + s^2)^{3/2}} ds =\sigma\left( \frac1{\sqrt{y^2 + 1}} - \frac1{\sqrt{y^2 + 9}}\right)##

    ##eE = ma = m y^{\prime\prime} = m v\dfrac{dv}{ dy}##

    ##\displaystyle {e\sigma \over m} \int_0^{\infty} \frac1{\sqrt{y^2 + 1}} - \frac1{\sqrt{y^2 + 9}} dy = \frac{v^2}{2}##

    Hence ##v = \sqrt{\dfrac{2e \sigma \ln 3}{m}}##


    Is this correct ? I think the second answer is not correct as I get ##2.16 \times 10^{7} m/s## as the answer, which is very huge.
     
  2. jcsd
  3. May 11, 2017 #2

    vela

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    You dropped a factor of 2, it seems, but it otherwise looks right.

    I'm not sure what this means. What ##\phi## supposed to represent?

    You seem to be making this part unnecessarily complicated. Use conservation of energy. What's the potential energy of a charge at the center in terms of the electric potential at the center?
     
  4. May 11, 2017 #3
    ##\displaystyle \int {s\over \sqrt(y^2 + s^2)} = \sqrt{s^2 + y^2}## ?

    I am sorry, this was vague.

    ##\phi(0,y,0)## is electrostatic potential which is ##2\pi \sigma (\sqrt{y^2 + 9} - \sqrt{y^2 + 1})##.

    The question to find the potential at the centre of the disc which is at ##y =0##

    So, ##\phi(0,0,0) = 2\pi \sigma (3 - 1) = 4 \pi\sigma##

    .

    Let me get first part correct because I can use that to find energy.
     
  5. May 11, 2017 #4

    vela

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    Sorry, you were right. I messed up the factor of 2.
     
  6. May 11, 2017 #5
    Is everything else correct ?
     
  7. May 12, 2017 #6

    haruspex

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    I think there are some errors in the field calculations.
    (It would be easier to follow if you had left the radii as variables instead of plugging in the values 1 and 3. That would have allowed the use of dimensional analysis to sanity-check the equations.)
    E should be obtainable by differentiating φ wrt y. If I do that there seems to be a missing factor 2πy.
    I certainly don't think you should be getting ln() terms in there when they do not turn up using the energy approach.

    As for the magnitude of the answer, I think you are in the right ballpark, but maybe that means you need to use relativistic equations.
     
  8. May 12, 2017 #7
    Yes you are correct.

    Then ##\displaystyle E = {2\pi y \sigma}\left({1\over \sqrt{1+y^2}} - {1\over \sqrt{9 + y^2}} \right)##

    and
    is
    ##\displaystyle {e\sigma \over m} \int_0^{\infty} {2\pi y }\left({1\over \sqrt{1+y^2}} - {1\over \sqrt{9 + y^2}}\right) dy = \frac{v^2}{2}##

    I got the improper integral as ##2##,

    So ##\displaystyle v = \sqrt{8\pi e \sigma \over m}## which is ##7.38 \times 10^7 m/s##
     
  9. May 12, 2017 #8

    haruspex

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    That's what I got by considering energy. But that is a quarter light speed, no? Doesn't that mean you need to use the relativistic formula for KE?
     
  10. May 12, 2017 #9
    Yes I guess you are correct but I don't know special relativity ;).
     
  11. May 12, 2017 #10

    haruspex

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  12. May 12, 2017 #11
  13. May 12, 2017 #12

    haruspex

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    I get ##\frac{v^2}{c^2}=1-\left(\frac{mc^2}{E+mc^2}\right)^2##.
    If v' is the velocity you calculated by Newtonian principles, E=mv'2/2.
    But looks like it's not quite fast enough to worry about. If e.g. v' is c/4 then v≈ 0.238c
     
    Last edited: May 12, 2017
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