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I Total energy of a shell charge

  1. May 2, 2017 #1
    Suppose a spherical shell is kept at the orgin. Its surface charge density is ##\sigma## and radius is ##R##.

    I think I remember the formula for its electric potential energy is ##U = \displaystyle {Q^2 \over 2R}##.

    Now I want to derive it. I used ##\displaystyle U =\displaystyle\dfrac{1}{8\pi}\int_{\text{Region}} \vec E\cdot \vec E\ \ dv##.

    I know that electric field by an sperical shell is ##Q\over R^2##.

    So I got,

    ##\displaystyle U =\displaystyle\dfrac{1}{8\pi}\int_{\text{Region}} {Q^2\over R^4}\ \ dv = \displaystyle\dfrac{Q^2 }{8\pi}\int_{\text{Region}} {1 \over R^4 }\ \ dv##

    Then I used ##v = {4\pi \over 3}R^3## to get ##\displaystyle \left({3v \over 4 \pi}\right)^{1/3} = R##

    Back in integral, ##U = \displaystyle\dfrac{Q^2 }{8\pi}\int_{\text{Region}} \left({4 \pi \over {3v} }\right)^{4/3}\ \ dv##

    For limits ##v \to \infty## I got the desired answer.

    Is this correct ? I have some confusions because the proofs I saw on internet involve triple integral and spherical coordinates.

    I think I am doing something wrong as why would anybody use spherical coordinates for such a simple proof ?
     
  2. jcsd
  3. May 2, 2017 #2
  4. May 2, 2017 #3
  5. May 2, 2017 #4

    Charles Link

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    You could write the integral as (a much simplified version of spherical coordinates) ## \int \limits_{R_o}^{+\infty} \frac{1}{r^4} (4 \pi r^2) dr ## and you get the same result. Meanwhile, a (editing: scratch "alternative") proof of the original formula is to add ## dQ ## from ## Q= 0 ## to ## Q_o ## with ## dU=(\frac{Q}{R_o}) dQ ## so that ## U=Q_o^2/(2 R_o) ##.
     
    Last edited: May 2, 2017
  6. May 2, 2017 #5
    I did not understand the integral.

    Isnt ##\displaystyle \int E\cdot E dv = \iiint E\cdot E dx dy dz##.
    How do you get ##R## in here ?
     
  7. May 2, 2017 #6

    Charles Link

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    I did two different integrals. The first one is one that you did, over ## dv ## where you replaced ## r^4 ## with an expression for ## v ##. The simpler way to do that is to use the surface area of a sphere is ## 4 \pi r^2 ## so that with spherical symmetry ## dv=4 \pi r^2 \, dr ##. (Otherwise, in polar coordinates ## dv=r^2 sin(\theta) d \theta \, d \phi \, dr ## where ## \theta ## integrates from ## 0 ## to ## \pi ## and ## \phi ## integrates from ## 0 ## to ## 2 \pi ##.) ## \\ ## The second integral was one where you assemble the charge on a sphere, and the potential energy ## dU ## for each ## dQ ## that you add is ## dU=(Q/R_o)dQ ##. The energy required depends on the charge that is already there. The charge ## Q ## that is already there determines the potential ## V ## at that instant which the charge ## dQ ## encounters when it gets added to the shell. (You gave the formula for this one at the very beginning, but without a derivation.)
     
    Last edited: May 2, 2017
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