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Hollow Spheres & potentials

  1. Aug 25, 2010 #1
    Hello all!
    I hope this question has not been raised previously, so that I wouldn't exhaust your time in vain, and if it had been, please forgive me, though extensive combing of the web yielded largely nothing.
    My problem concerns something extremely trivial: the derivation of the electric potential of a sphere with an isotropic surface charge [tex]\sigma[/tex].
    Let me declare head on: I am able to successfully compute the necessary result by first finding the Electric field, then integrating as necessary to unearth the Potential. But, as fate would have it, I am rather a plucky type, and I have always been intrigued about the direct approach("Nuke em'", if you will); Thus, I have inscribed the following formulae:
    [tex]\varphi=\int\frac{dq}{r'}[/tex]
    And naturally,
    [tex]dq=\sigma*dS[/tex]
    Whereby, dS, in Spherical coordinates:
    [tex]dS=2\pi r dr sin(\theta)d\theta[/tex]
    Where [tex]r'[/tex], as it were, being the distance from the charge [tex]dq[/tex], is of course:
    [tex]r'=\sqrt{r^2+r0^2-2*r*r0*cos(\theta)}[/tex];
    (Whereas [tex]r0[/tex] is my vector to the point on the axis of symmetry where I wish to measure my potential difference(Infinity is calibrated as Zero).
    I integrate r->R(Radius of the Sphere), Theta->(0, Pi)...
    But lo' and behold, the answer that emerges is very convoluted and does not at all match the one gained by incorporating the Electric field.
    What am I doing errantly here?
    Very grateful for your attention,
    Beholden,
    Daniel
     
    Last edited: Aug 25, 2010
  2. jcsd
  3. Aug 25, 2010 #2

    gabbagabbahey

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    Oh?...Are you under the impression that the radius of a spherical surface varies over the surface? You'll want to double check dS.:wink:
     
  4. Aug 25, 2010 #3
    Firstly, let me thank you for a very prompt response!
    Now, honestly, judging by the covariance of integral formulae, it doesn't really matter; After all, an integral over a function that should be invariant under [tex]rdr[/tex](as it is in this case), Will be, whether we like it or not, [tex]R^2/2[/tex].
    Second, not being overtly presumptuous, I had rewritten some of the conditions, Keeping as you suggested [tex]r=R[/tex] constant.
    Their new form would only constitute:
    [tex]dS^*=2\pi R^2sin(\theta)d\theta[/tex].
    And the distance Vector [tex]r'[/tex]:
    [tex]r'=\sqrt{R^2+r0^2-2*R*r0*cos(\theta)}[/tex].
    But, as you might imagine, since I am still harassing you(:D), the integration is still very flabby.
    Should this do?
    What else ought I to correct?
    Thank you very much again for all your help,
    Daniel
     
  5. Aug 25, 2010 #4

    gabbagabbahey

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    You seem to be very confused about surface integrals. The form of [itex]dS[/itex] depends entirely on what surface you are integrating over, since it is the differential are element for that particular surface.

    In the case of a spherical shell, the radius is clearly constant over the entire surface, but [itex]\theta[/itex] and [itex]\phi[/itex] vary. (At one point on the surface, [itex]\theta[/itex] may be zero and [itex]\phi=\pi[/itex], while at another point on the surface you could have [itex]\theta=\phi=\pi/2[/itex], for example). The differential are element [itex]dS[/itex] in this case will be the area subtended when you vary [itex]\theta[/itex] and [itex]\phi[/itex] by infinitesimal amounts [itex]d\theta[/itex] and [itex]d\phi[/itex]. There are many sites and calculus texts where that quantity is derived, and the well-known result is that [itex]dS=r^2\sin\theta d\theta d\phi[/itex].
     
  6. Aug 25, 2010 #5
    Thank you once again for a very rapid retort;
    Now, Unless I am VERY MUCH mistaken with an isotropic distribution of charges on the sphere, the integral of [tex]\int d\phi[/tex] will be perpetually [tex]2*\pi[/tex], no matter how we may spin it around.
    The only reason why I wrote, instantaneously that [tex]dS^*=2\pi R^2sin(\theta)d\theta[/tex], was that I had the "clairvoyance" to integrate [tex]d\phi[/tex] first, and so as not to clutter my expressions here, and give you a simpler, neater arrangement.
    I should also note, that these variables(Theta, and Phi) are independent of each other(in the Spherical coordinate system), thereby granting me the possibility of integrating them separately.
    What else should I do?
    Thank you very much again!
    Daniel
     
  7. Aug 25, 2010 #6

    gabbagabbahey

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    If you are having problems with integrating over [itex]\theta[/itex], just make the substitution [itex]u=R^2+r^2+2rR\cos\theta[/itex]. The [itex]\sin\theta[/itex] in the numerator makes the integral trivial.
     
  8. Aug 25, 2010 #7
    Once again, one has to appreciate the tantivy, so thanks for that.
    Now, let's consider the case of a halved sphere(where Theta would diverge from 0 to Pi/2).
    If you try and integrate it as such, the result will be(at r0=R, to encompass the entire bulk),
    The result would be:
    [tex]\varphi=2 \sqrt{2} k\pi r0 \sigma[/tex]
    (where [tex]k=\frac{1}{4 \pi \epsilon _0}[/tex]).
    However, If I were to integrate and find [tex]E[/tex] first, the elucidation would amount to:
    [tex]E= \int dE= -\pi k \sigma[/tex].
    Clearly by Integrating [tex]-\int Edr[/tex], I find that my Phi has a redundant factor of a two times the Square root of Two in it, which doesn't belong there.
    Where do you suggest it emanates from?
    Once again, I can't express my indebtedness for your help,
    Daniel
     
    Last edited: Aug 25, 2010
  9. Aug 25, 2010 #8

    gabbagabbahey

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    You'd have to post your calulations for the actual integration, but it seems like you might be forgetting that

    [tex]\sqrt{R^2+r^2\pm 2rR}=\left\{\begin{array}{lr}R\pm r, & R \pm r \geq 0 \\ -(R\pm r), & R\pm r <0\end{array}\right.[/tex]

    You always take the positive root ( [itex]\sqrt{\alpha^2}=|\alpha|[/itex] )
     
  10. Aug 25, 2010 #9
    Hi again,
    Not trusting my own(palpating) hand on this one, I actually turned to Mathematical software, something I normally never do, but just to demonstrate the severity of my desperation :).
    Anyway, the way I've calculated it is as follows:(knowing that R-r0>=0), I am looking at the potential at point r0=R:(this while running Theta from 0 -> Pi/2, in order to accomplish the 'Halved Sphere' projection)
    [tex]2 \pi R^2 \int^\frac{\pi}{2}_0 \frac{sin(\theta)}{\sqrt{R^2+r0^2-2Rr0cos(\theta)}}=[/tex]
    [tex]2 \pi \simga R^2*(\frac{\sqrt{R^2+r0^2}}{2Rr0} - \frac{\sqrt{(R-r0)^2}}{2Rr0})[/tex]
    Now at this point, I am at liberty(right?) to eliminate the term corresponding to (R-r0) since r0 = R, and I am left, as you can see, with [tex]\sqrt{2}[/tex]...
    What now?
    Thank you very much again,
    Daniel
    P.S, it's a trifle to evaluate E:
    [tex]\vec{E} = k\int^\frac{\pi}{2}_0 \int^{2\pi}_0 \frac{\sigma R^2sin(\theta)d \theta d \phi}{R^2}*(sin(\theta)*cos(\phi) \hat{x}+sin(\theta)*sin(\phi) \hat{y} + cos(\theta) \hat{z})[/tex]
    The first two elements of [tex]\hat {r}[/tex](i.e sin(theta)*cos(phi)+...), are immediately vaporized(the integral of dPhi over either Sin or Cos from(0 to 2*pi) is zero anyhow) and the third one give us: -Pi.
    And it's as simple as that, why can't the potential be so amenable?
     
    Last edited: Aug 26, 2010
  11. Aug 26, 2010 #10
    Any further advice guys?, all aid will be very appreciated!
    Thanks again,
    Daniel
     
  12. Aug 26, 2010 #11
    Anything folks?
    Thanks,
    Daniel
     
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