Deriving the Electric Potential of a Hollow Sphere with Isotropic Surface Charge

[danielakkerma]

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 \sigma.
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:
\varphi=\int\frac{dq}{r'}
And naturally,
dq=\sigma*dS
Whereby, dS, in Spherical coordinates:
dS=2\pi r dr sin(\theta)d\theta
Where r', as it were, being the distance from the charge dq, is of course:
r'=\sqrt{r^2+r0^2-2*r*r0*cos(\theta)};
(Whereas r0 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

 

[gabbagabbahey]

Whereby, dS, in Spherical coordinates:
dS=2\pi r dr sin(\theta)d\theta

Oh?...Are you under the impression that the radius of a spherical surface varies over the surface? You'll want to double check dS.

 

[danielakkerma]

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 rdr(as it is in this case), Will be, whether we like it or not, R^2/2.
Second, not being overtly presumptuous, I had rewritten some of the conditions, Keeping as you suggested r=R constant.
Their new form would only constitute:
dS^*=2\pi R^2sin(\theta)d\theta.
And the distance Vector r':
r'=\sqrt{R^2+r0^2-2*R*r0*cos(\theta)}.
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

 

[gabbagabbahey]

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 rdr(as it is in this case), Will be, whether we like it or not, R^2/2.
Second, not being overtly presumptuous, I had rewritten some of the conditions, Keeping as you suggested r=R constant.
Their new form would only constitute:
dS^*=2\pi R^2sin(\theta)d\theta.
And the distance Vector r':
r'=\sqrt{R^2+r0^2-2*R*r0*cos(\theta)}.
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

You seem to be very confused about surface integrals. The form of dS 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 \theta and \phi vary. (At one point on the surface, \theta may be zero and \phi=\pi, while at another point on the surface you could have \theta=\phi=\pi/2, for example). The differential are element dS in this case will be the area subtended when you vary \theta and \phi by infinitesimal amounts d\theta and d\phi. There are many sites and calculus texts where that quantity is derived, and the well-known result is that dS=r^2\sin\theta d\theta d\phi.

 

[danielakkerma]

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 \int d\phi will be perpetually 2*\pi, no matter how we may spin it around.
The only reason why I wrote, instantaneously that dS^*=2\pi R^2sin(\theta)d\theta, was that I had the "clairvoyance" to integrate d\phi 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

 

[gabbagabbahey]

If you are having problems with integrating over \theta, just make the substitution u=R^2+r^2+2rR\cos\theta. The \sin\theta in the numerator makes the integral trivial.

 

[danielakkerma]

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:
\varphi=2 \sqrt{2} k\pi r0 \sigma
(where k=\frac{1}{4 \pi \epsilon _0}).
However, If I were to integrate and find E first, the elucidation would amount to:
E= \int dE= -\pi k \sigma.
Clearly by Integrating -\int Edr, 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

 

[gabbagabbahey]

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:
\varphi=2 \sqrt{2} k\pi r0 \sigma
(where k=\frac{1}{4 \pi \epsilon _0}).
However, If I were to integrate and find E first, the elucidation would amount to:
E= \int dE= -\pi k \sigma.
Clearly by Integrating -\int Edr, 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

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

\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.

You always take the positive root ( \sqrt{\alpha^2}=|\alpha| )

 

[danielakkerma]

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)
2 \pi R^2 \int^\frac{\pi}{2}_0 \frac{sin(\theta)}{\sqrt{R^2+r0^2-2Rr0cos(\theta)}}=
2 \pi \simga R^2*(\frac{\sqrt{R^2+r0^2}}{2Rr0} - \frac{\sqrt{(R-r0)^2}}{2Rr0})
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 \sqrt{2}...
What now?
Thank you very much again,
Daniel
P.S, it's a trifle to evaluate E:
\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})
The first two elements of \hat {r}(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?

 

[danielakkerma]

Any further advice guys?, all aid will be very appreciated!
Thanks again,
Daniel

 

[danielakkerma]

Anything folks?
Thanks,
Daniel