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