Electric potential of a hollow sphere

[fluidistic]

Homework Statement

Consider a radial charge distribution of the form $$\rho = \rho _0 \frac{b}{r}$$, for $$a \leq r \leq b$$. Find the potential for all r.

Homework Equations

Not sure.

The Attempt at a Solution

From my memory (and I didn't find it in wikipedia), the potential at a point (x,y,z) is given by $$\phi (x,y,z)= \oint \frac{\rho (x',y',z') \vec r dV'}{r^2}$$ where (x',y',z') is the place where there's charges and $$\vec r$$ is the vector unifying (x,y,z) with (x',y',z'), though I'm only 43% sure.
So I set up $$\int _0 ^{2\pi} \int _0^{\pi} \int _a^b \rho _0 \frac{b}{r} \cdot r^2 \sin \phi dr d\phi d\theta=2\pi \rho _0 b (b^2-a^2)$$, which unfortunately does not depend on r... I'm 100% sure I made something wrong. Can you help me?
Thank you.

 

[ideasrule]

I don't think that equation is right.

If you can find the electric field for all r, you can easily find the potential (just integrate).

 

[fluidistic]

I don't think that equation is right.

If you can find the electric field for all r, you can easily find the potential (just integrate).

Ok. Should I take into account a negative sign? That is, $$\vec E=- \nabla V \Rightarrow V=-\int _a ^b \vec E \cdot d \vec l$$?. I ask this question because using the negative sign, I never reached the capacitance of a hollow spherical capacitor, precisely because the answer differed by a minus sign (see post #3 of this thread: https://www.physicsforums.com/showthread.php?t=363343).

 

[ideasrule]

In the other thread, you didn't need to worry about sign issues. Only the absolute value of the charge and the absolute value of the voltage difference matter in the definition C=Q/V. It doesn't make sense to talk about the sign on the charge anyhow; all capacitors have one positive and one negative plate.

For this question, yes, you should take into account the negative sign. It tells you that the potential at off-center positions is less than the potential of the center. Of course, you still need to find what the potential of the center is...

 

[fluidistic]

Thank you so much. I'm going to think about this and try my best tomorrow. It's almost 3 am now.

 

[fluidistic]

Here is my work. $$\vec E = \vec \nabla \phi$$. Thus, and I'm not 100% sure, $$\phi (r)= - \int _{\infty}^r \vec E \cdot d\vec l$$.
By Gauss's law, $$E=\frac{kQ_{\text{enclosed}}}{r^2}$$.
So for r<a, Q enclosed is worth 0. For $$a \leq r \leq b$$, it's worth $$\int _a^r \rho (r)dV=\rho _0 b \ln \left ( \frac{r}{a} \right )$$.
Thus $$E=kb \ln \left ( \frac{r}{a} \right ) \cdot \frac{1}{r^2}$$.
Hence $$\phi (r) = -kb \int _{\infty}^r \ln \left ( \frac{r}{a} \right ) \frac{1}{r^2}dr$$, because $$\vec E$$ and $$d\vec r$$ are parallel. Now I'm stuck on the integral. I don't see any u-substitution that could solve it. Maybe a trigonometric one? Also, I doubt I'm right until there! What do you say?

EDIT: and about the equation, you were right, I made an error. It's at page 48: http://books.google.com.ar/books?id...resnum=1&ved=0CAcQ6AEwAA#v=onepage&q=&f=false.

 

[ehild]

Check your volume integral for Q enclosed.

$$Q= \int _a^r \rho (r)dV=\int _a^r {\rho_0\frac{b}{r'}*4 \pi r'^2 dr'}$$

ehild

 

[fluidistic]

Check your volume integral for Q enclosed.

$$Q= \int _a^r \rho (r)dV=\int _a^r {\rho_0\frac{b}{r'}*4 \pi r'^2 dr'}$$

ehild

Oops, you're right. I get $$Q=\rho _0 b 2 \pi (r^2-a^2)$$.
Hence $$E=2\pi b \rho _0 k \left ( 1 -\frac{a^2}{r^2} \right )$$.
So $$\phi (r)=-\int _{\infty}^r \vec E d\vec l = k \rho _0 b 2\pi \int _r^{\infty} \left (1-\frac{a^2}{r^2}\right ) dr$$ which seems to diverge... I don't see what I'm doing wrong.

 

[ehild]

This expression for E is valid for a<r<b, but you need the potential anywhere, so you need E(r) from r=0 to infinity.

For r<a it is easy: There is zero enclosed charge, E=0.

For r>b, the enclosed charge is

$$Q=\rho _0 b 2 \pi (b^2-a^2)$$

and E=kQ/r².

Knowing the electric field everywhere, you can find the potential difference between two arbitrary points. You choose that the potential is zero at infinity. With that condition,

$$- \int _r^{\infty}{\nabla \phi d\vec l}=\int _r^{\infty} {\vec E d\vec l} \rightarrow \phi (r)=k Q/r$$

outside the shell, with Q the total charge of the shell.

The potential at r=b is

$$\phi(b)=kQ/b$$.


For a<r<b,

$$E=2\pi b \rho _0 k \left ( 1 -\frac{a^2}{r^2} \right )$$

You know the potential at r=b. The potential difference between any two points inside the shell is:


$$- \int _A^B{\nabla \phi d\vec l}=\int _A^B {\vec E d\vec l} \rightarrow \phi (A)-\phi (B)= k \rho _0 b 2\pi \int _A^B \left (1-\frac{a^2}{r^2}\right ) dl$$


Choosing B at the sphere of radius b and A at radius r, you get the potential function between a and b.

The potential is constant for r<a, and it is equal to phi(a).

ehild

 

[fluidistic]

Thank you very much ehild, that was very, very helpful. And I could follow you entirely.
I've some basic mathematics related questions I think: Is the electric potential function always a continuous function? If so, then I understand it would justify the step

The potential is constant for r<a, and it is equal to phi(a).

. I know the electric field function can be discontinuous, but I'm not 100% sure of the potential function.
Correct me if I'm wrong, but about the quoted sentence, the potential is constant for r<a because the E field there is null, right?

 

[ideasrule]

I've some basic mathematics related questions I think: Is the electric potential function always a continuous function? If so, then I understand it would justify the step .

Yes, the electric potential function cannot, even in theory, be discontinuous. The reason is that the electric field is the derivative (technically, the gradient) of the potential function, and if the potential function "jumps", the electric field would either be undefined or equal infinity.

Correct me if I'm wrong, but about the quoted sentence, the potential is constant for r<a because the E field there is null, right?

Correct, and that, in turn, is because of the shell theorem.

 

[fluidistic]

Thank you very much ideasrule, I'm done with the thread. (for now at least!)