# Electrical potential of a conducting spherical shell

1. May 16, 2016

### JulienB

1. The problem statement, all variables and given/known data

Hi everybody! I would like to clear up some doubts I have about my electromagnetism homework:

A positive point charge $q$ is placed in the center of an ideal conducting electrically neutral spherical shell, as shown in the attached picture.
a) Calculate the electrical potential $\phi(\vec{r})$ in all locations, that is $r < a$, $a < r < b$ and $r > b$.
b) Sketch the result.

2. Relevant equations

Gauss's law, equation for electrical potential

3. The attempt at a solution

First I have been attaching a second picture showing how I understand the situation. I think the point charge at the center attracts at the inner border of the shell its negative equivalent of charge and repelling the same amount of positive charge to the outer border. Is that a correct assumption? If yes, I assume there is an electric field going from the point charge $q$ towards the inner border of the shell and from the outer border of the shell outwards (as indicated in the picture).
Now for the math:

- For the case $r_1 < a$ :
$\oint \vec{E} \cdot d\vec{A} = \frac{q}{\varepsilon_0}$
$\iff E \int dA = \frac{q}{\varepsilon_0}$
$4 E \pi r_1^2 = \frac{q}{\varepsilon_0}$
$E = k \frac{q}{r_1^2}$
$\implies \phi (\vec{r}) = - \int_{\infty}^{r_1} \vec{E} \cdot d\vec{s}$
$= kq \int_{\infty}^{r_1} \frac{-1}{r^2} ds$
$= \frac{kq}{r_1}$

Is that correct? I'm not sure about the boundaries of my integral, since the path goes through the shell then.

I will wait for an answer before posting the two other results :)

Julien.

#### Attached Files:

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• ###### Photo 16-05-16 14 10 23.jpg
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2. May 16, 2016

### Staff: Mentor

Correct.
... where the electric field is different.
While the result is not directly wrong (the calculation is), using it would lead to a non-zero potential at infinity. Usually the potential is chosen to be zero at infinity. To fix that (later), you can add a constant to the potential, or start from the outside.

3. May 16, 2016

### JulienB

@mfb Thanks for your answer, that is what I was fearing. I could also start by calculating the potential for $r < b$, then $a < r < b$ and use those results with $b$ and $a$ instead of $r$ and add up the integrals? Like that:

$\phi(\vec{r}) = - (\int_{\infty}^{b} \vec{E_3} d\vec{s} + \int_{b}^{a} \vec{E_2} d\vec{s} + \int_{a}^{r_1} \vec{E_1} d\vec{s})$

That should work, right?

Julien.

4. May 16, 2016

### Staff: Mentor

That works.

5. May 16, 2016

### JulienB

@mfb Thanks a lot for your answer :) Then I publish my results starting from $r_3$ (for the case $r > b$):

$\oint \vec{E_3} d\vec{A} = \frac{q}{\varepsilon_0}$
$\implies E_3 = k \frac{q}{r_3^2}$

$\implies \phi(\vec{r}) = - \int_{\infty}^{r_3} \vec{E_3} d\vec{s} = kq \int_{\infty}^{r_3} \frac{-1}{r^2} d\vec{s} = k \frac{q}{r_3}$

Then for the case $a < r_2 < b$:

I would say $E_2 = 0$ because the negative charge of the inner surface of the conductive shell corresponds to the positive charge of the point mass at the center.

$\implies \phi(\vec{r}) = -\int_{\infty}^{b} \vec{E_3} d\vec{s} - \int_{b}^{r_2} \vec{E_2} d\vec{s} = k \frac{q}{b} - 0$

And finally for $r_1 < a$:

$E = k \frac{q}{r_1^2}$

$\implies \phi(\vec{r}) = - \int_{\infty}^{b} \vec{E_3} d\vec{s} - \int_{b}^{a} \vec{E_2} d\vec{s} - \int_{a}^{r_1} \vec{E_1} d\vec{s}$
$= k \frac{q}{b + r_1}$

Does that make sense? Thanks a lot for all your help!

Julien.

6. May 16, 2016

### Staff: Mentor

The last term is wrong, everything up to that is fine.

7. May 16, 2016

### JulienB

@mfb oops yeah I tried to go too fast. Of course it is:

$\phi(\vec{r}) = k q (\frac{1}{b} + \frac{1}{r_1})$

Thanks a lot.

Julien.

Last edited: May 16, 2016
8. May 16, 2016

### ehild

You can check the potential at r1=a. It should be the same as at r=b, Φ(b) = Φ(a) = kq/b.

9. May 16, 2016

### JulienB

@ehild Thanks for your answer! Mm that's interesting, especially since it doesn't seem to work in my case:

$\phi(\vec{r} = \vec{a}) = kq ( \frac{1}{b} + \frac{1}{a})$ if I take my last corrected equation

$\phi(\vec{r} = \vec{a}) = k \frac{q}{b}$ with the equation for the potential in between the radii.

10. May 16, 2016

### JulienB

Oh I did a big mistake in my last integral:

$\phi(\vec{r}) = -\int_{\infty}^{b} \vec{E_3} d\vec{s} -\int_{b}^{a} \vec{E_3} d\vec{s} - \int_{a}^{r_1} \vec{E_3} d\vec{s} = k \frac{q}{b} + kq \int_{a}^{r_1} \frac{- 1}{r^2} d\vec{s}$
$= k \frac{q}{b} + kq (\frac{1}{r_1} - \frac{1}{a})$

And then your method of checking the result works! Great!

Julien.

11. May 16, 2016

### ehild

It is correct now.

12. May 17, 2016

### JulienB

Great, thanks a lot everybody!

Julien.