Electrical potential of a conducting spherical shell

  • Thread starter JulienB
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  • #1
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Homework Statement



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.

Homework Equations



Gauss's law, equation for electrical potential

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 :)

Thanks a lot in advance for your answers!


Julien.
 

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Answers and Replies

  • #2
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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 qq towards the inner border of the shell and from the outer border of the shell outwards (as indicated in the picture).
Correct.
Is that correct? I'm not sure about the boundaries of my integral, since the path goes through the shell then.
... 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
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@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.
 
  • #5
408
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@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
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The last term is wrong, everything up to that is fine.
 
  • #7
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@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:
  • #8
ehild
Homework Helper
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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.
You can check the potential at r1=a. It should be the same as at r=b, Φ(b) = Φ(a) = kq/b.
 
  • #9
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@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
408
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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
ehild
Homework Helper
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1,915
It is correct now. :oldsmile:
 
  • #12
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Great, thanks a lot everybody!


Julien.
 

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