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Gauss's law spherical shell

  1. Dec 27, 2013 #1
    Just a quick general question in applying Gauss's law. Not exactly homework, more a general question so I can understand my other homeworks better.

    I have a spherical shell with inner radius [itex]R_1[/itex] and outer radius [itex]R_2[/itex] and a point charge [itex]Q[/itex] in its center. It is NOT a conducting sphere. In the region [itex]R_1 < r < R_2[/itex] there is another constant charge density [itex]\rho_0[/itex]. So total charge density could be expressed as:

    [itex]\rho(\vec{r}) = Q \delta(\vec{r}) + \rho_0 \Theta(r-R_1) \Theta(R_2-r)[/itex]

    Gauss's law:

    [itex]\int_{\partial V} \! \vec E \, d\vec{S} = \frac{1}{\epsilon_0} \int_V \! \rho(\vec{r}) \, d^3r[/itex]

    The right hand side is what interests me.

    I have to look at 3 different areas obviously.

    [itex]r < R_1[/itex]: In this are total charge is simply Q.

    The next part is where I'm insecure though.

    [itex]R_1 < r < R_2[/itex]:

    Is it [itex]\frac{4 \pi \rho_0 (r^3-R_1^3)}{3}[/itex] or is it [itex]Q + \frac{4 \pi \rho_0 (r^3-R_1^3)}{3}[/itex]?

    Does the point charge in the center add up or not for the total charge?
     
  2. jcsd
  3. Dec 27, 2013 #2

    TSny

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    Think about what the integral ##\int_V \! \rho(\vec{r}) \, d^3r## represents. It takes each volume element ##d^3r## within the volume V, multiplies by the charge density at the location of that volume element, and then adds this up for each volume element withing the volume V. In other words, it just sums up the charge within each volume element contained in the volume V. Each volume element within V is only included once.
     
  4. Dec 27, 2013 #3
    Okay. That reasoning I can follow. But here is the part I don't get. Outside the shell, for the region [itex]r > R_2[/itex], there is no charge density. There aren't any charges. Then the volume integral should be 0, right? There is only a charge inside and in the shell. Yet the result outside isn't 0, is it? When calculating the gravitational field outside of a planet in mechanics it isn't 0 either. It's the whole mass of the planet. So the equivalent result for the outside region here should be: [itex]Q + \frac{4 \pi \rho_0 (R_2^3-R_1^3)}{3}[/itex]

    ?
     
  5. Dec 27, 2013 #4

    TSny

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    When using Gauss's law, you have to pick the "Gaussian surface" through which you are calculating the flux (the left side of Gauss' law). The right hand side is then the total charge enclosed by the Gaussian surface divided by ##\epsilon_0##.

    Do you have a particular Gaussian surface in mind?
     
  6. Dec 27, 2013 #5
    I have a sphere in mind!
     
  7. Dec 27, 2013 #6

    TSny

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    Of what radius?
     
  8. Dec 27, 2013 #7
    Hmm, [itex]r > R_2[/itex] I guess so it encloses my entire spherical shell? Then [itex]Q + \frac{4 \pi \rho_0 (R_2^3-R_1^3)}{3}[/itex] is correct (total charge inside + total charge in the shell)?
     
  9. Dec 27, 2013 #8

    TSny

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    Yes, that's right if you choose r > R2.
     
  10. Dec 27, 2013 #9
    But then again why isn't Q included in region 2 as well? In that case I have a Gaussian spherical surface with radius [itex]R_1 < r < R_2[/itex] in mind. That does include the point charge Q in the center as well?

    Can you somewhat understand my understanding problem?
     
  11. Dec 27, 2013 #10

    TSny

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    As long as your Gaussian surface encloses the point charge Q, Q will be included in the charge enclosed by the Gaussian surface :wink:

    In a setup like this, you generally use Gauss' law to determine the electric field at some point P. The location of P will determine the choice of the Gaussian surface. If you want to find E for a point P located at distance r from Q, then you would choose a spherical Gaussian surface of radius r. The right hand side of Gauss' law would just be the total charge enclosed withing that spherical surface. Since Q would definitely be inside the enclosed region, you would need to include Q in the total charge enclosed.
     
  12. Dec 27, 2013 #11
    So to come back to my initial question, in region 2 it would be [itex]Q + \frac{4 \pi \rho_0 (r^3-R_1^3)}{3}[/itex]? :redface:
     
  13. Dec 27, 2013 #12

    TSny

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    Yes, if you pick the radius r to lie in region 2, then that would be the correct expression for the charge enclosed.
     
  14. Dec 27, 2013 #13
    Thank you. :smile:
     
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