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Solve Spherical Shell Gauss Problem with Differential Form Only

  1. Mar 17, 2013 #1
    1. The problem statement, all variables and given/known data
    A hollow spherical shell carries charge density [itex]\rho=\frac{k}{r^2}[/itex] in the region [itex]a<=r<=b[/itex], where a is the inner radius and b is the outer radius. Find the electric field in the region [itex]a<r<b[/itex].
    I'm not allowed to use integral form of Gauss's law, must use differential form.


    2. Relevant equations
    Relating charge density to electric field divergence [itex]\vec{∇}\cdot\vec{E}=\frac{\rho}{\epsilon_{0}}[/itex]


    3. The attempt at a solution
    Using the integral method, I believe the electric field is [itex]\vec{E}(\vec{r})=\frac{k}{\epsilon_{0}}\frac{r-a}{r^2}\hat{r}[/itex]
    I then went ahead and determined that spherical co-ordinates are the way to go, and that the E-field has only an [itex]\hat{r}[/itex] component and put together
    [itex]\vec{∇}\cdot\vec{E}=\frac{1}{r^2}\frac{∂(r^2\vec{E}(\vec{r}))}{∂r}=\frac{k}{r^2\epsilon_{0}}[/itex]

    By inspection I decided that I should have [itex]\vec{E}(\vec{r})=\frac{k}{\epsilon_{0}r}\hat{r}[/itex]. While this does satisfy [itex]\vec{∇}\cdot\vec{E}=\frac{\rho}{\epsilon_{0}}[/itex], it is not in agreement with the integral form.

    So my question is, am I using these relationships incorrectly? I'm very confused, since when I integrate ∇E from a to r, I get the same answer as the Gauss integral form, but not sure why this could be, since I understand that the integral of ∇E is equal to the charge enclosed by a Gaussian surface. Since I'm just interested in the E-field at a point in the shell, why would I need to integrate anyway?
     
  2. jcsd
  3. Mar 17, 2013 #2

    TSny

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    Hello, gavman.

    By inspection you found a particular solution to the differential equation. You can add to it any solution to the associated homogeneous differential equation. Then you can satisfy the appropriate boundary condition at r = a.
     
  4. Mar 18, 2013 #3
    Hi TSny
    Ok, I have [itex]\frac{k}{\epsilon_{0}r^2}[/itex] as a solution to the associated DE. To get this I used ∇E=0 then substituted my particular solution for E. Do I then just state that at the boundary, E=0 and then build an equation from my two E(r) equations which satisfies this condition? I'm just not sure if I have made an unreasonable assumption.
    thanks.
     
  5. Mar 18, 2013 #4

    TSny

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    Yes, that sounds right. Of course, you are invoking the boundary condition that E = 0 at r = a and you might ask where that comes from. It's easy to use Gauss' law to get E = 0 everywhere r<a. But since you are not to use Gauss' law, I'm not sure how you are supposed to get the boundary condition at r= a.
     
  6. Mar 18, 2013 #5

    ehild

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    In the inner range (r<a), ρ=0. The solution of ∇E=0 is E=C/r2, C is a constant. But there is no singularity at r=0, so C must be zero.

    ehild
     
  7. Mar 18, 2013 #6

    TSny

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    Ah. Good. That nails the boundary condition.
     
  8. Mar 18, 2013 #7
    Great, thanks for your help guys, appreciate it.
     
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