I don't understand Gauss' law

  • Thread starter hatsoff
  • Start date
  • #1
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Homework Statement



A hollow spherical shell carries charge density [tex]\rho=k/r^2[/tex] in the region [tex]a\leq r\leq b[/tex]. Use Gauss' Law in integral form to find the electric field in three regions: (i) r<a, (ii) a<r<b, (iii) r>b.

Homework Equations



Gauss' Law in integral form: [tex]\oint_{\text{surface}}\textbf{E}\cdot d\textbf{a}=\frac{1}{\epsilon_0}Q_{\text{enc}}[/tex]

where [tex]Q_{\text{enc}}=\int_{\text{volume}}\rho\;d\tau[/tex] is the total charge enclosed within the surface.



The Attempt at a Solution



I'm looking at (ii) to start. Due to symmetry we have [tex]\textbf{E}=|\textbf{E}|\hat r[/tex] and so we can pull this out of the left term of the integral in Gauss' law. But I'm confused on how to compute Qenc. I would expect to take the radius part of the integral from a to b, this way:

[tex]Q_{\text{enc}}=\int_{\text{volume}}\rho\;d\tau= \int_{\text{surface}}\int_a^b \rho\;dr\;d\textbf{a}[/tex]

However the solutions manual suggests that I should leave r as a variable and take the integral from a to r, this way:

[tex]Q_{\text{enc}}=\int_{\text{volume}}\rho\;d\tau= \int_{\text{surface}}\int_a^r \rho\;d\bar r\;d\textbf{a}[/tex]

I don't understand this. Why do we get to ignore the field between r and b?

Thanks.
 

Answers and Replies

  • #2
BruceW
Homework Helper
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well part ii) is asking for the electric field between a and b. So where will the Gaussian surface extend to?
 
  • #3
20
1
well part ii) is asking for the electric field between a and b. So where will the Gaussian surface extend to?
OH, I see now!

That was silly.

Thanks.
 
  • #4
BruceW
Homework Helper
3,611
119
haha, no worries!
 

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