How Does Gauss' Law Apply to a Non-Uniformly Charged Hollow Sphere?

[hatsoff]

Homework Statement

A hollow spherical shell carries charge density \rho=k/r^2 in the region a\leq r\leq b. 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: \oint_{\text{surface}}\textbf{E}\cdot d\textbf{a}=\frac{1}{\epsilon_0}Q_{\text{enc}}

where Q_{\text{enc}}=\int_{\text{volume}}\rho\;d\tau 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 \textbf{E}=|\textbf{E}|\hat r 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 Q_enc. I would expect to take the radius part of the integral from a to b, this way:

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

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

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

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

Thanks.

 

[BruceW]

well part ii) is asking for the electric field between a and b. So where will the Gaussian surface extend to?

 

[hatsoff]

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.

 

[BruceW]

haha, no worries!