# I don't understand Gauss' law

## 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 Qenc. 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.

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

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
Homework Helper
haha, no worries!