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

[hatsoff]

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 Q_enc. 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.

 

[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!

 

[asteriatic]

Hello,

Thank you for your question. Gauss' law can be a bit confusing at first, but I will try to explain it in a simple and clear way.

First, let's review the concept of Gauss' law. It states that the electric flux through a closed surface is equal to the total charge enclosed by that surface divided by the permittivity of the medium. This can be written mathematically as:

\oint_{\text{surface}}\textbf{E}\cdot d\textbf{a}=\frac{1}{\epsilon_0}Q_{\text{enc}}

where Q_{\text{enc}} is the total enclosed charge and \epsilon_0 is the permittivity of free space.

Now, let's look at the problem at hand. We have a hollow spherical shell with a charge density \rho=k/r^2 in the region a\leq r\leq b. This means that the charge density is not uniform, but varies with the distance from the center of the sphere. In order to find the electric field in the three regions, we need to use Gauss' law in integral form.

In region (ii), a<r<b, we have a spherical surface of radius r. The electric field at any point on this surface will be directed radially outwards, as you correctly stated. This means that we can write \textbf{E}=|\textbf{E}|\hat r, where \hat r is the unit vector in the radial direction. Now, let's look at the left-hand side of Gauss' law. We have an integral over the surface, which means that we need to consider all points on this surface. However, since the electric field is constant in magnitude and direction on this surface, we can pull it out of the integral. This is why we have |\textbf{E}| as a factor on the left-hand side of the equation.

Now, let's look at the right-hand side of Gauss' law. We have Q_{\text{enc}}=\int_{\text{volume}}\rho\;d\tau, which is the integral of the charge density over the enclosed volume. In this case, the enclosed volume is the region between r and b, so we need to integrate the charge density from r to b. This is why the solution manual suggests using \int_a^r \rho\;d\bar r instead of \