How Does Gauss's Law Apply to a Sphere with a Radial Charge Distribution?

[roam]

Homework Statement

A sphere of radius R has a volume charge distribution $\rho(r)$ given by:

$\rho(r)= \rho_0 \left( \frac{r}{R_0} \right) \ \ for \ 0 <r<R$
$\rho(r)=0 \ \ elsewhere$

Where $\rho_0$ is a constant. Use Gauss's law to find E field outside the sphere.

Homework Equations

Integral form of Gauss's law

The Attempt at a Solution

I am not sure how to find the enclosed charge. Here's what I did:

$\int \rho \ dV = \int (0) (r^2 \ sin \theta dr d \theta d\phi)$

$\int^R_0 (0) . r^2 dr \int^\pi_0 sin \theta \ d \theta \int^{2 \pi}_0 d \phi = C 4 \pi$

Did I do the integration correctly?

I'm not sure what to do with the constant, I tried to find $E_{in}$ and use boundary conditions:

$\frac{\rho_0}{R} \int r.r^2 \ sin \theta dr d \theta d \phi = \frac{\rho_0 r^4 \pi}{R_0} \implies E_{in} = \frac{\rho_0 r^2}{\epsilon_0 R_0}$

At the boundary R:

$4\pi C = \frac{\rho_0 R}{\epsilon_0} \implies C=\frac{\rho_0 R}{4 \pi \epsilon_0} \implies Q_{enc} = \frac{\rho_0 R}{\epsilon_0}$

So this gives:

$\oint E_{out} da= |E| 4 \pi r^2= \frac{\rho_0 R}{\epsilon_0^2} \implies \therefore E_{out} = \frac{\rho_0 R}{4 \pi \epsilon_0^2 r^2}$

It doesn't look right. Am I using the correct method here to solve the problem? Any explanation would be appreciated.

 

[SammyS]

Homework Statement

A sphere of radius R has a volume charge distribution $\rho(r)$ given by:

$\rho(r)= \rho_0 \left( \frac{r}{R_0} \right) \ \ for \ 0 <r<R$
$\rho(r)=0 \ \ elsewhere$

Where $\rho_0$ is a constant. Use Gauss's law to find E field outside the sphere.

Homework Equations

Integral form of Gauss's law

The Attempt at a Solution

I am not sure how to find the enclosed charge. Here's what I did:

$\int \rho \ dV = \int (0) (r^2 \ sin \theta dr d \theta d\phi)$

$\int^R_0 (0) . r^2 dr \int^\pi_0 sin \theta \ d \theta \int^{2 \pi}_0 d \phi = C 4 \pi$

Did I do the integration correctly?

I'm not sure what to do with the constant, I tried to find $E_{in}$ and use boundary conditions:

$\frac{\rho_0}{R} \int r.r^2 \ sin \theta dr d \theta d \phi = \frac{\rho_0 r^4 \pi}{R_0} \implies E_{in} = \frac{\rho_0 r^2}{\epsilon_0 R_0}$

At the boundary R:

$4\pi C = \frac{\rho_0 R}{\epsilon_0} \implies C=\frac{\rho_0 R}{4 \pi \epsilon_0} \implies Q_{enc} = \frac{\rho_0 R}{\epsilon_0}$

So this gives:

$\oint E_{out} da= |E| 4 \pi r^2= \frac{\rho_0 R}{\epsilon_0^2} \implies \therefore E_{out} = \frac{\rho_0 R}{4 \pi \epsilon_0^2 r^2}$

It doesn't look right. Am I using the correct method here to solve the problem? Any explanation would be appreciated.

The density inside the sphere is not zero.

 

[stedwards]

C seems to be on the wrong side of your equation for the integral. Is C the total charge?

For the integral, the area of a thin spherical shell is $4 \pi r^2$. It's volume is $4 \pi r^2 dr$. The charge within the volume element, $\rho(r) = \frac{r}{R_0}\rho_0$. The charge within the shell is $4 \pi r^2 \rho dr$.

 

[roam]

C seems to be on the wrong side of your equation for the integral. Is C the total charge?

For the integral, the area of a thin spherical shell is $4 \pi r^2$. It's volume is $4 \pi r^2 dr$. The charge within the volume element, $\rho(r) = \frac{r}{R_0}\rho_0$. The charge within the shell is $4 \pi r^2 \rho dr$.

C is a constant of integration. And this is a solid sphere, not a shell.

 

[roam]

The density inside the sphere is not zero.

No, as calculated it, density inside is not 0. But we are looking for density outside.

So do you mean the density outside is the sum density inside plus density outside:

$\frac{\rho_0}{R} \int^R_0 r.r^2 \ sin \theta dr d \theta d \phi + \frac{\rho_R}{r} \int^R_0 r.r^2 \ sin \theta dr d \theta d \phi$

$\frac{\rho_0 r^4 \pi}{R}+ 4 \pi C$

But then it's not possible to solve for C with the boundary conditions in my previous post.

 

[Nathanael]

C is a constant of integration.

There is no "constant of integration" for definite integrals.

And this is a solid sphere, not a shell.

I believe he was suggesting that you integrate over spherical shells to find the charge in the whole volume. This makes the integral a little nicer because it takes advantage of the fact that the charge-density is constant for each shell. But, if you want, you can use a triple integral over spherical coordinates. (Both methods become identical after you calculate the $d \theta d \phi$ integrals.)

No, as calculated it, density inside is not 0. But we are looking for density outside.

Why are you looking for the density outside of R? It's already given that it's zero.

$\frac{\rho_0}{R} \int^R_0 r.r^2 \ sin \theta dr d \theta d \phi + \frac{\rho_R}{r} \int^R_0 r.r^2 \ sin \theta dr d \theta d \phi$

These should be triple integrals, but I'll take that to be a typo.

What are these integrals supposed to mean? You are trying to calculate the enclosed charge, right? Your first integral is reasonable but your second one makes no sense to me at all. Can you explain it?Also I have a question about the problem statement; does R₀=R?

 

[roam]

There is no "constant of integration" for definite integrals.I believe he was suggesting that you integrate over spherical shells to find the charge in the whole volume. This makes the integral a little nicer because it takes advantage of the fact that the charge-density is constant for each shell. But, if you want, you can use a triple integral over spherical coordinates. (Both methods become identical after you calculate the $d \theta d \phi$ integrals.)

These should be triple integrals, but I'll take that to be a typo.

What are these integrals supposed to mean? You are trying to calculate the enclosed charge, right? Your first integral is reasonable but your second one makes no sense to me at all. Can you explain it?Also I have a question about the problem statement; does R₀=R?

Yes $R_0 =R$ radius of sphere.

I chose the integral for $\theta$ to run from 0 to pi, so the integral $\int^\pi_0 \ sin \theta d\theta$ becomes 2. The phi integral runs from 0 to $2 \pi$ which gives just $2 \pi$. I thought this is what we have to do in spherical. What is wrong?

Why are you looking for the density outside of R? It's already given that it's zero.

Because I need an expression for $\rho$ to place in the equation for $E_{out}$.

EDIT: So do you mean I must take the limits of the first integral to be between 0 and R? (i.e. over the entire sphere)

That gives $\rho_{enclosed} = \frac{\pi \rho_0 R^3}{4}$

 

[Nathanael]

I chose the integral for $\theta$ to run from 0 to pi, so the integral $\int^\pi_0 \ sin \theta d\theta$ becomes 2. The phi integral runs from 0 to $2 \pi$ which gives just $2 \pi$. I thought this is what we have to do in spherical. What is wrong?

That's right, but I was saying you should've written it as a triple integral: $\frac{\rho_0}{R} \int\limits^R_0\int_0\limits^{2\pi}\int_0\limits^{\pi} r.r^2 \ sin \theta d \theta d \phi dr$
The way you wrote it just doesn't make any sense but anyway you have the right idea so it's not important.

Can you please explain these integrals: (Especially the second one)

$\frac{\rho_0}{R} \int^R_0 r.r^2 \ sin \theta dr d \theta d \phi + \frac{\rho_R}{r} \int^R_0 r.r^2 \ sin \theta dr d \theta d \phi$

 

[Nathanael]

Sorry, didn't notice your edit.

That gives $\rho_{enclosed} = \frac{\pi \rho_0 R^3}{4}$

By $\rho_{enclosed}$ you mean $Q_{enclosed}$? It's not quite right, but almost. Show the integral that gives you that answer.

(It's hard to follow your methods because you're not really explaining yourself.)

 

[roam]

Sorry, didn't notice your edit.By $\rho_{enclosed}$ you mean $Q_{enclosed}$? It's not quite right, but almost. Show the integral that gives you that answer.

(It's hard to follow your methods because you're not really explaining yourself.)

Sorry, I was in a rush. Here's what I meant:

$Q_{enc} = \int^R_0 \frac{\rho r}{R_0} . r^2 dr . \int^\pi_0 sin \theta \ d \theta . \int^{2 \pi}_0 d \phi = \frac{\pi \rho_0 R^3}{4}$

Why is it not correct? I found the entire contribution due to the whole sphere, as you mentioned there are no charge densities outside it.

 

[Nathanael]

Sorry, I was in a rush. Here's what I meant:

$Q_{enc} = \int^R_0 \frac{\rho r}{R_0} . r^2 dr . \int^\pi_0 sin \theta \ d \theta . \int^{2 \pi}_0 d \phi = \frac{\pi \rho_0 R^3}{4}$

Why is it not correct? I found the entire contribution due to the whole sphere, as you mentioned there are no charge densities outside it.

The integral is correct but that's not what I get when I calculate it. (My answer is different by a factor of 4.)

 

[roam]

The integral is correct but that's not what I get when I calculate it. (My answer is different by a factor of 4.)

How?

Well, the $\phi$ and $\theta$ components become $4 \pi$ together. The r component is:

$\frac{\rho}{R} \int_0^R r.r^2 = \frac{\rho}{R} \left[ \frac{r^4}{4} \right]^R_0 = \frac{\rho R^3}{4} \implies Q_{enc} =\frac{\rho R^3 . 4 \pi}{4} =\pi \rho R^3$

 

[roam]

Alrigh, I got it (I think). Thank you so much for your help.

 

[stedwards]

C is a constant of integration. And this is a solid sphere, not a shell.

No, $4 \pi r^2 dr$ is a volume, there is no constant of integration and $Q={\int_0}^{R_0} 4\pi r^2 \rho(r) dr$ is the total charge, or do it the hard way using angles.