What Is the Total Charge in a Non-Uniform Spherical Charge Distribution?

[henrybrent]

Homework Statement

A Non-Uniform but spherically symmetric charge distribution has a charge density:

$\rho(r)=\rho_0(1-\frac{r}{R})$ for $r\le R$
$\rho(r)=0$ for $r > R$

where $\rho = \frac{3Q}{\pi R^3}$ is a positive constant

Show that the total charge contained in this charge distribution is Q

Homework Equations

$Q_{total} = \int \rho(r)dV$ with limits 0 and R
$dV = 4 \pi r^2 dr$

The Attempt at a Solution

[/B]
I have tried so many solutions it is driving me insane.

Is my dv wrong?

my main method is substituting $\rho_0$ in and then trying to take the constants out of the integral but then I'm stuck with r^3/R or something like that...

This is a 4 mark question, so that usually indicates it's a 4 step process, but this is taking me many steps to get even close..

 

[Fightfish]

Would you mind showing more details of your working? I'm afraid I can't really tell what problem you are facing.

 

[henrybrent]

$\displaystyle\int^R_0 \rho_0(1- \frac{r}{R}) 4 \pi r^2 dr = \displaystyle\int^R_0\ \frac{3Q}{\pi R^3}(1- \frac{r}{R}) 4 \pi r^2 dr$

 

[Fightfish]

That looks reasonable, just carry on evaluating the integral.

 

[henrybrent]

Got it!

Q_total = Q that took like 3 hours when it it was, was a very basic integration mistake.

 

[henrybrent]

That looks reasonable, just carry on evaluating the integral.

I am just wondering, what is the proper definition of r and R?

R is the radius of the sphere? then what is r ?

 

[henrybrent]

That looks reasonable, just carry on evaluating the integral.

Also, how would I go about deriving an expression for electric field in the region r≤R

 

[Fightfish]

R is the radius of the sphere? then what is r ?

r is the variable that describes the distance of the point from the origin of the coordinate system. R is just a constant.

Also, how would I go about deriving an expression for electric field in the region r≤R

Have you learned about Gauss's law yet?

 

[henrybrent]

r is the variable that describes the distance of the point from the origin of the coordinate system. R is just a constant.Have you learned about Gauss's law yet?

For a uniform sphere yes I could find the electric field, but not for a non uniform sphere, that has not been taught to us. I have used Gauss' Law = Q_enc/E_0,
I think the integral limits change from 0 to R, to now 0 to r. But I'm not sure

 

[Fightfish]

Yup that is correct; because the quantity you want is the charge enclosed within the spherical Gaussian surface that you are using.

 

[henrybrent]

Yup that is correct; because the quantity you want is the charge enclosed within the spherical Gaussian surface that you are using.

would you be able to guide me a bit further please?

Atm, I have integral of p0(1-r/R)4pi*r^2 dr all divided by e_0 = E4pi*r^2 limits 0 to r

Sorry it's not in latex I have had to type this in a rush!

 

[Fightfish]

Well it actually doesn't look like you need a lot of guidance, really. You have all the correct ideas - you just need to have a bit more confidence, sit down and evaluate your integrals!

 

[henrybrent]

Yup that is correct; because the quantity you want is the charge enclosed within the spherical Gaussian surface that you are using.

Well it actually doesn't look like you need a lot of guidance, really. You have all the correct ideas - you just need to have a bit more confidence, sit down and evaluate your integrals!

I still don't get it! If my limits are 0 and r, I am just subbing r into r again for the same thing?I get $\frac{4 \pi*p_0( \frac{r^3}{3} - \frac{r^4}{4R}){\episilom} =E4\pi*r^2 [\itex]$

 

[Fightfish]

The r in the integral is a dummy variable, not to be confused with the limit r that defines the size of the Gaussian surface that you have chosen. If you want to be careful, then maybe you want to write
$$Q_{enc} = \int_{0}^{r} \rho_{0}\left(1 - \frac{r'}{R}\right) 4 \pi r'^{2} dr'$$
to 'distinguish' the two of them.

 

[henrybrent]

The r in the integral is a dummy variable, not to be confused with the limit r that defines the size of the Gaussian surface that you have chosen. If you want to be careful, then maybe you want to write
$$Q_{enc} = \int_{0}^{r} \rho_{0}\left(1 - \frac{r'}{R}\right) 4 \pi r'^{2} dr'$$
to 'distinguish' the two of them.

$\frac{4 \pi*p_0( \frac{r^3}{3} - \frac{r^4}{4R})}{\epsilon_0} =E4\pi*r^2$

That's where I'm at? not sure how to progress.

 

[Fightfish]

Yeah, you're there already, just make E the subject!

 

[henrybrent]

Fair enough, I had already done that and then substituted P_0 in and then things got silly. If that's all there is to it then that's fine, it doesn't say I have to include it in terms of Q so i'll leave it as P_0, cheers for the help.

 

[Fightfish]

Fair enough, I had already done that and then substituted P_0 in and then things got silly.

Silly? How so? Quick way to check your answer is to set r = R, you should recover the expected $\frac{Q}{4\pi R^{2}}$

 

[henrybrent]

Silly? How so? Quick way to check your answer is to set r = R, you should recover the expected $\frac{Q}{4\pi R^{2}}$

I'm still all over the place. Right,

$E4 \pi r^2 = \frac{\displaystyle \int \rho_0(1-\frac{r}{R})4\pi r^2 dr}{\epsilon_0}$

Can't I just take p_0 out of the integral and then divide by $4\pi r^2$ ?
I'm still not getting the right answer!

 

[Fightfish]

The answer you posted earlier, $$\mathbf{E} = \frac{\rho_0( \frac{r}{3} - \frac{r^2}{4R})}{\epsilon_0} \hat{r}$$
(I rearranged and added in the direction vector for completeness)
was correct. So I don't kind of get why you are returning to the integral.

 

[henrybrent]

The answer you posted earlier, $$\mathbf{E} = \frac{\rho_0( \frac{r}{3} - \frac{r^2}{4R})}{\epsilon_0} \hat{r}$$
(I rearranged and added in the direction vector for completeness)
was correct. So I don't kind of get why you are returning to the integral.

Well I decided to re-write my method more neatly, then it occurred to me, hold on, why can't I just divide by $4\pi r^2$ it saves multiplying out.

 

[Fightfish]

Ah, because as I mentioned in an earlier post, the r in the LHS (representing the radius of the Gaussian surface) is different from the r under the integral, which is just a dummy variable. If you want to avoid confusion, then you can do what I suggested by replacing the dummy variable r you are integrating over with r' instead, so as to distinguish the two more clearly.

 

[henrybrent]

Ah, because as I mentioned in an earlier post, the r in the LHS (representing the radius of the Gaussian surface) is different from the r under the integral, which is just a dummy variable. If you want to avoid confusion, then you can do what I suggested by replacing the dummy variable r you are integrating over with r' instead, so as to distinguish the two more clearly.

Ofcourse! Apologies for the trivial questions! I have not been thinking straight the last few days! thank you for your help.