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Spherical Charge distribution

  • Thread starter henrybrent
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  • #1
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Homework Statement




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

[itex] \rho(r)=\rho_0(1-\frac{r}{R}) [/itex] for [itex] r\le R[/itex]
[itex] \rho(r)=0 [/itex] for [itex] r > R[/itex]

where [itex] \rho = \frac{3Q}{\pi R^3} [/itex] is a positive constant

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

Homework Equations



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

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 [itex] \rho_0 [/itex] 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..
 
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Answers and Replies

  • #2
954
117
Would you mind showing more details of your working? I'm afraid I can't really tell what problem you are facing.
 
  • #3
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[itex] \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 [/itex]
 
  • #4
954
117
That looks reasonable, just carry on evaluating the integral.
 
  • #5
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Got it!

Q_total = Q that took like 3 hours when it it was, was a very basic integration mistake.
 
  • #6
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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 ?
 
  • #7
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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
 
  • #8
954
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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 learnt about Gauss's law yet?
 
  • #9
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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 learnt 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
 
  • #10
954
117
Yup that is correct; because the quantity you want is the charge enclosed within the spherical Gaussian surface that you are using.
 
  • #11
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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!
 
  • #12
954
117
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!
 
  • #13
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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, im just subbing r into r again for the same thing?


I get [itex] \frac{4 \pi*p_0( \frac{r^3}{3} - \frac{r^4}{4R}){\episilom} =E4\pi*r^2 [\itex]
 
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  • #14
954
117
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
[tex]Q_{enc} = \int_{0}^{r} \rho_{0}\left(1 - \frac{r'}{R}\right) 4 \pi r'^{2} dr'[/tex]
to 'distinguish' the two of them.
 
  • #15
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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
[tex]Q_{enc} = \int_{0}^{r} \rho_{0}\left(1 - \frac{r'}{R}\right) 4 \pi r'^{2} dr'[/tex]
to 'distinguish' the two of them.
[itex] \frac{4 \pi*p_0( \frac{r^3}{3} - \frac{r^4}{4R})}{\epsilon_0} =E4\pi*r^2 [/itex]

That's where I'm at? not sure how to progress.
 
  • #16
954
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Yeah, you're there already, just make E the subject!
 
  • #17
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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.
 
  • #18
954
117
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 [itex]\frac{Q}{4\pi R^{2}}[/itex]
 
  • #19
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Silly? How so? Quick way to check your answer is to set r = R, you should recover the expected [itex]\frac{Q}{4\pi R^{2}}[/itex]
I'm still all over the place.


Right,

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

Can't I just take p_0 out of the integral and then divide by [itex]4\pi r^2[/itex] ?
I'm still not getting the right answer!
 
  • #20
954
117
The answer you posted earlier, [tex]\mathbf{E} = \frac{\rho_0( \frac{r}{3} - \frac{r^2}{4R})}{\epsilon_0} \hat{r}[/tex]
(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.
 
  • #21
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The answer you posted earlier, [tex]\mathbf{E} = \frac{\rho_0( \frac{r}{3} - \frac{r^2}{4R})}{\epsilon_0} \hat{r}[/tex]
(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 occured to me, hold on, why can't I just divide by [itex]4\pi r^2[/itex] it saves multiplying out.
 
  • #22
954
117
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.
 
  • #23
57
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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.
 

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