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Homework Help: Determine n when you know the charge density (Electromagnetism)

  1. Mar 19, 2010 #1
    1. The problem statement, all variables and given/known data
    You are informed that the charge density is:

    [tex]$\rho \left( r \right)=k{{r}^{-n}},r<a$[/tex],
    where k is a constant and n is a whole number.

    Determine n.

    2. Relevant equations

    Besides that, I know that:

    [tex]\[E=\frac{Q}{4\pi {{\varepsilon }_{0}}}\frac{1}{{{r}^{2}}},r<a\][/tex]

    (E is spherically distributed btw)


    3. The attempt at a solution
    When I have the charge density I know that:

    [tex]dQ=4\pi {{r}^{2}}\rho \left( r \right)$[/tex]

    So by integrating from 0 to a that I get that:

    [tex]Q=\frac{4\pi {{r}^{3-n}}k}{3}$[/tex]

    Gauss' Law states (For my case):

    [tex]E\left( r \right)\left( 4\pi {{r}^{2}} \right)=\frac{Q}{{{\varepsilon }_{0}}}$[/tex]

    So when I put this together I just get the one side equal the other side :S

    And I know that n = 1.

    What am I doing wrong ?
     
  2. jcsd
  3. Mar 19, 2010 #2

    kuruman

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    How do you know that this is the case? Is that given by the problem? What happens at r = 0?
     
  4. Mar 19, 2010 #3
    Yes, it's given in the problem.

    I also know that:

    [tex]\[E=\frac{Q}{4\pi {{\varepsilon }_{0}}}\frac{1}{{{r}^{2}}},r>a\][/tex]

    Oh, and of course, my mistake, for E when r < a (As I already said) is:

    [tex]\[E=\frac{Q}{4\pi {{\varepsilon }_{0}}}\frac{1}{{{a}^{2}}},r<a\][/tex],
    not r2.

    And Q > 0 always.
     
  5. Mar 19, 2010 #4

    gabbagabbahey

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    Double check that integration :wink:
     
  6. Mar 19, 2010 #5

    gabbagabbahey

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    That looks reasonable. Assuming, of course that [itex]Q[/itex] represents the total charge of the distribution.

    What happens when you take the divergence of this field?
     
  7. Mar 19, 2010 #6
    Exhange r3 with a3 ?

    Hmmm, after some checking I get:

    [tex]Q=\frac{-4\pi {{a}^{3-n}}k}{n-3}$[/tex]

    But does that change anything since Q is on both sides ?
     
    Last edited: Mar 19, 2010
  8. Mar 19, 2010 #7
    Well, for a -> 0 E becomes 0, if thats what you mean ?
     
    Last edited: Mar 19, 2010
  9. Mar 19, 2010 #8

    gabbagabbahey

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    No, that's not what I mean. I meant exactly what I said: calculate the divergence of the electric field you are given...what do you get? Compare your result to Gauss' Law in differential form.
     
  10. Mar 19, 2010 #9

    kuruman

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    Denver Dang, I think the time has come for you to post the entire problem exactly as it is given to you. Then we will sort out what is given, what is known and what is assumed. If it is just what your first posting says, then you cannot say that you know that the field inside is either Q/4πεr2 or Q/4πεa2; you need to derive the field inside using Gauss's Law correctly.
     
  11. Mar 19, 2010 #10
    Ok, sorry it's so confusing...

    Here goes:

    Consider a spherically charge distribution with radius a. The charge distribution gives an electric field that is directed radially from the distribution center. In the distance r from the center the electric field has the strength:

    [tex]\[E=\frac{Q}{4\pi {{\varepsilon }_{0}}}\frac{1}{{{a}^{2}}},r<a\][/tex]

    and

    [tex]\[E=\frac{Q}{4\pi {{\varepsilon }_{0}}}\frac{1}{{{r}^{2}}},r>a\][/tex]

    where Q is the total charge: Q >0. The dielectricconstant K is 1 everywhere.

    You are informed that the charge density is:

    [tex]$\rho \left( r \right)=k{{r}^{-n}},r<a$[/tex],
    where k is a constant and n is a whole number.

    Now determine n.
     
  12. Mar 19, 2010 #11

    kuruman

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    Suppose you constructed a spherical Gaussian surface of radius r < a. Can you use Gauss's Law to find the electric field? It must have a dependence on n. Then, since the electric field inside is ostensibly
    [tex]
    \[E=\frac{Q}{4\pi {{\varepsilon }_{0}}}\frac{1}{{{a}^{2}}}
    [/tex]
    you can set it equal to the expression that you found from Gauss's Law and solve for n. That is probably what the author of this problem intended you to do.

    Warning: Making sense of the answer is another issue because this does not appear to be a well-posed problem. As gabbagabbahey indicated, what is the divergence of E inside the spherical distribution and how does it relate to the charge density?
     
  13. Mar 20, 2010 #12
    If I do what you say, won't I just end up with what I usually posted ?
    I mean, by Gauss Law I have:

    [tex]E\cdot A=\frac{{{Q}_{encl}}}{{{\varepsilon }_{0}}}$[/tex]

    Qencl I can find by saying:

    [tex]{{Q}_{encl}}=\rho \left( r \right)V=\left( k{{r}^{-n}} \right)\left( \frac{4}{3}\pi {{r}^{3}} \right)=\frac{4\pi {{r}^{3-n}}k}{3}$[/tex]

    And when I put that into Gauss Law, I can an electric field of:

    [tex]\[E=\frac{{{r}^{1-n}}k}{3{{\varepsilon }_{0}}}\][/tex]

    Setting that equal to the expression I know, I get, well nothing I can get even close to 1 :S


    And I haven't had Vectoranalysis yet, so I'm not quite sure what you mean about the divergence tbh...
     
  14. Mar 20, 2010 #13

    kuruman

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    That's almost what you should get given what the problem says. If you are looking for the E field inside, you need to find the charge enclosed by the Gaussian surface, not the total charge in the distribution. What is the charge enclosed by a shell of radius r < a?
    Note that the electric field that the problem claims to be the correct expression inside is a constant and does not depend on r. Assuming that you get the correct expression for the field inside, what must n be to get a constant field inside?
    OK, forget the divergence if you haven't had that yet. Answer my question above and see what form the electric field inside takes for the value of n that makes it a constant. You will see that it is not what the problem claims it to be. In other words, the field inside cannot be what the problem claims. That's what I meant earlier when I said it is an ill-posed problem.
     
  15. Mar 20, 2010 #14
    Isn't the charge enclosed by the Gaussian surface the Qencl I found ?
     
  16. Mar 20, 2010 #15

    kuruman

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    No it is not. You need to redo the integral
    [tex]\int_0^r \rho (r) 4 \pi r^2 dr[/tex]
    correctly. Neither of the two expressions you have shown above is correct.
     
  17. Mar 21, 2010 #16
    Oh, my mistake...

    I get:

    [tex]\[{{Q}_{encl}}=-\frac{4\pi {{r}^{3-n}}k}{n-3}\][/tex]

    So now what ? Inserting it into:

    [tex]EA=\frac{{{Q}_{encl}}}{{{\varepsilon }_{0}}}$[/tex] ?

    And then setting it equal the other E I have ? Won't that just give me the one side equals the other again ?
     
    Last edited: Mar 21, 2010
  18. Mar 21, 2010 #17

    kuruman

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    First use

    [tex]
    EA=\frac{{{Q}_{encl}}}{{{\varepsilon }_{0}}}$
    [/tex]

    to find E as a function of r inside the distribution. What do you get?
     
  19. Mar 21, 2010 #18
    I get:

    [tex]\[E=-\frac{k{{r}^{1-n}}}{\left( n-3 \right){{\varepsilon }_{0}}}\][/tex]
     
  20. Mar 21, 2010 #19

    gabbagabbahey

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    Right, so comparing that to the field inside the distribution that you you are given in the problem statement, you get

    [tex]-\frac{kr^{1-n}}{\left( n-3 \right)\varepsilon_0}=\frac{Q}{4\pi\epsilon_0}\frac{1}{a^2}[/tex]

    You can solve this both for [itex]n[/itex] and for [itex]Q[/itex]
     
  21. Mar 21, 2010 #20
    But how will I end up with n = 1 in that case ?
    I have 4 unknowns: k, r, Q and a. So I will not be able to remove them from the equation no matter what I do. I just end up with an expression that involves those 4 unknowns. Or am I missing something :S
     
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