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Homework Help: Ohm's Law in a sphere with current density

  1. Mar 18, 2010 #1
    1. The problem statement, all variables and given/known data
    Two concentric metal spherical shells of radius a and b, respectively, are separated by a weakly conducting material of conductivity sigma.
    (a) If they are maintained at a potential difference V, what current flows from one to the other?
    (b) What is the resistance between the shells?
    (c)Notice that of b>>a the outer radius (b) is irrelevant. How do you account for that? Exploit this observation to determine the current flowing between the two metal spheres, each of radius a, immersed deep in the sea and held quite far apart, if the potential difference between the is V.


    2. Relevant equations
    Spherical Gradient:
    [tex]\nabla t=\frac{\partial t}{\partial r}\hat{r}+\frac{1}{r}\frac{\partial t}{\partial\theta}\hat\theta+\frac{1}{r\sin\theta}\frac{\partial t}{\partial\phi}\hat\phi[/tex]
    The rest are attached to the homework itself.


    3. The attempt at a solution
    Attached. I attached it having typed it up once so the prof could read it. The negative current in part a doesn't bother me as it is a vector in this case. That r in all of the answers and the negative answers in part b and c do.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Mar 18, 2010 #2

    vela

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    You're right that the current shouldn't depend on r. Your mistake was when you went from

    [tex]\nabla V = \frac{\partial V}{\partial r} \hat r[/tex]

    to

    [tex]\nabla V = \frac{\Delta V}{\Delta r} \hat r = \frac{V}{b-a}\hat r[/tex]

    You've assumed the electric field is constant, but it's not.

    Your approach is a bit backwards. You're trying to go from V to E to J to I. Instead, try assuming you have a current I flowing between the two shells and going the other direction.
     
  4. Mar 18, 2010 #3
    That works much better. The two currents and resistance are not dependant on position and they are not negative.
    I decided to assume a charge Q instead of a potential V and found the E-field using Gauss's Law in integral form.
    1a) [tex]\vec{I}=\frac{\sigma Q}{\epsilon}\hat{r}[/tex]
    1b) [tex]R=\frac{V\epsilon}{\sigma Q}=\frac{1}{\epsilon_r4\pi a\sigma}[/tex]
    1c) [tex]I=\frac{\sigma Q}{2\epsilon}=2Va\epsilon_r\pi\sigma[/tex]
     
  5. Mar 18, 2010 #4

    vela

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    That's not correct. For one thing, R should depend on b.

    If there's a current I flowing from the inside sphere to the outside sphere, what is the current density j at the surface of the inside sphere, at the surface of the outside sphere, and at some distance r between the two spheres?
     
  6. Mar 19, 2010 #5
    As noted in the attachment:
    [tex]\vec{J}=\frac{Q\sigma}{4r^2\pi\epsilon}[/tex]
    But the current is constant through any sphere between a and b, inclusive. Failure to do so suggests that the current curls. But because the current follows the E-field, which doesn't curl in this case, and there is no B-field, the current can't curl either.
    [tex]\nabla\times\vec{E}=0\Rightarrow\nabla\times\vec{J}=0[/tex]

    I used the electric displacement because the E-field is going through a material of unknown nature other than "weakly conductive" which made me think that the electric permittivity of the material might be important to the results.

    According to my work, which uses Gauss's Law in integral form, the presence of b is wholly irrelevant because it does not effect the E-field or D-field internal to itself.
     

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  7. Mar 19, 2010 #6

    ideasrule

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    No vector calculus--or vectors for that matter--is needed for this problem. Are you familiar with the equation E=rho*J where rho is the resistivity? Since you know that J=I/(4*pi*r^2), you can easily find the relationship between V and I.
     
  8. Mar 19, 2010 #7

    vela

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    You've contradicted yourself. If E doesn't depend on b as you have claimed, then surely the potential difference V between the two spheres does. A larger separation between them but with the same E field will result in a larger V.

    In any case, you're making this problem unnecessarily complicated. There's no need to use Gauss's law. You have a current I which is spread out over a sphere of radius r. What is the current density J in terms of I and r? (This is what I was getting at with my questions.) To find the potential difference between the spheres, integrate E along a path that goes from one sphere to the other.
     
  9. Mar 19, 2010 #8
    I am familiar with it in the form of [tex]\vec{J}=\sigma\vec{E}[/tex] where sigma is the conductivity which is equal to the multiplicative inverse of rho, which was used. I'm not sure what you are purposing as I don't have the current density to find the current with. The current density was found from the electric displacement and it's relation to the electric field.

    Actually, I accidentally made it simpler than it is. 1a is correct, but 1b is not. The use of [tex]V=\frac{RQ}{\epsilon_0A}[/tex] for the whole V is a bad choice because it is only true at a and not true of the difference between a and b. As I didn't use potential to find the current, I should use the electric field which was derived from charge to find the potential difference that will be affected by a and b.

    The current in 1a is unchanged as [tex]\frac{Q\sigma}{\epsilon}\hat{r}[/tex]
    The potential difference is [tex]\frac{Q(a-b)}{4ab\epsilon\pi}[/tex]
    The resistance is [tex]\frac{b-a}{4\pi\sigma ab}[/tex]
    The current in 1c is [tex]I=2Va\sigma\pi[/tex]
     

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  10. Mar 19, 2010 #9

    vela

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    You don't have Q either, but that didn't stop you from using it. It didn't matter because it cancels out in the end. The same thing happens with I.

    You never answered 1a. You still haven't indicated what current would flow if the potential difference is V. Your answer is still in terms of unknowns like Q.
     
  11. Mar 19, 2010 #10
    There is nothing I can do to remove the electric permittivity of the "weak conductor." http://en.wikipedia.org/wiki/Permittivity#Classification_of_materials" says that a material is considered a poor conductor when "[tex]\frac{\sigma}{\omega\epsilon}<<1[/tex]" meaning the electric permittivity could be large. Yes, I do realize that that relation is for AC which I don't have. The characterization of the material as a "weak conductor" says that the electric permittivity may have a significant affect on the answer.

    But [tex]V=\frac{Q}{4a\pi\epsilon_0}\Rightarrow Q=4aV\pi\epsilon_0[/tex] from that previous example in the book. PnP (Plug and Play). [tex]\frac{4aV\pi\epsilon_0\sigma}{\epsilon}=4aV\pi\sigma\epsilon_r=I[/tex]
     
    Last edited by a moderator: Apr 24, 2017
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