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Given a current, calculate the charge distribution

  1. Aug 2, 2015 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    A ring of radius R has a current density ##\vec J=J(r, \theta) \sin \phi \hat \phi## where phi is the azimuthal angle in spherical coordinates. Calculate the charge distribution considering that it was initially null.

    2. Relevant equations
    Not sure. Maybe ##\nabla \cdot \vec J + \frac{\partial \rho}{\partial t}=0##.
    The divergence theorem.
    3. The attempt at a solution
    So my idea was to maybe use the continuity equation that I wrote above. From it, I am not sure what to do. Maybe integrate in space so that I can use the divergence theorem, in other words I can reach that ##\int _S \vec J \cdot d\vec A + \int \frac{\partial \rho}{\partial t}dV=0##. But I am stuck there because I don't know how to calculate ##\vec J \cdot d\vec A##.

    Then my other idea is to integrate the continuity equation with respect to time, but again I'm not sure how to do this...
    I'd appreciate a little push in the right direction, thanks!
     
  2. jcsd
  3. Aug 2, 2015 #2

    TSny

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    I'm not sure what "ring of radius R" means. Maybe it's a ring as shown below. Anyway, I don't think R will play a role.

    My guess is that they want you to come up with an expression for ##\rho(r, \theta, \phi, t)##. The continuity equation seems like a good approach. What do you get explicitly for ##\frac{\partial \rho}{\partial t}## by evaluating the divergence of ##\vec{J}##?
     

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  4. Aug 3, 2015 #3

    fluidistic

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    Yes that's exactly it and what they ask for.
    I took the divergence in spherical coordinates, I reached ##\nabla \cdot \vec J = J(r, \theta ) \frac{\cot \theta}{r}=-\frac{\partial \rho}{\partial t}##.
    That would make ##\rho = - \frac{J(r,\theta) \cot (\theta ) t}{r} + f(r, \theta)## where f is an arbitrary function appearing when I integrated ##\partial \rho##... The result doesn't look right to me, especially this dependence on t, which seems to grow up infinitely.
     
  5. Aug 3, 2015 #4

    TSny

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    I don't get your expression for the divergence. Make sure to distinguish ##\theta## from ##\phi##.

    At t = 0 you want ##\rho## to be zero everywhere.

    As long as this peculiar current flows, the charge density will grow (positive in some places and negative in others).
     
  6. Aug 3, 2015 #5

    fluidistic

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    [tex]\nabla \cdot \vec J ={1 \over r^2}{\partial \left( r^2 J_r \right) \over \partial r}
    + {1 \over r\sin\theta}{\partial \over \partial \theta} \left( J_\theta\sin\theta \right)
    + {1 \over r\sin\theta}{\partial J_\phi \over \partial \phi}[/tex] but ##J_r=J_\theta=0## because ##\vec J = J_\phi \hat \phi##. I found that divergence formula in https://en.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinates, and the convention used is theta is zenithal while phi is azimuthal, same convention that I use.
     
  7. Aug 3, 2015 #6

    TSny

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    How do you get a cotangent of theta out of this? Shoudn't the numerator end up with a cosine of phi instead of a cosine of theta?
     
  8. Aug 3, 2015 #7

    fluidistic

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    My bad, you are correct. I reach ##\rho = - J(r, \theta ) \frac{\cos \phi}{r\sin \theta}t##.
     
  9. Aug 3, 2015 #8

    TSny

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    That looks correct.
     
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