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Charge velocity and current density

  1. Jul 2, 2008 #1

    Defennder

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    1. The problem statement, all variables and given/known data
    Current density is given in cylindrical coordinates as [tex]\vec{J} = -10^6z^{1.5} \hat{z} Am^{-2} \ \mbox{in the region} \ 0 \leq r \leq 20\mu m , \mbox{and for} \ r \geq 20 \mu m, \ \vec{J} = 0[/tex]

    If the volume charge density at z=0.15 m is -2000C/m^3, find the charge velocity there.


    2. Relevant equations
    [tex]\nabla \cdot \vec{J} = -\frac{\partial \rho_v}{\partial t} [/tex]


    3. The attempt at a solution
    Okay so this seems pretty straightforward. Given that J is only in the z-direction, then isn't it simply possible to find v_z by dividing J_z(0.15) by -2000 there? But this gives me 29 which isn't the answer. The answer is -2900.

    Another method I tried was solving for [tex]\rho_v[/tex] using the relevant equations. This gives me [tex]\rho_v = 1.5x10^6 \int \sqrt{z} dt = 1.5x10^6\sqrt{z}t + g(z)[/tex]. But how do I find what g(z) is? And I have to solve for t as well, since it's not stated what value of t I should evaluate the charge velocity for at z=0.15.
     
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  3. Jul 2, 2008 #2
    I don't think that answer is right. You are looking for dQ/dt and you know that [itex]\rho=Q/V[/tex] for a uniform charge, so then you can solve for dQ/dt from the continuity equation.
     
  4. Jul 2, 2008 #3

    Defennder

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    I don't think the charge is uniform. Why should it be?
     
  5. Jul 2, 2008 #4
    Oh, I guess I read it wrong. I suppose all the problem is really saying that only at the point z=.15 that it is uniform.

    I mean, you have to know something about the charge density, or else you are shooting in the dark when it comes to finding the charge.
     
  6. Jul 3, 2008 #5

    Defennder

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    The question is as stated. I didn't omit anything.
     
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