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Charge from field, cylindrical symmetry

  1. Sep 10, 2012 #1
    The problem states:

    From the field with a radial cylindrical component only given by the following equations:

    E(r)= (ρ0*r3)/(4 * ε0*a2) for r<=a

    E(r)= (ρ0*a2)/(4*ε0*r2) for r > a

    obtain the corresponding charge distribution in free space in which the equation is:

    ρ(r) = ρ0*(r2/a2) (0<=r<=a)

    So I know that dE(r)/dr = p(r)/ε0

    After differentiating the first E(r) equation I come to (3/4)*ρ0*r2/a2.

    It would be correct if the 3/4 weren't there but I'm not sure where I'm going wrong. Any help appreciated.
     
    Last edited: Sep 10, 2012
  2. jcsd
  3. Sep 10, 2012 #2

    Simon Bridge

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    In cylindrical coordinates [itex]\vec{p}=(r,\theta,z)[/itex], the div vector is
    [tex]\vec{\nabla}=\left ( \frac{1}{r}\frac{\partial}{\partial r}r, \frac{1}{r}\frac{\partial}{\partial \theta}, \frac{\partial}{\partial z} \right )[/tex]

    Since [itex]E(r,\theta,z)=E(r)[/itex], you needed: [tex]\frac{1}{r}\frac{\partial}{\partial r}\big ( rE(r) \big ) = \frac{1}{\epsilon_0}\rho(r)[/tex]
     
    Last edited: Sep 10, 2012
  4. Sep 10, 2012 #3
    Very helpful! Thanks!
     
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