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Electric field

  1. Apr 9, 2005 #1
    I have to find the electric field everywhere using Gauss’ law in differential form. Charge density is [tex]\rho = \rho_{0}r^{3}[/tex] for a<r<b and 0 otherwise in spherical symmetry and then in cylindrical coordinates

    [tex] \nabla \cdot D=\rho [/tex]
    I have look for D and then just get E = D/epsilon. D is where I need help. Thanks
  2. jcsd
  3. Apr 9, 2005 #2
    Well, just apply Gauss' law on a sphere and on a cilinder. this should not be that difficult. You don't need this D. You will only need to be careful when the radial coordinate r is inside the sphere or cilinder. However, you have been given a charge density, so that is no problem.

    What have you done so far ?


    Just as an example : Suppose you have a sphere of radius a in which there is an uniform chargedensity. The total charge Q is then equal to [tex]\frac{4 \pi a^3}{3} \rho[/tex] Then Gauss's law says : [tex]E 4 \pi r^2 = \epsilon_0 Q[/tex]

    This yields the electric field at distance r from the center of the sphere. Keep in mind that r must be BIGGER then a in this case
    Last edited: Apr 9, 2005
  4. Apr 9, 2005 #3
    So for [tex]0\leq r\leq a[/tex] Qencl=0

    for [tex]a < r \leq b[/tex]

    Qencl = [tex]\int_{a}^{r} \rho dv[/tex]

    for r>b
    Qencl = [tex]\int_{a}^{b} \rho dv [/tex]
    and I just have to solve that,
    Is this ok?

    so for cylindrical I have to used 2pi*r*L
    Last edited: Apr 9, 2005
  5. Apr 9, 2005 #4


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    It's [tex]E 4 \pi r^2 =~electric~flux~=~EA= \frac{q}{\epsilon_0}[/tex]

    I'm sure you can take it from here
  6. Apr 9, 2005 #5
    To get E I just have to solve Q and substitute to the equation.
    But isn't D = epsilon*E and I can get E from that
    Last edited: Apr 9, 2005
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