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Finding E field inside a sphere with charge proportional to radius

  1. Jan 28, 2010 #1
    First time poster here! EDIT: SOLVED!!

    Thanks, I figured out from the related links at the bottom of the page. >_>b

    1. The problem statement, all variables and given/known data
    Find the electric field inside a sphere which carries a charge density proportional
    to the distance from the origin, [tex]\rho[/tex] = kr, for some constant k.

    2. Relevant equations
    [tex]\oint E.da[/tex]
    [tex]a = 4 \pi r^2/3[/tex]
    [tex]da = 4 \pi r^2[/tex]
    [tex]\rho = kr[/tex]
    [tex]E = q/(r^2 4 \pi \epsilon _0)[/tex]
    Where q = charge inside

    3. The attempt at a solution
    [tex]\oint E \bullet da = \int q/(r^2 4 \pi \epsilon _0) \bullet 4 \pi r^2[/tex]

    The 4 pi r^2 terms cancel, leaving on the right

    [tex]q/ \epsilon_0[/tex]

    Substitute rho into the eqn. as to integrate all dimensions of the sphere

    [tex] \int \rho d \tau / \epsilon_0[/tex]

    Here's where I get stuck, I know that

    [tex] \rho = kr [/tex]

    What do I do with [tex]d \tau[/tex]? I'd imagine that it'd be easiest to do in spherical coordinates, so do I just add dr, dtheta, drho?

    Also... How do I put a dot into this LaTex thing?

    Thank you!
    Last edited: Jan 28, 2010
  2. jcsd
  3. Jan 28, 2010 #2


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    Homework Helper

    You do not know the electric field, but you can calculate it by applying Gauss' law.

    Because of symmetry, E depends only on r and is the same along a sphere of radius R. The surface integral for a sphere of radius R is equal to Q/epszilon, where Q is the charge confined inside the sphere

    [tex]4 \pi R^2 E(R) = \frac{4 \pi}{\epsilon}\int_0^R{kr*r^2dr=\frac{4\pi}{\epsilon}k R^4/4 \rightarrow E(R)=\frac{kR^2}{4\epsilon}[/tex]

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