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Gauss's Law for Spherical Symmetry

  1. Oct 3, 2013 #1
    Find the electric field for a non-conducting sphere of radius R = 1 meter that is surrounded by air in the region r > 1. The interior of the sphere has a charge density of ρ(r) = r.

    The answer is k(pi)/r^2, but I can't seem to get that. My problem is with finding the enclosed charge. I've tried getting the answer many ways, but I keep getting it wrong. Generally, I know you have to set up an integral of the from ∫ρ(r)4(pi)r^2 dr to get the enclosed charge and then plug it into E(r) = kQ/r^2, but something keeps going wrong somewhere. I would greatly appreciate any help/advice anyone could offer. Thanks.
  2. jcsd
  3. Oct 3, 2013 #2


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    Post your calculation so we can see where you're going awry.
  4. Oct 3, 2013 #3
    So from the center to the edge of the sphere (R = 1), I know the enclosed charge should be:

    from 0 to 1, ∫ρ(r)4(pi)r^2 dr = 4pi[(1/4)r^4] from 0 to 1
    = pi

    When put into E(r), this does give k(pi)/r^2, but I'm confused about this result because this is supposed to be the final answer to the region r > 1, not 0 to R = 1.

    From 1 to r > 1, its surrounded by air, which means the charge density should be constant, so from 1 to r, the integral to find the enclosed charge is:

    ∫4(pi)r^2 dr = 4pi[(1/3)r^3] from 1 to pi
    = (4/3)(pi)[r^3 - 1]

    This is the integral that I though would give the charge enclosed for r > 1, but it yields a really ugly solution. The only thing I can think of (and I'm not sure if it makes any sense) is that my first result would be the charge on the surface of the sphere, and since its surrounded be air, that charge would remain constant as you move away from the surface of the sphere.
  5. Oct 3, 2013 #4


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    You've calculated the total charge on the sphere to be ##\pi## Coulombs. This is correct. Gauss' law tells us we can treat a spherically symmetric charge distribution as a point charge when we're outside of it, so the external field for r > 1 obeys Coulombs law, which is what you've arrived at.
    Clearly there are two distinct regions to deal with, one where the charge density is nonzero (within the non-conducting sphere), and the other where the charge density is zero (within the surrounding air). So your charge density function is discontinuous at r = 1.
    \rho (r) = \left \{
    r & : 0 \leq r \leq 1 \\
    0 & : r > 1
    \right .
    mathematically you deal with this by breaking the sum into two separate integrals. Since the charge density function is uniformly zero for the second region, that integral must vanish.
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