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Gauss' Law

  1. Dec 5, 2011 #1
    An insulating solid sphere of radius a = 1.2 m is uniformly charged with charge Q = 4.5 x 10^-6 C. Point P inside the sphere is at a distance r = 0.60 m from the sphere center C. What is the magnitude of the electric field at point P?

    So I'm using Gauss' Law:

    q,enc = Q[(pi * r^2)/ (pi * a^2)]

    int(E dot dA) = q,enc / epsilon,naut
    = Q[(pi * r^2)/ (pi * a^2)] / epsilon,naut

    E * (pi * r^2) = Q(r^2 / a^2) / epsilon,naut

    Thus, E = (Q) / (pi * epsilon,naut * a^2) = 1.12 * 10^5 V/m.

    This doesn't seem to be the answer. Anyone know what I did wrong?
     
  2. jcsd
  3. Dec 5, 2011 #2

    gneill

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    Staff: Mentor

    Volume varies as the cube of the radius.
     
  4. Dec 5, 2011 #3
    Now I'm doing E * pi * r^2 = (Qr^3) / (epsilon,naut * a^3), but E is turning out to be 5.62E4 V/m, which is 4 times the correct answer.
     
  5. Dec 5, 2011 #4

    gneill

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    Staff: Mentor

    Rather than trying to do everything at once, why not take logical steps? First, what fraction of the total charge Q is contained in the inner spherical volume of radius r?
     
  6. Dec 5, 2011 #5
    r^3 / a^3
     
  7. Dec 6, 2011 #6

    gneill

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    Okay, so given that you know the total charge Q, you now know the fraction of that charge that's within the spherical region with radius r.

    Now, given a charge q within that sphere of radius r, what's the resulting electric field at distance r from its center?
     
  8. Dec 6, 2011 #7
    E = (Qr) / (pi * a^3)
     
  9. Dec 6, 2011 #8
    Edit:
    E = (Qr) / (pi * a^3 * epsilon,naut)
     
  10. Dec 6, 2011 #9
    Actually, E = kq / r^2 = (kQr) / (a^3) works, but I'm trying to use Gauss' Law.
     
  11. Dec 6, 2011 #10

    gneill

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    So choose your Gaussian surface and write Gauss' law for for it. You'll find that you need the total charge enclosed by the surface (done above), and the volume of the Gaussian sphere (done above). I suppose it's just a matter of putting the steps in an order that shows the appropriate progression :smile:
     
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