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Charged Sphere with a Hole - Check my work?

  1. Jan 11, 2013 #1
    Charged Sphere with a Hole -- Check my work?

    1. The problem statement, all variables and given/known data

    You have a spherical shell of radius a and charge Q. Your sphere is uniformly charged except for the region where θ<= 1° (which has σ = 0).
    Imagine that your fi eld point is somewhere on the positive z-axis (so z could be larger or smaller than a). Determine E as a function of z.

    I believe I can represent this as a uniformly charged sphere without a hole and a thin disk with a charge density of -σ. Then the law of superposition lets me add the two together. I think I did it right, but before I go on to the computer program portion of the assignment, I'd love if somebody would double-check my logic and work. If you see an error, please let me know. If you think it's correct, let me know that, too.

    2. Relevant equations

    sin(1°) = r/a Where r is the radius of the disk. r = sin(1°)a = 0.017a

    E field of a sphere: E(r) = Q/(4∏r2ε0) = σa2/(r2ε0)

    E field of a disk: E(z) = q/(2∏ε0(0.017a2)*(1-z2/(√z2+(0.017a)2))

    3. The attempt at a solution

    Along the z-axis, the E field of the sphere can be written as E(z) = σa2/(z2ε0)

    Therefore, Etotal = σa2/(z2ε0) + q/(2∏ε0(0.017a2)*(1-z2/(√z2+(0.017a)2))

    or

    Etotal = σ/ε0*(a2/z2 - 1/2 + z/√(z2 + (0.017a)2)

    Is this correct? I'm sorry if it's messy and thank you, thank you in advance.
     
  2. jcsd
  3. Jan 11, 2013 #2

    mfb

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    Re: Charged Sphere with a Hole -- Check my work?

    Both equations (for disk+sphere) are true in a specific range of z (inside or outside?) only.
    It might be useful to calculate the field for both cases individually.
     
  4. Jan 11, 2013 #3
    Re: Charged Sphere with a Hole -- Check my work?

    I know the equation I have for the sphere contribution only works outside of the sphere, because the electric field inside a (normal) uniformly charged sphere is zero. The field around the disk, though, I thought would look the same in both directions. That would make the Etotal in my initial post the Etotal outside the sphere, while inside the sphere it would be

    Etotal = -σ/(2ε0)*(1-z/√(z2+(0.017a)2))
     
  5. Jan 11, 2013 #4

    mfb

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    Re: Charged Sphere with a Hole -- Check my work?

    The magnitude is the same, but the direction is not.
     
  6. Jan 12, 2013 #5
    Re: Charged Sphere with a Hole -- Check my work?

    Right, it would be in the opposite direction...but now I'm confused. Going by Gauss' Law, inside the sphere the E field will still be zero, won't it? Even with the little disk there's still no enclosed charge inside the sphere and therefore no E field...
     
  7. Jan 12, 2013 #6
    Re: Charged Sphere with a Hole -- Check my work?



    This holds if the disk's center is at the origin, at z=0. But this disk is at z = a , so in your equation make the substitution z → z-a . I think that the rest are correct...
     
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