Charged Sphere with a Hole - Check my work?

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Charged Sphere with a Hole -- Check my work?

Homework Statement



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.

Homework 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))

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.
 

Answers and Replies

  • #2
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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.
 
  • #3
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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))
 
  • #4
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The magnitude is the same, but the direction is not.
 
  • #5
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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...
 
  • #6
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E field of a disk: E(z) = q/(2∏ε0(0.017a2)*(1-z2/(√z2+(0.017a)2))



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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