Charged Sphere with a Hole - Check my work?

AI Thread Summary
The discussion revolves around calculating the electric field (E) of a charged spherical shell with a hole, specifically addressing the impact of the hole on the field at a point along the positive z-axis. The user proposes using the principle of superposition by treating the system as a uniformly charged sphere and a negatively charged disk to find the total electric field. They express concern about the validity of their equations, particularly regarding the behavior of the electric field inside and outside the sphere. Clarifications are made about the application of Gauss' Law, noting that the electric field inside a uniformly charged sphere remains zero despite the presence of the hole. The conversation emphasizes the need for careful consideration of the disk's position in the equations to ensure accurate calculations.
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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 let's 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.
 
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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.
 


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))
 


The magnitude is the same, but the direction is not.
 


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...
 


Raen said:


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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