Charged Sphere with a Hole - Check my work?

[Raen]

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.

 

[mfb]

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.

 

[Raen]

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))

 

[mfb]

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

 

[Raen]

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...

 

[cosmic dust]

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...