# Charged Sphere with a Hole - Check my work?

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.

mfb
Mentor

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))

mfb
Mentor

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...

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...