A flat ring is uniformly charged

[hitemup]

Homework Statement

A flat ring (inner radius R_0, outer radius 4R_0) is uniformly charged. In terms of the total charge Q, determine the electric field on the axis at points

a) 0.25R_0
b) 75R_0

from the center of the ring. [Hint: The ring can be replaced with two oppositely charged superposed disks.]

Homework Equations

Gauss's Law
Coulomb's Law

The Attempt at a Solution

I'm okay with part b. The ring will be like a point charge, so using Coulomb's law would lead to the correct result.
But I'm not even sure what I'm supposed to imagine for part a. I guessed that it would be zero eventually because there cannot be electrical field inside a conductor, but didn't do anything like hint says I should.

 

[gneill]

But I'm not even sure what I'm supposed to imagine for part a. I guessed that it would be zero eventually because there cannot be electrical field inside a conductor, but didn't do anything like hint says I should.

You should be able to make an argument from symmetry to back up your guess.

 

[hitemup]

I also have a PDF in which a solution exists to this problem.

"We treat the source charge as a disk of positive charge of radius concentric with a disk of negative charge of radius R_0 . In order for the net charge of the inner space to be 0, the charge per unit area of the source disks must both have the same magnitude but opposite sign. The field due to the annulus is then the sum of the fields due to both the positive and negative rings."

But again, I can't understand why we are doing this.

 

[gneill]

I also have a PDF in which a solution exists to this problem.

"We treat the source charge as a disk of positive charge of radius concentric with a disk of negative charge of radius R_0 . In order for the net charge of the inner space to be 0, the charge per unit area of the source disks must both have the same magnitude but opposite sign. The field due to the annulus is then the sum of the fields due to both the positive and negative rings."

But again, I can't understand why we are doing this.

Presumably because the solution for the field due to a charged disk has been presented previously and you can use that result (cleverly) to solve this problem.