# Homework Help: Electric field of disk vs ring problem

1. Aug 24, 2014

### Crush1986

1. The problem statement, all variables and given/known data
Suppose you design an apparatus in which a uniformly charged disk of radius R is to produce an electric field. The field magnitude is most important along the central perpendicular axis of the disk, at a point P at distance 2R from the disk. Cost analysis suggests that you switch to a ring of the same outer radius R but with inner radius R/2. Assume that the ring will have the same surface charge density as the original disk. If you switch to the ring, by what percentage will you decrease the electrric feield magnitude at P?

2. Relevant equations

Electric field due to disk = σ/2ε*(1-z/(z^2+r^2)

3. The attempt at a solution

Now what I tried was finding the electric field of the big disk, then finding the electric field of the ring by subtracting the electric field of the missing disk that was taken out of radius R/2.

(Edisk-Ering)/Edisk.

For E of the large disk I get σ/2ε(1-2R/(5R^2)^(1/2)) for the E of the ring I get
σ/2ε((1-2R/(5R^2)^1/2)-(1-2R/(17R^2/4)^(1/2)). Which I simplify to
σ/2ε(2R/(17R^2/4)^(1/2)-2R/(5R^2)^(1/2)). It seems I'm making a mistake in calculating (Edisk-Ering. I've seen then answer and my Edisk is correct. Am I making an algebra mistake somewhere? I did this problem correct two days ago with no problem so I don't know how I'm getting it wrong now, I lost the paper :(.

Sorry for the ugliness of all the type. I tried to use this itex but I wasn't able to input the equations in correct apparently.

I'd really appreciate any help as this problem is driving me nuts.

2. Aug 24, 2014

### Zondrina

You seem to be using the equation for the disk to calculate the electric field of the ring.

The electric field of a charged ring is given by:

$$\frac{kqz}{(z^2 + R^2)^{3/2}}$$

3. Aug 24, 2014

### Crush1986

OK wait never mind. My answer works in a way... I found the percentage that the Electric field of the ring is in relation to the bigger disk. I just need to do 1-(Edisk-Ering)/Edisk to get the answer that my book wants.

4. Aug 24, 2014

### Crush1986

They say ring in the book but I guess it's really a washer heehee. I found the E field of the washer by subtracting the missing disk's E field.

The other day I found it by adding up rings by integrating from R/2 to R. It looks like both methods work just fine.

Luckily I just found out I needed to do one more little step to get the answer my book asked for.

5. Aug 24, 2014

### Zondrina

I see, the book intended it as a disk with a "hole" in it. If there is an internal cavity in the disk, then there is no electric field there.

Simply subtracting the "lack of" electric field from the electric field of the whole disk will give $E_{Disk_{net}}$.

The rest follows.