# E&M Charged disk

1. Oct 4, 2012

### hansbahia

1. The problem statement, all variables and given/known data
A charged disk of total charge "Q" and radius "a" lies in the xy-plane, centered at the origin. The surface-charge density distribution is nonuniform, having the surface-density, at any point inside the disk at distance "r" from the center of the form
σ(r)= m x r^2 , m being a constant
a) Evaluate the indicated constant in terms of Q and a, and express s(r) and your answers below in terms of these parameters.
b) At what value of "r" (relative to a) is s(r) equal to its average value on the disk? (Use only Gaus laws equation! Don't use Intensities)
c) Derive aformula for the charge q(r) contained within a circle of any radius r, and graph this function
d) Express the electric field "Ez(z)" at any point on the +z-axis as an integral over the source0charge distribution. (Start with he result for a charged ring -- draw a diagram, and explain briefly. Be sure to define every symbol you introduce.
e) Grap the function Ez(z) for -∞≤ z≤∞

2. Relevant equations
σ=Q/A
A=4π^2
∫E.ndA=(1/εo)∫pdV
∇.E=ρ/εo

3. The attempt at a solution
a) σ=Q/A=mr^2=Q/4πa^2
m=Q/(4πa^2r^2)

σ(r)=mr^2=Q/(4πa^2r^2)=Q/(4πa^2)
right?

b) I got lost here

I know the average σ= Q/A=Q/(4πa^2)
so I would assume r is equal 1?

c) σ=q/A→q=σA→dq=from 0 to r ∫σA=∫(Q/(4πa^2))(2πr)dr=Q/2a^2(from 0 to r ∫r dr)
=Q/2a^2(r^2/2)=Qr^2/4a^2
right?
d)

e)

I'm willing to go letter by letter (a) than (b)... in details

2. Oct 4, 2012

### vela

Staff Emeritus
What is s(r) supposed to be? It just appears here with no definition.

3. Oct 4, 2012

### hansbahia

I'm sorry, s(r) is the surface charge density σ= s(r). Sometimes is a pain trying to keep looking for sigma

4. Oct 4, 2012

### vela

Staff Emeritus
In part a), you can't have $\sigma=Q/A$ because that would be a constant, but you know that the charge density varies with the radius. Also, you're not using the right expression for the area.

What you need to do is write down an integral that corresponds to the total charge on the disk and set it equal to Q. This will give you an equation you can solve for m.

5. Oct 4, 2012

### hansbahia

dQ=σ(r)dA

Q=(from 0 to a) ∫σ(r)A
Q=(from 0 to a) ∫mr^2 (2πr)dr
Q=2πm (from 0 to a) ∫r^3 dr
Q=2πm (a^4/4)
m=2Q/(πa^4)

σ(r)=mr^2=2Qr^2/(πa^4) ?

6. Oct 4, 2012

### vela

Staff Emeritus
Yup, perfect!

In part b), like you said earlier, the average is Q/A, where A is the total area. You need to use the right formula for A, however.

You should be able to handle part c) as well.

7. Oct 4, 2012

### hansbahia

σ(r)=Q/A
2Qr^2/(πa^4)=Q/πr^2
2r^2/a^4=1/r^2
r^4=a^4/2
r=a/2^(1/4)
r≈0.84a

It seems small to me

Part c) I integrated from 0 to r therefore my first attempt steps were right?

σ=dq/dA
dq=σdA
q=from 0 to r ∫σA
q=from 0 to r ∫(2Qr^2/(πa^4))(2πr)dr
q=from 0 to r ∫ 4Qr^3/a^4dr
q=(4Q/a^4) from 0 to r ∫r^3
q=(4Q/a^4) (r^4/4)
q=Qr^4/a^4?

8. Oct 4, 2012

### hansbahia

I really don't know how to graph these functions

Last edited: Oct 5, 2012
9. Oct 4, 2012

### hansbahia

This is my diagram for part d)

Last edited: Oct 5, 2012
10. Oct 4, 2012

### vela

Staff Emeritus
This isn't quite correct. Think carefully about what A represents.

This looks fine. When r=a, it yields q=Q, as it should.

You'll need to fix your answer to part c) first. Surely, you've graphed functions before.

Use the hint. The electric field due to a ring of charge is probably worked out in your textbook somewhere.

11. Oct 4, 2012

### vela

Staff Emeritus
You have the right idea, but there are some math errors. A couple of comments:
1. You're given a coordinate system in the problem. You should use it. The x-axis in your diagram should be the z-axis.
2. dEy isn't 0; however, Ey is 0, due to symmetry. It's when you sum all the non-zero contributions do you get the cancellations that result in Ey=0.
3. What you labeled dEz in your diagram is actually dE.
4. Your formula for dE at the top of the page should have r2 in the bottom, not r3.
5. Your integral for Ex should be
$$E_x = \int dE_x = \int dE\cos\theta = \int \left(\frac{dq}{4\pi\epsilon_0 r^2}\right)\frac{x}{r}.$$ This is just straight substitution of the various formulas you have written down. Note, in particular, that there's no $dl$; you have $dq$ in the integral instead. I'll let you finish evaluating the integral.

12. Oct 4, 2012

### hansbahia

I thought the σ=Q/total area of the disk
the total area of the disk is πr^2.

Wait its the total area of the disk, and the radius of the disk is "a" therefore is πa^2

Last edited: Oct 5, 2012
13. Oct 4, 2012

### hansbahia

Before, when I was solving for a spherical charged ball, I came across the same equation for q(r)=Qr^4/a^4 and in class they just draw the graph and I didn't get it

Last edited: Oct 5, 2012
14. Oct 4, 2012

### hansbahia

Woww. I totally forgot that the charged disk was on the xy-plane. My book actually doesn't show the steps. It doesn't really help a lot. I'm using Foundations of Electromagnetic Theory by Reitz, Milford and Christy. It doesn't go in depth.
My question really is, do I do the steps same steps for uniform and nonuniform?

Last edited: Oct 5, 2012
15. Oct 4, 2012

### hansbahia

Whats wrong with part c?

Did you mean part b)?

16. Oct 4, 2012

### vela

Staff Emeritus
Yeah, I meant part b).

17. Oct 4, 2012

### vela

Staff Emeritus
Yeah, pretty much.