# Homework Help: Charge distribution of a uniformly charged disk

1. Mar 28, 2012

### fluidistic

1. The problem statement, all variables and given/known data
The problem can be found in Jackson's book, I think in chapter 1 problem 3 or something like this.
I must determine the charge distribution of a uniformly charged disk of radius R in spherical coordinates (I've done it in cylindrical coordinates and had no problem). The total charge is Q.
I've found a solution on the internet but the answer is different from mine.
I forgot to mention that I have to use Dirac's delta.

2. Relevant equations
$\int _{\mathbb{R}^3} \rho (\vec x )=Q$.

3. The attempt at a solution
Since the charges are over a 2d surface, there will be 1 Dirac's delta in the expression for rho, the charge density. I will use Heaviside's step function because the surface is limited.
Let $(r, \theta , \phi )$ be the coordinates. I make the ansatz/educated guess that $\rho$ is of the form $C \delta \left ( \theta - \frac{\pi }{2} \right ) \Theta (r \sin \theta -R)$.
Integrating this distribution in all the space, I reach that C is worth $\frac{3Q}{2\pi R^3}$.
Therefore $\rho (r, \theta )=\frac{3Q\delta \left ( \theta - \frac{\pi }{2} \right ) \Theta (r \sin \theta -R)}{2\pi R^3}$.
However the solution provided on the internet is $\rho (\vec x )=\frac{q }{\pi R^2r} \delta \left ( \theta - \frac{\pi }{2} \right ) \Theta (r -R)$.
Are they both equivalent (I doubt it), if not, did I do something wrong? If so, what did I do wrong? Thanks a lot!

2. Mar 28, 2012

### dikmikkel

So your charge diverges at theta = Pi/2 ? that seems odd to me. If the disc is uniformly charged then the delta function seems to much to me.
Can't you use the delta functions most useful prop: $Q(r) = \int_{-\infty}^{\infty} \delta(r-r')Q(r')dr'$
But i think that uniformly means that Q = rho * A (integrate constant)

3. Mar 28, 2012

### fluidistic

Not really. My charge distribution says that it's confined in the plane z=0, or in spherical coordinates, $\delta \left ( \theta - \frac{\pi }{2} \right )$. This isn't a function (does not diverge). This notation just tells me that there's no charge outside the z=0 plane. The charge doesn't diverge in the z=0 plane. To get the total charge, I have to integrate the charge density (rho) over all the space. The integral converges, thankfully :)

4. Mar 28, 2012

### sunjin09

a delta function in spherical coordinates is δ(r,r')=δ(r-r')δ(θ-θ')δ(ψ-ψ')/(r^2sinθ)

Last edited: Mar 28, 2012
5. Mar 28, 2012

### fluidistic

Hmm I see but how this can help me?
Also I've solved the exact same problem with cylindrical coordinates and my answer was exactly the same as the internet's one, and I'm guessing that δ(r,r')=δ(r-r')δ(θ-θ')δ(z-z')/r in cylindrical coordinates. I didn't use this fact to get the answer.

6. Mar 29, 2012

### sunjin09

The δ(θ-pi/2) part of your expression is missing a 1/r factor, if you think about the difference between dθ and dz. What you want is an infinitesimal displacement in the z direction, i.e., dz=r*dθ at pi/2.

BTW in cylindrical system δ(r,r')=$\delta(r-r')\frac{\delta(\phi-\phi')}{r}\delta(z-z')$, where the 1/r is associated with the delta of $\phi$ coordinate, for the same consideration, but z coordinate is unaffected.

7. Mar 29, 2012

### fluidistic

What I don't understand is why do I have to use "δ(r,r')"?
For instance in cylindrical coordinates, I made the ansatz that the solution to the problem is $\rho (\vec x)=C \Theta (r-R) \delta (z)$. When integrated over all the space, this gave me $C=\frac{Q}{\pi R^2}$ leading to $\rho (r,z)=\frac{Q}{\pi R^2}\Theta (r-R) \delta (z)$ which coincides with the answer found on the Internet. I tried the exact same method in spherical coordinates but for some unknown reason to me, the answer found on the Internet differs from mine.
In either cases, I didn't use the expression for δ(r,r').

8. Mar 29, 2012

### sunjin09

Your ansatz for the spherical coordinates is incorrect, the delta function on θ should be δ(θ-π/2)/r, which corresponds to the infinitely thin x-y plane, upon which you multiply the Heaviside function. δ(θ-π/2) alone corresponds to an infinitely thin SOLID ANGLE around the x-y plane.

9. Mar 29, 2012

### fluidistic

Ok thank you very much, I'm going to think about it. If I have any problem, I'll post here.