Charge distribution of a uniformly charged disk

In summary, the conversation discusses determining the charge distribution of a uniformly charged disk in spherical coordinates using Dirac's delta. The problem can be found in Jackson's book and a solution is found on the internet, but the answers differ. The individual discussing the problem solved it in cylindrical coordinates and found the same answer as the internet, but when attempting to solve it in spherical coordinates, the answer was different. The conversation then delves into the correct use of Dirac's delta in spherical coordinates.
  • #1
fluidistic
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Homework Statement


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.

Homework Equations


[itex]\int _{\mathbb{R}^3} \rho (\vec x )=Q[/itex].

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 [itex](r, \theta , \phi )[/itex] be the coordinates. I make the ansatz/educated guess that [itex]\rho[/itex] is of the form [itex]C \delta \left ( \theta - \frac{\pi }{2} \right ) \Theta (r \sin \theta -R)[/itex].
Integrating this distribution in all the space, I reach that C is worth [itex]\frac{3Q}{2\pi R^3}[/itex].
Therefore [itex]\rho (r, \theta )=\frac{3Q\delta \left ( \theta - \frac{\pi }{2} \right ) \Theta (r \sin \theta -R)}{2\pi R^3}[/itex].
However the solution provided on the internet is [itex]\rho (\vec x )=\frac{q }{\pi R^2r} \delta \left ( \theta - \frac{\pi }{2} \right ) \Theta (r -R)[/itex].
Are they both equivalent (I doubt it), if not, did I do something wrong? If so, what did I do wrong? Thanks a lot!
 
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  • #2
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: [itex]Q(r) = \int_{-\infty}^{\infty} \delta(r-r')Q(r')dr'[/itex]
But i think that uniformly means that Q = rho * A (integrate constant)
 
  • #3
dikmikkel said:
So your charge diverges at theta = Pi/2 ?
Not really. My charge distribution says that it's confined in the plane z=0, or in spherical coordinates, [itex]\delta \left ( \theta - \frac{\pi }{2} \right )[/itex]. 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
a delta function in spherical coordinates is δ(r,r')=δ(r-r')δ(θ-θ')δ(ψ-ψ')/(r^2sinθ)
 
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  • #5
sunjin09 said:
a delta function in spherical coordinates is δ(r,r')=δ(r-r')δ(θ-θ')δ(ψ-ψ')/(r^2sinθ)
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
fluidistic said:
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.

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')=[itex]\delta(r-r')\frac{\delta(\phi-\phi')}{r}\delta(z-z') [/itex], where the 1/r is associated with the delta of [itex]\phi[/itex] coordinate, for the same consideration, but z coordinate is unaffected.
 
  • #7
sunjin09 said:
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')=[itex]\delta(r-r')\frac{\delta(\phi-\phi')}{r}\delta(z-z') [/itex], where the 1/r is associated with the delta of [itex]\phi[/itex] coordinate, for the same consideration, but z coordinate is unaffected.

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 [itex]\rho (\vec x)=C \Theta (r-R) \delta (z)[/itex]. When integrated over all the space, this gave me [itex]C=\frac{Q}{\pi R^2}[/itex] leading to [itex]\rho (r,z)=\frac{Q}{\pi R^2}\Theta (r-R) \delta (z)[/itex] 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
fluidistic said:
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 [itex]\rho (\vec x)=C \Theta (r-R) \delta (z)[/itex]. When integrated over all the space, this gave me [itex]C=\frac{Q}{\pi R^2}[/itex] leading to [itex]\rho (r,z)=\frac{Q}{\pi R^2}\Theta (r-R) \delta (z)[/itex] 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').

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
sunjin09 said:
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.

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

1. What is the formula for the charge distribution of a uniformly charged disk?

The charge distribution of a uniformly charged disk is given by the formula: σ = Q/πR2, where σ is the surface charge density, Q is the total charge on the disk, and R is the radius of the disk.

2. How is the electric field calculated at a point above the center of a uniformly charged disk?

The electric field at a point above the center of a uniformly charged disk can be calculated using the formula: E = σ/2ε0, where E is the electric field, σ is the surface charge density, and ε0 is the permittivity of free space.

3. Can the charge distribution of a uniformly charged disk be non-uniform?

No, a uniformly charged disk, by definition, has a constant surface charge density. If the charge distribution is not constant, then the disk is not uniformly charged.

4. How does the electric field vary at different distances from the center of a uniformly charged disk?

The electric field decreases as the distance from the center of a uniformly charged disk increases. It follows an inverse square law, meaning that the electric field is inversely proportional to the square of the distance from the center.

5. Can the charge distribution of a uniformly charged disk affect the motion of charged particles?

Yes, the electric field created by the charge distribution of a uniformly charged disk can exert a force on charged particles and affect their motion. This is the principle behind devices such as the cathode ray tube and the electron microscope.

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