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Charge distribution of a uniformly charged disk 
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#1
Mar2812, 01:39 PM

PF Gold
P: 3,225

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


#2
Mar2812, 04:26 PM

P: 169

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(rr')Q(r')dr'[/itex] But i think that uniformly means that Q = rho * A (integrate constant) 


#3
Mar2812, 04:41 PM

PF Gold
P: 3,225




#4
Mar2812, 08:15 PM

P: 312

Charge distribution of a uniformly charged disk
a delta function in spherical coordinates is δ(r,r')=δ(rr')δ(θθ')δ(ψψ')/(r^2sinθ)



#5
Mar2812, 10:07 PM

PF Gold
P: 3,225

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(rr')δ(θθ')δ(zz')/r in cylindrical coordinates. I didn't use this fact to get the answer. 


#6
Mar2912, 02:01 PM

P: 312

BTW in cylindrical system δ(r,r')=[itex]\delta(rr')\frac{\delta(\phi\phi')}{r}\delta(zz') [/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
Mar2912, 02:51 PM

PF Gold
P: 3,225

For instance in cylindrical coordinates, I made the ansatz that the solution to the problem is [itex]\rho (\vec x)=C \Theta (rR) \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 (rR) \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
Mar2912, 03:07 PM

P: 312




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