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Compute the volume charge density of a charged spherical conductor 
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#1
Aug2109, 03:54 PM

P: 2,158

In this problem we assume that the photon mass is m > 0. We take a conducting sphere of radius R and put a charge of Q on it. Some fraction of the charge will then reside at the surface and a fraction will move into the bulk. Evaluate the volume charge density in the bulk.
I was shocked to learn in another thread that some people believe that the fact that there are no charges inside a conductor has nothing to do with Coulomb's law (apart from the mere fact that it leads to a repulsion between like charges). So, I invented this problem, solved it, and because it turned out to be quite easy, posted it here. B.t.w., what is even easier is to simply see that not all of the charge can reside at the surface. 


#2
Aug2109, 08:18 PM

P: 1,133

I forgot a bit of electrodynamics, so I'll ask:
For a conductor, any charge applied to it can by uniformly distributed across the volume of the conductor, right? 


#3
Aug2609, 10:52 AM

P: 2,158




#4
Aug2609, 10:54 AM

P: 2,158

Compute the volume charge density of a charged spherical conductor
If no one solves the problem by Monday, I'll write up the answer then.



#5
Aug2609, 11:02 AM

P: 666

Are you assuming idealized pointlike charge carriers and a perfect conductor, etc.? If so, then doesn't Gauss's Law ensure that the charges will all accumulate a the surface? I'm just imagining a spherical Gaussian surface concentric with the conducting sphere and with a radius less than but arbitrarily close to the radius of the conductor.



#6
Aug2609, 11:18 AM

P: 2,158




#7
Aug3009, 08:56 AM

P: 2,158

Deadline is 30 hours from now. An F minus grade will be awarded to PF if there is no response by then.



#8
Aug3009, 01:00 PM

P: 1,133




#9
Aug3009, 01:33 PM

Mentor
P: 16,166

No, he means photon. This is leftover from another thread.
What I suspect is going to happen is that Count Iblis will note that [itex]\nabla \cdot E = \rho/\epsilon_0  m^2 \phi [/itex], and then argue that with a surface charge the field isn't zero because of the [itex]m^2 \phi [/itex]. There will then be a long drawnout discussion about what a conductor is: is it a material where the electric field is zero everywhere, or is it a material where charges are free to move? In Maxwell electrodynamics, the two are the same. In Proca electrodynamics, they are not. 


#10
Aug3009, 04:04 PM

P: 1,133




#11
Aug3009, 04:06 PM

P: 2,158

A conductor is an object in which charges are free to move and that implies (also in Proca theory) that the electric field is zero in the interior of the conductor.



#12
Aug3009, 04:51 PM

P: 1,133

Where do photons come into play?



#13
Aug3009, 06:08 PM

P: 4,513




#14
Aug3009, 06:24 PM

P: 2,158

V(r) = q/(4 pi epsilon_0) exp(m c/hbar r)/r 


#15
Aug3009, 10:08 PM

P: 1,133

Looks like there is more involved with this problem than I thought there was.



#16
Aug3109, 10:49 AM

P: 2,158




#17
Sep209, 11:22 AM

P: 2,158

Answer:
[tex]\rho = \frac{Q}{4\pi R \lambda^{2}}\frac{1}{1+\frac{R}{\lambda}+\frac{R^2}{3\lambda^2}}[/tex] where [tex]\lambda=\frac{\hbar}{mc}[/tex] is the photon Compton wavelength 


#18
Sep1909, 09:58 PM

P: 2,158

Sadly, I have had to award PF an Fminus grade, for not having submitted any work



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