- #1
MrAlbot
- 12
- 0
Hello, I've been trying to understand how the fact of grounding a conductor affects its charge distribution.
So, for example, let's assume there are three spherical shells with radius R1 R2 and R3. Supose I charge the R1 shell with q and the R3 shell with -q , and I connect the R2 shell to the ground and now I want to find out what is the charge distribution over the inside and outside of these shells.
My quick answer would be:
As far as I can see, the middle shell acts like a perfect faraday cage, and the distribution of charge of R3 of -q will scatter through the outside surface.
so the Electric field should only exist betwen R1 and R2 and out of the bigger shell (r>R3)
The charge distribution would be:
R1- = 0
R1+ = q
R2- = -q
R2+ = 0
R3- = 0
R3+ = -q
Based in the fact that if the cage acts and contains the electric field inside the larger (R3) shell then the Gauss law would work (* now that I see it I think here is the problem, because the potential must somehow go back to zero *)
Although I've seen people say that the distribution should be:
R1- = 0
R1+ = q
R2- = -q
R2+ = q'
R3- = -q'
R3+ = -q+q'
and then be solved (knowing that The potential from the midle sphere must be zero)
calculating the potential from infinity to R2 must be zero, so The potential from infinity to R3 plus the potential from R3 to R2 must be zero, and this 2 potentials must be simetric.
In the meantime, while I was writting all of this down, I came to the conclusion thaat the second option is the correcto one, but I was hoping I could have someone else's aproval on this.
Thanks in advance,
Pedro
So, for example, let's assume there are three spherical shells with radius R1 R2 and R3. Supose I charge the R1 shell with q and the R3 shell with -q , and I connect the R2 shell to the ground and now I want to find out what is the charge distribution over the inside and outside of these shells.
My quick answer would be:
As far as I can see, the middle shell acts like a perfect faraday cage, and the distribution of charge of R3 of -q will scatter through the outside surface.
so the Electric field should only exist betwen R1 and R2 and out of the bigger shell (r>R3)
The charge distribution would be:
R1- = 0
R1+ = q
R2- = -q
R2+ = 0
R3- = 0
R3+ = -q
Based in the fact that if the cage acts and contains the electric field inside the larger (R3) shell then the Gauss law would work (* now that I see it I think here is the problem, because the potential must somehow go back to zero *)
Although I've seen people say that the distribution should be:
R1- = 0
R1+ = q
R2- = -q
R2+ = q'
R3- = -q'
R3+ = -q+q'
and then be solved (knowing that The potential from the midle sphere must be zero)
calculating the potential from infinity to R2 must be zero, so The potential from infinity to R3 plus the potential from R3 to R2 must be zero, and this 2 potentials must be simetric.
In the meantime, while I was writting all of this down, I came to the conclusion thaat the second option is the correcto one, but I was hoping I could have someone else's aproval on this.
Thanks in advance,
Pedro