# Grounded conductor charge distribution

1. Oct 28, 2013

### MrAlbot

Hello, I've been trying to understand how the fact of grounding a conductor affects its charge distribution.

So, for example, lets 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.

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.

Pedro

2. Oct 29, 2013

### Simon Bridge

Grounding a conductor provides a large amount of charges to balance out whatever the conductor has.

The effect is that any excess charge on the conductor gets conducted away to ground - unless there is some other reason for them to stick around. (It also sets the potential of the surface that got grounded to the same as the ground - usually "zero".) This means that attracted charges get to stay, repelled charges get conducted away.

R1<R2<R3?

If it were not grounded, then the charges in the shell would be able to redistribute to cancel out any external field.

Taking R1<R2<R3:

Ignore the grounding for a bit - there would be an electric field everywhere except inside R1.
The neutral shell at R2 has no effect on the field.
The charge at R3 has no contribution to the field for r<R3.
For r>R3 there are equal and opposite contributions from R1 and R3... so there is no field.

Now you want to look carefully at the (induced) charge at R3.

Once you have that - then you can figure out what happens when R2 is grounded.