Charged conductor in electrostatic equilibrium

[yoyo311]

I have read that in electrostatic equilibrium, their are no electric fields or else the charges would be moving. So given a positively charged spherical shell, the positive charges would repel each other and reside on the outside, causing the shell to be in electrostatic equilibrium.

But if we now put a positively charged particle in the inside the shell, some negative charges from the shell would gather on the inner lining of the shell thus making a same amount of positively charged particles reside on the outermost part of the shell.

What about the remaining initial positive charge? Does it reside on the outermost lining of the shell because the charges repel each other (ie. charges in a conductor gather as far away from each other on the lining of the conductor), or are they attracted towards the inner side of the shell because the negative charges already there attract them and the positive charges already on the outside repel them?

I hope my question makes sense. This is not exactly a "homework problem", but a question arising from a problem.

 

[ehild]

Hi yoyo311, welcome to PF!

I have read that in electrostatic equilibrium, their are no electric fields or else the charges would be moving. So given a positively charged spherical shell, the positive charges would repel each other and reside on the outside, causing the shell to be in electrostatic equilibrium.

You meant a metal shell. There is no electric field in the metal.

But if we now put a positively charged particle in the inside the shell, some negative charges from the shell would gather on the inner lining of the shell thus making a same amount of positively charged particles reside on the outermost part of the shell.

What about the remaining initial positive charge? Does it reside on the outermost lining of the shell because the charges repel each other (ie. charges in a conductor gather as far away from each other on the lining of the conductor), or are they attracted towards the inner side of the shell because the negative charges already there attract them and the positive charges already on the outside repel them?

Do not forget that the electrons can move in a metal, the positive charge is associated to the stationary metal ions.

Putting a positively charged particle inside the void, it generates electric field there. There is non-zero electric field at the inner wall of the shell, and that field attracts the electrons of the metal to the surface. Those electrons are missing inside the metal, and their place is filled with other electrons, producing positively charged regions farther outward, till this positive charge gathers at the outward surface of the shell, adding to the positive charge already there.

The electric field lines in the void connect the positive particle and the electrons on the inner surface. In stationary state, there is no electric field inside the metal. The positive charge on the surface does not "feel " the negative charge on the inner surface.

ehild

 

[rude man]

I have read that in electrostatic equilibrium, their are no electric fields or else the charges would be moving. So given a positively charged spherical shell, the positive charges would repel each other and reside on the outside, causing the shell to be in electrostatic equilibrium.

But if we now put a positively charged particle in the inside the shell, some negative charges from the shell would gather on the inner lining of the shell thus making a same amount of positively charged particles reside on the outermost part of the shell.

What about the remaining initial positive charge? Does it reside on the outermost lining of the shell because the charges repel each other (ie. charges in a conductor gather as far away from each other on the lining of the conductor), or are they attracted towards the inner side of the shell because the negative charges already there attract them and the positive charges already on the outside repel them?

I hope my question makes sense. This is not exactly a "homework problem", but a question arising from a problem.

Use successively larger Gaussian surfaces to track the charge distribution on the inside and outside of your shell: ∫∫D*dA = Q.