Gauss' law for concentric circles

[gracy]

NO. Why do you think so?

 

[Doc Al]

View attachment 89159

Once again, that +2q charge is on the surface of the outer conductor, not in the space between conductors.

From r > b to r < c there is no charge, only empty space.

 

[gracy]

that +2q charge is on the surface of the outer conductor, not in the space between conductors.

Ok.I finally understand.Can we proceed?

 

[gracy]

What is the electric field at radius r in the region between the shells?

It would be - 2q/4πε0r^2

 

[gracy]

Consider a sphere inside the outer pink region. It is conductor, the field is zero. How much is the enclosed charge then?

zero.

 

[gracy]

This sphere encloses the small shell and the inner surface of the big shell. How much is the charge that compensates the charge of the inner shell?

+2q

 

[gracy]

It can not leave the shell. Where is it?

on the outer surface.

 

[gracy]

If there is some charge on the inner surface of the big shell, how much is on the outer surface?

+2q.Now I understood the complete setup of the problem.

 

[ehild]

I am happy

 

[gracy]

Wait.
what is the distance of -2q (distributed throughout the outer surface of inner conductor) from the center of inner shell?How can it be b?

 

[gracy]

Because -2q is distributed how it can have only one distance from the center of inner shell?

 

[ehild]

Wait.
what is the distance of -2q (distributed throughout the outer surface of inner conductor) from the center of inner shell?How can it be b?

The outer radius of the inner shell is b. The charge distributes on the outer surface of the inner shell. It is "surface charge". How far is the surface of a sphere from the centre of the sphere?

 

[gracy]

How far is the surface of a sphere from the centre of the sphere?

The answer is b.

 

[ehild]

Because -2q is distributed how it can have only one distance from the center of inner shell?

It is distributed on the outer surface of the shell. Again: at what distance are the points of the surface of a sphere from the centre of the sphere?

 

[gracy]

I was asking we take the +2q as a whole in spite of the fact that they are distributed.

 

[ehild]

The answer is b.

So all charge on the surface is at b distance from the centre. Are you satisfied?

 

[gracy]

I was asking we take the +2q as a whole in spite of the fact that they are distributed.

 

[ehild]

I was asking we take the +2q as a whole in spite of the fact that they are distributed.

The whole charge is -2q, and it is distributed evenly on the outer surface of the inner shell.
Remember "charge" is not an object but a property. There are particles with charge. That -2q charge can belong to a lot of excess electrons, which are distributed on the surface.

 

[gracy]

But at r= c why Electric field s not zero?Charge enclosed is zero.

 

[gracy]

Charge enclosed is zero.

+2q &-2q cancel each other.

 

[ehild]

But at r= c why Electric field s not zero?Charge enclosed is zero.

There is discontinuity of the field at r=c. E is not zero for r<c and E= at r>c

 

[gracy]

There is discontinuity of the field at r=c. E is not zero for r<c and E= at r>c

so how to deal with it?with that formula ;given by blueleaf 7?but I have something to ask about it.

 

[ehild]

+2q &-2q cancel each other.

The 2q and -2q cancel each other inside a Gaussian sphere which radius is greater than c. It might by greater only by the size of an atom, but it must be grater.

 

[gracy]

The 2q and -2q cancel each other inside a Gaussian sphere which radius is greater than c

you mean +2q is at distance which might by greater only by the size of an atom, but it must be grater than c

 

[ehild]

so how to deal with it?with that formula ;given by blueleaf 7?but I have something to ask about it.

Blueleaf said that E=0 inside the metal shell, for<r<d. r = c is the boundary between the empty space and the body of the metallic shell. E is defined at both sides of this boundary, but these values are not equal. E is not a continuous function.

 

[ehild]

you mean +2q is at distance which might by greater only by the size of an atom, but it must be grater than c

NO. That 2q is at distance c from the centre, but the Gaussian surface must be a bit farther inside the shell. Its radius is r>c. It can be greater than c by the size of an atom, but it must be greater.

 

[gracy]

but the Gaussian surface must be a bit farther inside the shell

why?

 

[ehild]

Electrodynamics treats the materials as if they were continuous, but they are corpuscular, consist of atoms.
The metal shell consists of positive ions and the free electrons around them. The +2q charge comes from the ions which miss their electrons, as the -2q charge of the inner shell repel them. These "lonely" ions are distributed on the inner surface of the outer shell, they are among the surface ions. So the surface layer of the metal shell is different from the inner part of the metal, which is electrically neutral. The charge is distributed in that surface layer which has thickness of about an atom. What is true for the inside of the shell, it is not true for the surface layer. You can apply Gauss' Law on a region where the electric field is "regular", inside the metal, where it is zero. A sphere with radius greater than c by at least the size of one atom encloses the whole 2q charge.

You can also say that the electric field is not defined at r=c, also the enclosed charge is not defined for a sphere with r=c.

 

[gracy]

electric field is not defined at r=c

But in my graph #19 it has been given magnitude of 1/4πε0.-2q c^2

 

[ehild]

Your formula is wrong. Use parentheses.
E is something at one side of the boundary and zero at the other side. You can calculate the limits of E at c at both sides, but they are not equal.