Gauss' law for concentric circles

[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.

 

[gracy]

1/4πε0.(-2q )c^2

 

[ehild]

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

If something is written by you it need not be right.
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.

 

[gracy]

If something is written by you it need not be right.

It's in my book.Not by me.

 

[gracy]

I don't know how to right mathematical equations/formula here.Don't know about latex.

 

[ehild]

It can not be so in your book. You wrote $\frac{1}{4} π ε_0 (-2q) c^2$. Is the electric field proportional to the square of the radius?

 

[gracy]

$1/4πε0.(-2q)/c^2$

 

[ehild]

I don't know how to right mathematical equations/formula here.Don't know about latex.

Use parentheses.

 

[ehild]

$1/4πε0.(-2q)/c^2$

better. But πε0 is still at the wrong place. You should divide by it.

 

[gracy]

But πε0 is still in the numerator.

But I think (can see) it is denominator .

 

[ehild]

But I think (can see) it is denominator .

It is not. 1/ab means $\frac{1}{a}b$

 

[ehild]

It is not. 1/ab means $\frac{1}{a}b$

try to input to your calculator 10/2*5

 

[gracy]

$1πε0/4.(-2q)c^2$

 

[ehild]

$1πε0/4.(-2q)c^2$

I give up.

 

[gracy]

I give up.

Well,thanks for the answers you have given me so far.

 

[gracy]

@Doc Al have you also given up?Want to ask a question.

 

[gracy]

I now understand that electric field just inside the conductor is zero and just outside the conductor it is $σ/ε0$ so at at the r=c electric field has two values zero & $σ/ε0$ .As the graph in post 19 shows.
Right?

 

[BvU]

Ehild has been so patient with you. You really should repair that !

I notice that in the figure in post #19 (also a figure you reconstructed from what you have in the book ?) the same problem exists.

The expressions there should read, e.g. for the region b < r < c :$$E = {1\over 4\pi\epsilon_0}\, {-2q\ \over r^2} $$
Elizabeth really bent over backwards to make that clear to you.

In your post #136 you are either derailing badly or saying the right thing, depending on choices of $\sigma$ and the direction of $\vec E$. At the risk of triggering another 100+ posts:

If we take the r+ direction as positive, and proceed from 0 to b, the field is 0.

At b there is a jump from 0 to $\sigma\over \epsilon_0$. Here $\sigma ={ -2q\ \over 4\pi b^2 }$ with q a positive number of Coulombs. So the field jumps discontinuously from 0 to ${1 \over 4\pi \epsilon_0}\,{ -2q\ \over b^2 }$

From b to c, so b < r < c you have ${1 \over 4\pi \epsilon_0}\,{ -2q\ \over r^2 }$

At c there is another discontinuity. Approaching from r < c the value is as above, with the limit value ${1 \over 4\pi \epsilon_0}\,{ -2q\ \over c^2 }$.

In fact, the surface charge density at c is such that the jump at c ends up with E = 0 for r>c.
This is simply required from the fact that there is no E-field inside a conductor.
So this surface charge at c is induced by the charge on the inner sphere.
​

for c < r < d (d the radius of the outer sphere) the field is 0.

At d there is yet another discontinuity. Approaching from r < d the value is 0, and the jump is from 0 to $\sigma\over \epsilon_0$. Here $\sigma ={ +2q\ \over 4\pi d^2 }$ with q a positive number of Coulombs. So the field jumps discontinuously from 0 to ${1 \over 4\pi \epsilon_0}\,{ +2q\ \over d^2 }$

Finally, for d < r the field is like the field from a +2q point charge at the origin: ${1 \over 4\pi \epsilon_0}\,{ +2q\ \over r^2 }$.
with the limit value ${1 \over 4\pi \epsilon_0}\,{ -2q\ \over d^2 }$ for $r \downarrow d$.

No further comments.

---

​

 

[gracy]

No further comments.

You are locking this thread!(Although it's a mentor's job.)

 

[BvU]

No, I can't do that (lock a thread). And there is no reason to. But I urge you to read things through carefully and check if you have understood everything. Some digestion before firing of a bunch of "why ?"s and "right?"s. Did the considerable amount of assistance you received at least help you improve your mastership over this exercise ?

(And I appreciate your effort to pick up some TeX know-how!).

 

[gracy]

No, I can't do that (lock a thread)

I know it's a mentor's job.