Gauss' law for concentric circles

[blue_leaf77]

That equation physically means that if in the interface between two media there is surface charge density $\sigma$, the electric field components perpendicular to this interface in the two media separated are discontinuous by the amount $\sigma/\epsilon_0$. Let's take an example for the spherical surface of radius $b$ in your problem. This surface separates the conducting medium below ($r<b$) and the vacuum outside ($r>b$). This spherical surface also carries a surface charge density of
$$
\sigma = \frac{-2q}{Area\hspace{1mm}of\hspace{1mm} the\hspace{1mm} surface} = \frac{-2q}{4\pi b^2}
$$
We know that the field in immediately below the surface is zero since it's inside the a conductor, so $E_{1\perp}=0$. The perpendicular component of the field immediately outside our spherical surface is then
$$
E_{2\perp} = E_{1\perp} - \frac{\sigma}{\epsilon_0} = 0 - \frac{-2q}{4\pi \epsilon_0 b^2} = \frac{2q}{4\pi \epsilon_0 b^2}
$$
But since the E field vector itself in our problem is perpendicular on our spherical surface, we have that $E(b) = E_{2\perp} = \frac{2q}{4\pi \epsilon_0 b^2}$.

 

[gracy]

E2⊥=2q4πϵ0b2E(b) = E_{2\perp} = \frac{2q}{4\pi \epsilon_0 b^2}.

But It is (negative) -2q/4πϵ0b^2 in the graph.

 

[BvU]

So it's a matter of interpreting blue leaf's 1 and 2 in post #29 and in post #31, right ?
My guess is (and this guy backs me up in his (634) ) he mixed them up. If you go from 1 to 2 over a surface charge $\sigma$ you have $E_{2\perp}-E_{1\perp} = \frac{\sigma}{\epsilon_0}$.

This way the field in $b < r < c$ is the field of -2q at the origin -- as it should be.

Re your

But for b &c E is zero.

that is true coming from the conductor side. In his post #27 he wants you to come from the other side, so you get a value for the jump in E at the discontinuity.

 

[gracy]

But for b &c E is zero

I think I am confusing the point P2 with b.Electric field at r=b was never zero.

 

[gracy]

And sign confusion is still there .Why for r=d it's positive and for r=b &r=c it's negative.What pattern does it follow?from outside to inside or from inside to outside?

 

[Doc Al]

And sign confusion is still there .Why for r=d it's positive and for r=b &r=c it's negative.What pattern does it follow?from outside to inside or from inside to outside?

Consider the sign of the charge enclosed by Gaussian spheres at those points.

 

[gracy]

Consider the sign of the charge enclosed by Gaussian spheres at those points.

But it also depends on

E2⊥−E1⊥

E1⊥−E2⊥

 

[Doc Al]

But it also depends on

That's an alternate way of looking at it. Mine is easier.

 

[gracy]

That's an alternate way of looking at it

So which formula are you using?

 

[Doc Al]

So which formula are you using?

I don't need a 'formula' to tell me the direction of the electric field! If the enclosed charge is positive, the field points outward; if negative, inward.

 

[gracy]

I don't need a 'formula' to tell me the direction of the electric field

E2⊥−E1 or E1⊥−E2

 

[blue_leaf77]

DocAl's way is actually more practical. But if you insist on knowing how to really decide between above - below or the other way around, I will go into a slightly deeper math, which I hope you can still grasp. See the picture I attached. The gray plane surface is actually a multiple zoom in result of a certain portion of your spherical conducting surface. Make a cylindrical volume such that part of it lies above the conductor, other part lies below it. We have Gauss law $\oint \mathbf{E}\cdot d\mathbf{a} = Q_{enc}/\epsilon_0 = \int \mathbf{E}_{above}\cdot d\mathbf{a} + \int \mathbf{E}_{below}\cdot d\mathbf{a}$, the integral over the side surface of the cylinder should be zero. To answer your confusion, we actually really have to know the direction of the fields in each region. So, the minus sign in $E_{2\perp} - E_{1\perp}$ can even be a positive sign, depending on the relative directions of the fields in the two regions. The source I was using when I brought up that equation for the first time assumes that the field in both regions point in the same direction, so the minus sign appears. As a conclusion, to determine the correct sign you need to know the direction of the fields in each region. In your problem, the charge on surface r=b is negative, so the fields outside it must point inward.

 

[gracy]

See the picture I attached

where is it?

 

[gracy]

DocAl's way is actually more practical.

But I did not understand his way.

 

[gracy]

I think I am confusing the point P2 with b.Electric field at r=b was never zero.

But for b &c E is zero.

right?

 

[gracy]

E1⊥

it does not have any magnitude so no direction.And E2⊥ will have negative sign.

 

[gracy]

If the enclosed charge is positive, the field points outward; if negative, inward.

if no charges are enclosed?how am I going to decide direction?

 

[blue_leaf77]

if no charges are enclosed?how am I going to decide direction?

If there is no surface charge density, the normal field components are continuous.

 

[Doc Al]

if no charges are enclosed?how am I going to decide direction?

If no charges are enclosed within your Gaussian sphere (and there is symmetry), then what will the field at the surface be?

 

[gracy]

then what will the field at the surface be?

Zero.

 

[Doc Al]

But I did not understand his way.

It's easy, if you use Gauss' law.

For example, say you want the field in the space between the two conducting shells. So draw a Gaussian sphere in that region. What is the charge contained within that Gaussian surface? Answer that and we can proceed.

 

[Doc Al]

Zero.

Right!

 

[gracy]

What is the charge contained within that Gaussian surface?

The charge enclosed by one of the(inner) conductors.

 

[Doc Al]

The charge enclosed by one of the(inner) conductors.

Give me the charge in terms of q.

 

[gracy]

Give me the charge in terms of q.

Shall I refer picture in op.

 

[Doc Al]

Shall I refer picture in op.

Yes, that's the problem we are discussing. The one you started the thread with.

 

[gracy]

Zero.Because there is +2q and -2q.

 

[Doc Al]

Zero.Because there is +2q and -2q.

No.

I suspect the diagram is throwing you off. For some reason, there is a blue circle drawn in the space between the two conducting shells (where p3 is). That circle is labeled as "+2q", implying that there is some charge just floating in space. I doubt that's what you meant. I think you meant to show that the inner surface of the outer shell has a charge of +2q.

Until this is straightened out there is little point in continuing.

 

[gracy]

But +2q is in between the two conductive shells.

For example, say you want the field in the space between the two conducting shells.

 

[Doc Al]

But +2q is in between the two conductive shells.

Really? Why then doesn't the problem statement say that? If true, that changes everything! (But I doubt it's true.)