Electric field for a spherical shell when r=R?

[laser1]

Homework Statement

N/A

Relevant Equations

N/A

Griffith's keeps mentioning the case r<R and r>R blah blah blah. Ok. But what if r = R?

My attempt at an answer: when I take the limit from outside the sphere, I get the case when r > R. When I take the limit from inside the sphere, I get the case when r < R. Left and side hand side limits do not agree, so is the electric field undefined when r=R? But how is this possible to be undefined in reality? Although to be fair, in reality, a 2D shell does not exist (I think) as everything has 3 dimensions...

 

[kuruman]

Have you studied boundary conditions on the electric field blah blah?

 

[PeroK]

A charged shell of zero thickness is not a real scenario.

 

[laser1]

Have you studied boundary conditions on the electric field blah blah?

not yet :(

 

[Orodruin]

A charged shell of zero thickness is not a real scenario.

This. A ”real” shell would have some thickness over which the field changes from the inside solution to the outside one.

(Of course, that’s just a model as well. Going small enough that model is going to break down as well.)

 

[BvU]

is the electric field undefined when r=R

Yes. The electric field strength is discontinuous. The potential is not.
Mathematically there is no problem.
Physically there is no problem either (you can't really be at $r=R$ with anything useful)

$\$

 

[kuruman]

Homework Statement: N/A
Relevant Equations: N/A

as everything has 3 dimensions

That may be so, but mathematical descriptions of 3d objects can be one-dimensional. Consider the case of a charged, conducting sphere of radius $R$ carrying total charge $Q$. By symmetry, as far as the electric field is concerned, any direction away from the center of the sphere is equivalent to any other direction. In other words, the electric field vector points radially away from the center of the sphere and is a function only of the distance $r$ from the center. In that sense, the electric field here is one-dimensional.

There are two regions in space. In region I, $r<R$, you have conducting material. In that region the electric field is zero so $$E_{\text{I}}= 0.$$ In region II, $r>R$, you have vacuum. In that region the (magnitude of the) electric field is $$E_{\text{II}}=\frac{Q}{4\pi \epsilon_0 r^2}.$$At $r=R$ the two mathematical expressions become
$$E_{\text{I}}(R)=0~;~~E_{\text{II}}(R)=~\frac{Q}{4\pi \epsilon_0 R^2}.$$Here is where the physics comes in. The boundary condition that you have not studied yet, is an equation that tells you what happens at $r=R$. It says that the difference (discontinuity) of the normal component of the electric field at the conductor-vacuum interface is equal to the surface charge density $\sigma$ divided by $\epsilon_0.$ $$\frac{\sigma}{\epsilon_0}=E_{\text{II}}(R)-E_{\text{I}}(R)=E_{\text{II}}(R)-0=\frac{Q}{4\pi \epsilon_0 R^2}\implies \sigma=\frac{Q}{4\pi R^2}.$$ This says that the charge distribution is uniform over the surface as one would expect. When you get to boundary conditions, you will see how the one described above is a direct consequence of Gauss's Law. I am sure you can find its derivation in Griffiths if you look for it.

 

[laser1]

I am sure you can find its derivation in Griffiths if you look for it.

Ah yes I see. Around 10 pages in front of me, so I will come across it soon!

 

[kuruman]

Ah yes I see. Around 10 pages in front of me, so I will come across it soon!

Yes, you cannot learn everything together and at the same time. Just out of curiosity, are you self-studying out of Griffiths?

 

[laser1]

Just out of curiosity, are you self-studying out of Griffiths?

For now, yes. I will have to take 2 semesters of electromagnetism in 2025, though.

 

[laser1]

When you get to boundary conditions, you will see how the one described above is a direct consequence of Gauss's Law. I am sure you can find its derivation in Griffiths if you look for it.

I don't exactly follow this equation. "Upward" is the positive direction for both by definition. Fine. Gauss' law states that flux = E dot A. For the flux above $\sigma$, both E and A are positive. So E_above is positive, which agrees with the text. However, for the flux below $\sigma$, both E and A are negative. So why isn't E_below positive?

The reason why I think it should be E_above + E_below is: recall when you have a flat sheet of surface charge. The total flux is E_above + E_below. Why isn't in the same here?

 

[laser1]

Ah I think E_below is negative so it's fine?

 

[Orodruin]

Ah I think E_below is negative so it's fine?

No. The negative sign comes from the surface normal being in the negative direction for the surface below.

 

[laser1]

No. The negative sign comes from the surface normal being in the negative direction for the surface below.

The reason why I think it should be E_above + E_below is: recall when you have a flat sheet of surface charge. The total flux is E_above + E_below. Why isn't in the same here?

Are they not equivalent situations?

 

[kuruman]

You are missing the information that when you have a Gaussian surface, the normal to it is always outward. Consider the pillbox shown on the right. Only the flux through the top and bottom flat surfaces need to be considered in the limit that the pillbox shrinks to zero thickness. Note that the unit normals point in opposite directions, $\mathbf n_{\text{II}}=-\mathbf n_{\text{I}}.$ Then $$\Phi_{\text{total}}=\Phi_{\text{top}}+\Phi_{\text{bottom}} =\mathbf E_{\text{I}}\cdot \mathbf n_{\text{I}}~\Delta A+\mathbf E_{\text{II}}\cdot \mathbf n_{\text{II}}~\Delta A=(\mathbf E_{\text{I}}-\mathbf E_{\text{II}})\cdot \mathbf n_{\text{I}}~\Delta A=(E_{\text{I,n}}-E_{\text{II,n}})\Delta A. $$ That's how the minus sign comes into the picture.

 

[laser1]

$$(\mathbf E_{\text{I}}-\mathbf E_{\text{II}})\cdot \mathbf n_{\text{I}}~\Delta A=(E_{\text{I,n}}-E_{\text{II,n}})\Delta A. $$

I don't follow this step. Why isn't $\mathbf E_{\text{II}} \cdot \mathbf n_{\text{I}} = -E_{\text{II}}$? In your diagram, $\mathbf E_{\text{II}}$ points down. And $\mathbf n_{\text{I}}$ points up.

 

[Orodruin]

It is $E_{II,n}$ not $E_{II}$ - the component of $\vec E_{II}$ in the direction of $\vec n_I$. This is by definition $\vec E_{II}\cdot \vec n_I$.

 

[laser1]

It is $E_{II,n}$ not $E_{II}$ - the component of $\vec E_{II}$ in the direction of $\vec n_I$. This is by definition $\vec E_{II}\cdot \vec n_I$.

Ah, I see. This does mean that $E_{II,n}$ is a negative number, right? As they are always in opposite directions.

 

[Orodruin]

Ah, I see. This does mean that $E_{II,n}$ is a negative number, right? As they are always in opposite directions.

Huh? No. There is no requirement that the field points in any specific direction.

 

[laser1]

Huh? No. There is no requirement that the field points in any specific direction.

Ah I think I see my misconception. Usually when doing these Gauss' law questions the flux is just $2EA=\frac{\sigma A}{\epsilon_0}$. But when doing this I assumed that sigma was positive. However, I see now that sigma could just as well be negative, leading to $-2EA=\frac{\sigma A}{\epsilon_0}$, which is also valid, as $\sigma$ would be negative.

When writing $2EA=\frac{\sigma A}{\epsilon_0}$ in future, should I be more careful and say that this equation is only true if $\sigma > 0$? And if $\sigma < 0$, then $-2EA=\frac{\sigma A}{\epsilon_0}$.

Edit: I guess we could also just treat E as a signed quantity and use one formula.

 

[Orodruin]

Ah I think I see my misconception. Usually when doing these Gauss' law questions the flux is just $2EA=\frac{\sigma A}{\epsilon_0}$. But when doing this I assumed that sigma was positive. However, I see now that sigma could just as well be negative, leading to $-2EA=\frac{\sigma A}{\epsilon_0}$, which is also valid, as $\sigma$ would be negative.

When writing $2EA=\frac{\sigma A}{\epsilon_0}$ in future, should I be more careful and say that this equation is only true if $\sigma > 0$? And if $\sigma < 0$, then $-2EA=\frac{\sigma A}{\epsilon_0}$.

No, you have some serious misconceptions here. You seem to be thinking only about a field caused solely by an infinite flat sheet of constant surface charge. However, this is originally about a sphere and, more generally, about the field boundary conditions near any surface charge. You cannot assume a particular form of the field and the result is a discontinuity in the field component normal to the surface and proportional to the surface charge density at that point.