Unanswered Query: Is E(r) Undefined at r=R?

[quasar987]

I asked the following question to my teacher through email a week ago, and he hasn't replied.

Consider a uniformly charged spherical shell of radius R and areal density \sigma and consider also a cartesian system whose origin coincide with the center of the shell.

Clearly if we attempt to calculate the field at r = R directly through Coulomb's law, we obtain that the field is undefined, because we'd have a (r' - r') at the denominator (equ. 2.7 in Griffiths)

But, as Griffiths remarks (pp.88), if we apply Gauss' Law, we obtain that the field at r = R is of magnitude \sigma \epsilon_0. So what should we think? Is the sclalar function E(r) undefined at r = R or not ?!

 

[dextercioby]

It is defined...Naturally,since it has a finite value...Compute the field inside the sphere and outside and then plot the graph...Do you see anything interesting??

Daniel.

P.S.How do you compute the field using Coulomb's law?

 

[quasar987]

Well it's not so simple I think, since, like I said, using coulomb's law, it DO comes out undefined. Both assistant teachers said it was undefined, but I don't trust them.

Griffith's equ. 2.7 I have written in post #13 of the infamous Gauss' Law proof thread. I would copy it but for some reason, clicking the formula does nothing so I can't retrieve the code.

You can see, in the integral, if \vec{r} = \vec{r'}, it is undefined.

 

[dextercioby]

Well it's not so simple I think, since, like I said, using coulomb's law, it DO comes out undefined. Both assistant teachers said it was undefined, but I don't trust them.

Griffith's equ. 2.7 I have written in post #13 of the infamous Gauss' Law proof thread. I would copy it but for some reason, clicking the formula does nothing so I can't retrieve the code.

You can see, in the integral, if \vec{r} = \vec{r'}, it is undefined.

Are u referring to poste #6 and its first formula?
If so,then I have 2 objections...
a)That is not Coulomb's law...
b)You cannot use that formula...

Which formula need you use?

Daniel.

 

[quasar987]

Yes, I'm referring to the scalar equation of the field from post #6. That is the expression Griffiths gives for the field produced by a continuous distribution of charge.

 

[dextercioby]

Yes,that's true.But that one is valid for \rho (\vec{r'}) which is A VOLUME DENSITY OF CHARGE.You need another formula for a SURFACE DENSITY OF CHARGE...

Daniel.

 

[quasar987]

Oh. Right. Well. This one is given right before. It has exactly the same form, except

1) The integral is over a surface.
2) \rho(\vec{r'}) is replaced by \sigma(\vec{r'})
3) dV is replaced by dA

 

[krab]

Clearly if we attempt to calculate the field at r = R directly through Coulomb's law, we obtain that the field is undefined, because we'd have a (r' - r') at the denominator (equ. 2.7 in Griffiths)

It is undefined. It is related to the fact that an infinitely thin shell is unphysical. In the limit, the field has a discontinuity at r=R, but in real life, the shell does have a thickness and the field varies smoothly from zero inside the shell to its max value just outside.

 

[quasar987]

Mmh, but how should one interpret the result provided by Gauss' Law:

But, as Griffiths remarks (pp.88), if we apply Gauss' Law, we obtain that the field at r = R is of magnitude \sigma \epsilon_0. So what should we think? Is the sclalar function E(r) undefined at r = R or not ?!

 

[Galileo]

At theboundary of a charged surface there's always a discontinuity in the electric field. Specifically in the normal component of the electric field (normal to the surface). The parallel component is continuous.

I think the relation was:

(\vec E_1 - \vec E_2)\cdot \vec n = 4\pi \sigma
or
(\vec E_1 - \vec E_2)\cdot \vec n = \sigma/\epsilon_0
depending on your choice of units.
The n is the unit normal to the surface.

Therefore strictly the field is not defined. You always have a singularity in E precisely at the point where charge is.
Since this is an approximation, as krab pointed out, in reality E is always smooth.