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

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In summary, the conversation discusses the calculation of the electric field at r = R for a uniformly charged spherical shell using Coulomb's law and Gauss' Law. It is pointed out that using Coulomb's law directly results in an undefined field due to the (r' - r') term in the denominator. However, applying Gauss' Law gives a finite value for the field at r = R. There is a discussion about the interpretation of this result and a mention of a discontinuity in the electric field at the boundary of a charged surface. It is also noted that in reality, the field is always smooth and the singularity at r = R is due to the approximation of an infinitely thin shell.
  • #1
quasar987
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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 [itex]\sigma[/itex] 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 [itex]\sigma \epsilon_0[/itex]. So what should we think? Is the sclalar function E(r) undefined at r = R or not ?!
 
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  • #2
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?
 
  • #3
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 [itex]\vec{r} = \vec{r'}[/itex], it is undefined.
 
  • #4
quasar987 said:
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 [itex]\vec{r} = \vec{r'}[/itex], 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.
 
  • #5
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.
 
  • #6
Yes,that's true.But that one is valid for [itex] \rho (\vec{r'}) [/itex] which is A VOLUME DENSITY OF CHARGE.You need another formula for a SURFACE DENSITY OF CHARGE...

Daniel.
 
  • #7
Oh. Right. Well. This one is given right before. It has exactly the same form, except

1) The integral is over a surface.
2) [itex]\rho(\vec{r'})[/itex] is replaced by [itex]\sigma(\vec{r'})[/itex]
3) dV is replaced by dA
 
  • #8
quasar987 said:
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.
 
  • #9
Mmh, but how should one interpret the result provided by Gauss' Law:
quasar987 said:
But, as Griffiths remarks (pp.88), if we apply Gauss' Law, we obtain that the field at r = R is of magnitude [itex]\sigma \epsilon_0[/itex]. So what should we think? Is the sclalar function E(r) undefined at r = R or not ?!
 
  • #10
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:

[tex](\vec E_1 - \vec E_2)\cdot \vec n = 4\pi \sigma[/tex]
or
[tex](\vec E_1 - \vec E_2)\cdot \vec n = \sigma/\epsilon_0[/tex]
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.
 

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

What does it mean for E(r) to be undefined at r=R?

When a function, such as E(r), is undefined at a certain point (in this case r=R), it means that the value of the function at that point is not defined or does not exist. In other words, the function does not have a specific or unique value at that particular point.

What factors can cause E(r) to be undefined at r=R?

There are several factors that can cause E(r) to be undefined at r=R. These can include mathematical limitations, physical constraints, or special cases where the function does not have a well-defined value at that point. It is important to carefully consider the context and assumptions of the problem to determine the cause of the undefined value.

Can E(r) still be used for calculations even if it is undefined at r=R?

It depends on the specific problem and the reason for the undefined value. In some cases, the undefined value may not affect the overall calculations and can be ignored. However, in other cases, the undefined value may indicate a problem with the calculations and should be addressed. It is important to carefully consider the implications of the undefined value before using the function for calculations.

How can E(r) be defined or resolved at r=R?

In some cases, the undefined value of E(r) at r=R may be a result of a mathematical or physical limitation. In these cases, it may not be possible to define or resolve the function at that point. However, if the undefined value is due to an incorrect assumption or calculation, it may be possible to redefine or resolve the function by addressing the underlying issue.

How can I determine if E(r) is undefined at r=R?

To determine if E(r) is undefined at r=R, you can evaluate the function at that point and see if it produces a unique value or if it is undefined (such as division by zero or taking the square root of a negative number). You can also check for any mathematical or physical constraints that may result in an undefined value. Additionally, carefully reviewing the problem and its assumptions can help identify any potential issues with the function's definition at r=R.

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