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Regarding electric field at the center of a uniformly charged disc?

by f2009049
Tags: field at the center
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f2009049
#1
Jan18-10, 09:58 AM
P: 3
The electric filed at a distance x from the center of a uniformly charged disc of radius R,along the axis passing through the center is given by E = sigma/2e(1-x/rt(x^2+R^2)) where sigma is surface charge density and e is permittivity of free space .Putting x=0 in this eqn gives filed at the center of the disc =sigma/2e ?? How is this possible ??? By symmetry considerations and considering the ring to be made up of concentric rings; for each such ring field at the center will be zero and thus the net field must be zero?? Can anyone explain why it is so ??
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elect_eng
#2
Jan18-10, 10:55 AM
P: 370
Quote Quote by f2009049 View Post
The electric filed at a distance x from the center of a uniformly charged disc of radius R,along the axis passing through the center is given by E = sigma/2e(1-x/rt(x^2+R^2)) where sigma is surface charge density and e is permittivity of free space .Putting x=0 in this eqn gives filed at the center of the disc =sigma/2e ?? How is this possible ??? By symmetry considerations and considering the ring to be made up of concentric rings; for each such ring field at the center will be zero and thus the net field must be zero?? Can anyone explain why it is so ??
Are you sure that expression is correct. I quickly did the calculation and came up with an expression that yields zero for x=0.

I obtained the following: E=sigma*x*R^2/(4*e*(R^2+x^2)^1.5)
clem
#3
Jan18-10, 12:14 PM
Sci Advisor
P: 1,256
Your expression is correct for positive x and for the limit x--> from above.
For negative x, the field is in the opposite direction. This gives the correct discontinuity in E when crossing a charged disk.

clem
#4
Jan18-10, 12:15 PM
Sci Advisor
P: 1,256
Regarding electric field at the center of a uniformly charged disc?

Quote Quote by elect_eng View Post
Are you sure that expression is correct. I quickly did the calculation and came up with an expression that yields zero for x=0.

I obtained the following: E=sigma*x*R^2/(4*e*(R^2+x^2)^1.5)
Check your calculation.
elect_eng
#5
Jan18-10, 01:03 PM
P: 370
Quote Quote by clem View Post
Check your calculation.
I did check it. If you feel his expression is correct then how do you answer his question. His question is the correct question to ask because the electric field must be zero at the center of the disk. At the center the radial component of the field must be zero because the charge is symmetrical. The axial component of field must be zero because there are no charges outside the plane of the disk.

EDIT: OK, I checked it again, and found my mistake. I get the same answer as given by the OP. Still would like to know the answer to the OP's question.
elect_eng
#6
Jan18-10, 02:11 PM
P: 370
I'll throw out my guess at answering this paradox.

In the real world you would not have a perfectly thin disk. If you imagine that the disk is two separate thin surfaces with small separation, and each with half of the charge, the boudary condition for the normal component of electric field is met on both sides of the disk, and the center of the disk has zero field.

This kind of thing could happen with a good conducting surface, although the problem does not restrict us to this case. The field would be nearly zero inside and the surface charge would be on a thin outer (i.e. molecular layers) shell of the conductor. The surface charge density would be half on the one side and half on the other of the disk. This idea should work until the disk is only a few atomic layers thick.

If we are not talking about conductors, but fixed charges embedded in an insulator, then the model of the inside of the disk is too simplistic, and we need to go beyond the concept of surface charge and use true volume based charge density.
clem
#7
Jan18-10, 07:37 PM
Sci Advisor
P: 1,256
You have to read an EM textbook. A conducting disk cannot have a uniform surface charge. Do you see anything wrong about my answer directly after the question?
The E field just has a discontinuity, no matter how thin the disk is.
elect_eng
#8
Jan18-10, 08:16 PM
P: 370
Quote Quote by clem View Post
You have to read an EM textbook. A conducting disk cannot have a uniform surface charge. Do you see anything wrong about my answer directly after the question?
The E field just has a discontinuity, no matter how thin the disk is.
I have read many EM textbooks. There is no need to make a dig like that. I understand EM theory very well, but being a little a rusty and human, I do make mistakes. You make a good point about the the conducting disk not having uniform charge. I wasn't really thinking about an exact solution, but putting forth a couple of ideas for the OP to think about.

I see nothing wrong with the discontinuity no matter how thin the disk is. The boundary conditions are met with no problems. I didn't mean my comments to imply that your comments were not right. I'm just trying to address the good paradoxical point that the OP has made. Your answer is good and correct, but it may not help the OP resolve the paradox in his mind.
f2009049
#9
Jan19-10, 06:15 AM
P: 3
I asked my friend : This is what he says:
" Mathematically , we are arriving at a result which makes us perceive that for any disk of finite radius having uniform charge distribution,as we approach the centre of the ring;it is same as we are approaching infinity.It is the radius x/r which matters.where r->infinity or x-->0 => the same"
I think as if he is just giving me a answer on mathematical basis.
Please someone throw some light on this question.
elect_eng
#10
Jan19-10, 10:14 AM
P: 370
Quote Quote by f2009049 View Post
Please someone throw some light on this question.
I think that my second suggestion above is the way to explain this.

Quote Quote by elect_eng
"If we are not talking about conductors, but fixed charges embedded in an insulator, then the model of the inside of the disk is too simplistic, and we need to go beyond the concept of surface charge and use true volume based charge density."
clem correctly noted that a conductive disk won't maintain a uniform surface charge density. So, we can model this disk as an insulator with fixed charges. I've attached a PDF with a calculation of the on-axis field when the disk is not perfectly thin. Here I show that the value of the field is indeed zero at the center of the disk. I then take the limit as the thickness of the disk goes to zero, and the answer is shown to agree with your stated solution for a perfectly thin disk.

Basically, your original simplistic model with an infinitely thin disk gives you an indeterminate value of the field at the center. Note that that approach shows equal, but opposite value of field when you use the solution from one side of the disk, or the solution from the other side. A simple minded thing to do is to guess that the indeterminate value is likely zero (i.e. the average of the two values). The attached computation demonstrates that this is correct if you monitor the value in the limit as the disk thickness goes to zero.

EDIT: Sorry, but I noticed a typo (sign error in one term) in the posted PDF, I've updated the PDF.
Attached Files
File Type: pdf Charged Disc.pdf (19.9 KB, 50 views)
elect_eng
#11
Jan19-10, 02:25 PM
P: 370
Just to elaborate a little further; I provided a normalization of the field equation and a plot. This plot makes it clearer what is happening. The value of the field keeps rising as you get closer to the disk, if the disk is not infinitely thin, you eventually hit the outer surface of the disk before you get to the center. Once you hit the disk, the field decreases as you penetrate, eventually reaching zero at the center. Modeling the thin disc as infinitely thin does not allow you to see this behavior. It can only be seen in the limiting process.
Attached Files
File Type: pdf Charged Disc2.pdf (12.2 KB, 34 views)


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