Electricfield of a Disk using Gauss's law

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Discussion Overview

The discussion revolves around the calculation of the electric field of a disk with a given charge density using Gauss's law. Participants explore the applicability of Gauss's law in this context, particularly considering the disk's thickness and symmetry. The conversation also touches on the relationship between electric fields and electric potential.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant proposes using Gauss's law to find the electric field of a disk, breaking it into concentric rings and integrating Coulomb's law.
  • Another participant points out that the assumption of a uniform electric field across the Gaussian surface is incorrect, noting that the field would only be uniform for an infinite disk.
  • There is a suggestion to treat the disk as having infinitesimal layers if considering its thickness, which complicates the integral.
  • Some participants argue that Gauss's law cannot be effectively applied to a disk due to its lack of symmetry.
  • Others clarify that Gauss's law is valid in all cases but is most useful in situations with sufficient symmetry, such as spherical or cylindrical distributions.
  • There is a discussion about the meaning of uniform charge density and whether it implies a three-dimensional object or can be treated as a two-dimensional surface charge.
  • One participant questions the equivalence of two equations for calculating electric potential, seeking clarification on their applicability.
  • Another participant emphasizes that Gauss's law is always valid but finding the electric field using it is limited to specific symmetrical cases.

Areas of Agreement / Disagreement

Participants generally agree that Gauss's law is valid but express differing views on its applicability to the disk scenario, with some asserting that its symmetry is inadequate for effective use. The discussion remains unresolved regarding the best approach to calculate the electric field of the disk.

Contextual Notes

Participants note the complexity introduced by the disk's thickness and the implications of charge density definitions. There is also mention of the need for sufficient symmetry for Gauss's law to be useful, indicating potential limitations in the mathematical application of the law in this case.

perplexabot
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Electricfield of a Disk using Gauss's law!

Hi all. Trying to find electric field of a disk with a charge density of D and a radius R (let us assume it has a thickness h). I know you can do so by breaking up the disk into concentric rings and integrating coulombs law. However, i would like to go about using gauss's law if that is possible. So:

Let k = permittivity of free space (usual symbol: epsilon)

(integral of) (E*dA) = QEnc/k

gonna use a cylinder as a gaussian surface (pillbox).

E * (integral of) (dA) = QEnc/k

QEnc = D*V​
(integral of) (dA) = (2*pi*R^2) + (2*pi*R*h) = 2*pi*R * (R + h)​

=> E * 2*pi*R*(R+h) = (D*V)/k
=> E = (D*V) / (k*2*pi*R*(R+h))

V = h*pi*R^2​

=> E = (D*h*pi*R^2) / (k*2*pi*R*(R+h))
=> E = (D*R*h) / (2*k*(R+h))

this is my final answer, i know it's probably wrong but i have no idea why. Also using coulombs law as mentioned above does not leave an h variable (thickness of the cylinder) in the final equation. Can someone please shed some light as to what my problem is?

P.S: My final goal is to find the electric potential of a point on the disk's axis a distance x from the middle of the disk. My book uses V = (integral of) dQ/r but I'm trying to use Va - Vb = (integral of E*dL), So that is why I'm trying to find the E-field. Help provided for calculating electric Potential of disk will be greatly appreciated.
 
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perplexabot said:
this is my final answer, i know it's probably wrong but i have no idea why.
You assume that the field is uniform across your Gaussian surface, but it's not. (It would be uniform--or zero--if the disk were infinite.)

Also using coulombs law as mentioned above does not leave an h variable (thickness of the cylinder) in the final equation.
Treating the disk as having a non-zero thickness just makes the integral harder to do. If you really want to do it, break the disk into infinitesimal layers.
 


Doc Al said:
You assume that the field is uniform across your Gaussian surface, but it's not. (It would be uniform--or zero--if the disk were infinite.) <----------[1st quote]


Treating the disk as having a non-zero thickness just makes the integral harder to do. If you really want to do it, break the disk into infinitesimal layers. <----------[2nd quote]

Reply to 1st quote:
wow that makes great sense. Thank you. But let's say i still want to go on and use Gauss's law with non-uniform e-field, what would i have to do? or does Gauss's law only apply to symmetrical objects.

Reply to 2nd quote:
I got this question from a problem in a book. And it says that the disk has a uniform DENSITY charge, which i assume means that the cylinder has 3-dimensions, which led me to use h. Why is it that we can disregard an objects thickness?

Once again thank you for the reply.
 


Gauss's law can't be used for a disc, because it has the wrong symmetry.
 


perplexabot said:
Reply to 1st quote:
wow that makes great sense. Thank you. But let's say i still want to go on and use Gauss's law with non-uniform e-field, what would i have to do? or does Gauss's law only apply to symmetrical objects.
Gauss's law applies in all cases, but is only useful in cases with sufficient symmetry. As clem said, a finite disk has the wrong kind of symmetry.

Reply to 2nd quote:
I got this question from a problem in a book. And it says that the disk has a uniform DENSITY charge, which i assume means that the cylinder has 3-dimensions, which led me to use h. Why is it that we can disregard an objects thickness?
They probably meant a uniform surface charge density, usually symbolized as σ. (Unless of course they specifically said it has thickness h.) If the disk is thin enough, treating it as just a charged surface is good enough.
 


rcgldr said:
The math for electric field of disk is shown here (scroll up to see math for ring and line):

http://hyperphysics.phy-astr.gsu.edu/%E2%80%8Chbase/electric/elelin.html#c3

Potentials here:

http://hyperphysics.phy-astr.gsu.edu/%E2%80%8Chbase/electric/potlin.html#c3

It's also possible to do this for an infinite plane by segmenting the plane into thin infinitely long rectangles that can be treated as infinitely long lines for an integral. The result is the same as an infinitely wide disc.

thanks for the links. i got one question though.

For calculating Electric potential you can use either of the following equations (assuming all required variables are given)?:
Va-Vb = (integral of) E*dL​
(equation 1)
V = (integral of) dQ/r​
(equation 2)

in the link you gave for calculating electric potential they used equation 2, i just used equation 1 with the E-field (e-field from link) and achieved the same answer. Is this always the case?

clem said:
Gauss's law can't be used for a disc, because it has the wrong symmetry.


ok, so gauss's law only works for specific symmetries? And when gauss's law fails i should just resort to coulomb's law.

Thank you all
 
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Doc Al said:
Gauss's law applies in all cases, but is only useful in cases with sufficient symmetry. As clem said, a finite disk has the wrong kind of symmetry.


They probably meant a uniform surface charge density, usually symbolized as σ. (Unless of course they specifically said it has thickness h.) If the disk is thin enough, treating it as just a charged surface is good enough.

ok so gauss's concept always applies but is hard to apply mathematically with bad symmetries.
And no the problem says nothing about thickness. Thank you for clarifying that.
 


Gauss's law is ALWAYS valid. However, finding the electric field using that law is only possible in three cases:
a) spherically symetric charge distribution. The charge density may vary with r but not with theta and phi.
b) Infinitely long cylindrical distribution. Again, the charge density may depend on r but not on z or phi.
c) Uniformly charged plane.
d) A clever combination of the above.
 


Gordianus said:
Gauss's law is ALWAYS valid. However, finding the electric field using that law is only possible in three cases:
a) spherically symetric charge distribution. The charge density may vary with r but not with theta and phi.
b) Infinitely long cylindrical distribution. Again, the charge density may depend on r but not on z or phi.
c) Uniformly charged plane.
d) A clever combination of the above.

thank you for clearing that out!
 

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