I Why can't we apply Gauss's law to a circular disk?

AI Thread Summary
Gauss's law can be applied to any electrostatic situation, including charged disks, but it is most effective in cases with high symmetry, such as infinite planes or spheres. For a finite charged disk, the lack of symmetry complicates the analytical solutions, often requiring numerical methods instead. When the radius of the disk approaches infinity, it behaves like an infinite plane, allowing the use of Gauss's law to derive similar results. Graduate-level electromagnetism emphasizes solving these problems using the Poisson equation and Green's functions rather than solely relying on Gauss's law. Ultimately, while Gauss's law is applicable, its utility diminishes in less symmetric configurations like finite disks.
r0ss
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We apply Gauss's law to find electric field at a point due to chaged plane or plate. But what's wrong when applying to circular disk which can also be considered as a plane?
 
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Hi Ross, :welcome:
In those example applications (infinite plate, sphere, ... ) there is a symmetry being exploited. For a disk (many disks are circular :wink: ) of charge there is no such symmetry.
 
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r0ss said:
But what's wrong when applying to circular disk which can also be considered as a plane?
Nothing is wrong with it. You can always apply Gauss’ law. However, it only gives you simple easy formulas when you can exploit a high degree of symmetry. Otherwise you would just use it numerically.
 
r0ss said:
We apply Gauss's law to find electric field at a point due to chaged plane or plate. But what's wrong when applying to circular disk which can also be considered as a plane?

Try it. Are you able to solve it analytically?

Gauss's law applies to ANY electrostatic situation. However, it doesn't mean that it can be solve analytically or easily in all those situations. Only in high-symmetric cases can this be done. Otherwise, you will have to solve it numerically.

Zz.
 
Thank You. I studied and understood that the plane in the question was an infinite plane but in case of the disk it was defined by finite radii R. I tested the equation derived from such disk assuming R is infinity which gave me the same equation derived assuming infinite charged plate using Gauss's law. When we consider R is infinite, we can exploit symmetry.
 
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BvU said:
For a disk (many disks are circular :wink: ) of charge there is no such symmetry.
In case of disk having finite radii right?
 
Yes. You can consider limiting cases: close by above the surface the disk looks like ifinite and very far away it looks like a point charge.
 
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r0ss said:
I studied and understood that the plane in the question was an infinite plane but in case of the disk it was defined by finite radii R. I tested the equation derived from such disk assuming R is infinity which gave me the same equation derived assuming infinite charged plate using Gauss's law.

I don't understand what this means.

A disk with "infinite radius" is an infinite plane. The shape, whether it is a disk, a rectangular plate, an oval plane, etc... no longer matters. So when you said that you "tested the equation" by using R → ∞, aren't you just applying Gauss's law to the usual infinite plane of charge? You haven't "tested" anything.

What we do in graduate level E&M (i.e. when you start using texts such as Jackson's Classical Electrodynamics), is that you solve for things like this, i.e. a finite disk of charge, and then, you see if (i) the off-axis solution matches the on-axis solution, and (ii) at field points very far away, the series solution approaches that of a point charge solution.

But none of these can be easily solved using Gauss's law. In fact, the whole point of solving these types of problems is to get the students to solve the Poisson equation and using Green's function method.

Zz.
 
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ZapperZ said:
I don't understand what this means.

A disk with "infinite radius" is an infinite plane. The shape, whether it is a disk, a rectangular plate, an oval plane, etc... no longer matters. So when you said that you "tested the equation" by using R → ∞, aren't you just applying Gauss's law to the usual infinite plane of charge? You haven't "tested" anything.

What we do in graduate level E&M (i.e. when you start using texts such as Jackson's Classical Electrodynamics), is that you solve for things like this, i.e. a finite disk of charge, and then, you see if (i) the off-axis solution matches the on-axis solution, and (ii) at field points very far away, the series solution approaches that of a point charge solution.

But none of these can be easily solved using Gauss's law. In fact, the whole point of solving these types of problems is to get the students to solve the Poisson equation and using Green's function method.

Zz.
There was two question. 1. find electric field due to infinite charged plate. In the solution, Gauss's Law was applied.
2. Find electric field due to charged disk having radii R at a point on the axis. In this case, a lengthy calculation had made to find the field. So I thought why can't we use Gauss's law. I tried to apply Gauss's law in the same manner in case of charged plane and failed to derive the equation in the solution for disk problem. Then I understood in case of finite value of R, we can't exploit symmetry hence Gauss's law. Finally I put value of R = infinity in the equation and found that it brcoming equation for infinite charged plane.
That was the whole scenario...
 
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BvU said:
Yes. You can consider limiting cases: close by above the surface the disk looks like ifinite and very far away it looks like a point charge.
Amazing! Thank you BvU... ☺
 
  • #13
rcgldr said:
Links that lead to an example using an infinite number of infinitely thin rings:
http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elelin.html#c3
r0ss said:
Yes. I saw it and to skip those lengthy calculation, I was to try applying Gauss's Law. But we can't right?
If R is infinite, then the fraction part goes to zero, leaving just the 1 as a multiplier, simplifying the answer to $$E = k \ \sigma \ \ 2 \ \pi$$
For a infinite rectangular plane you need to treat it as an infinite number of infinitely long lines with zero width (the result ends up the same).
 

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