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

• I
• r0ss
In summary, when applying Gauss's law to a charged disk, you can only do so if the disk is considered to be a plane. If the disk is considered to be a circular shape, then Gauss's law still applies. However, because there is no symmetry, it is a more difficult problem to solve.
r0ss
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?

Hi Ross,
In those example applications (infinite plate, sphere, ... ) there is a symmetry being exploited. For a disk (many disks are circular ) of charge there is no such symmetry.

r0ss
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.

Dale
BvU said:
For a disk (many disks are circular ) 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.

r0ss
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.

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...

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... ☺

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).

## 1. Why can't we apply Gauss's law to a circular disk?

We cannot apply Gauss's law to a circular disk because Gauss's law only applies to closed surfaces, and a disk is an open surface.

## 2. What is the difference between a closed surface and an open surface?

A closed surface is one that completely encloses a volume, while an open surface does not enclose any volume and has an opening or boundary.

## 3. Why does Gauss's law only apply to closed surfaces?

Gauss's law is based on the principle of flux, which states that the net electric flux through a closed surface is equal to the charge enclosed by that surface. Therefore, Gauss's law can only be applied to closed surfaces that enclose a volume and have a defined amount of charge enclosed.

## 4. Can Gauss's law be applied to any other shapes besides closed surfaces?

Yes, Gauss's law can be applied to other shapes, such as spheres, cubes, and cylinders, as long as they are closed surfaces.

## 5. Is there an alternative law that can be applied to open surfaces like a circular disk?

Yes, there is an alternative law called Ampere's law, which can be applied to open surfaces such as a circular disk. Ampere's law states that the line integral of the magnetic field around a closed loop is equal to the current enclosed by that loop.

• Classical Physics
Replies
8
Views
804
• Classical Physics
Replies
20
Views
1K
• Classical Physics
Replies
17
Views
541
• Classical Physics
Replies
5
Views
1K
• Classical Physics
Replies
16
Views
837
• Classical Physics
Replies
1
Views
676
• Classical Physics
Replies
24
Views
714
• Classical Physics
Replies
5
Views
2K
• Classical Physics
Replies
2
Views
4K
• Classical Physics
Replies
7
Views
1K