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?
The discussion revolves around the application of Gauss's law to a circular disk of charge, comparing it to the application of the law to infinite planes or plates. Participants explore the implications of symmetry in these scenarios and the challenges of deriving electric fields analytically for finite geometries.
Participants generally agree that Gauss's law applies to various electrostatic situations but disagree on its effectiveness for circular disks due to the lack of symmetry. The discussion remains unresolved regarding the best approach to analytically derive the electric field for a finite disk.
Limitations include the dependence on the definitions of symmetry and the challenges of applying Gauss's law to non-symmetric configurations. The discussion also highlights the complexity of transitioning from finite to infinite geometries.
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:But what's wrong when applying to circular disk which can also be considered as a plane?
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?
In case of disk having finite radii right?BvU said:For a disk (many disks are circular) of charge there is no such symmetry.
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.
There was two question. 1. find electric field due to infinite charged plate. In the solution, Gauss's Law was applied.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.
Yes. I saw it and to skip those lengthy calculation, I was to try applying Gauss's Law. But we can't right?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
Amazing! Thank you BvU... ☺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.
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
If R is infinite, then the fraction part goes to zero, leaving just the 1 as a multiplier, simplifying the answer to $$E = k \ \sigma \ \ 2 \ \pi$$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?