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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?
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.But what's wrong when applying to circular disk which can also be considered as a plane?
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?
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.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?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
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 $$E = k \ \sigma \ \ 2 \ \pi$$Yes. I saw it and to skip those lengthy calculation, I was to try applying Gauss's Law. But we can't right?