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BvU

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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.

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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?

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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.

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I don't understand what this means.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 infinitywhich gave me the same equation derived assuming infinite charged plate using Gauss's law.

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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rcgldr

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http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elelin.html#c3

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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.

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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Yes. I saw it and to skip those lengthy calculation, I was to try applying Gauss's Law. But we can't right?

http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elelin.html#c3

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rcgldr

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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?

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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