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Why is Gauss's law used to calculate electric field only for "infinite" cases such as an infinite sheet, infinite wire, etc.? Why does the equation of calculating electric field near an "infinitely" non conductor charged sheet which is equal to $\sigma/2 \epsilon$ not work for finite sheet (with uniformly charge distribution)?Likewise why we do not use Gauss's law to compute electric field near an "finitely" long wire instead of an infinite wire?(Above description only for the view of my 12th class textbook,l don't know about any higher study book which may compute the electric field near a finite object using Gauss's law and advanced Calculus).[1]: https://i.stack.imgur.com/X9rd9.jpg

Abu sayed
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  • Formulas for infinite cases are derived from finite cases, usually by integrating from $-\infty$ to $\infty$. Infinite cases are generally helpful to consider since they can provide good approximations for very large (comparatively) finite sheets, wires, etc. – zh1 Aug 27 '16 at 19:55
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    @zhutchens1 Infinite objects are considered because you can get the exact solution from symmetry arguments (i.e. you avoid doing the integral entirely). – DanielSank Aug 27 '16 at 20:30
  • @DanielSank is correct. The question is about Gauss' Law. However I'll note that another one of Maxwell's equations, Ampere's Law, gives incorrect results when applied to some finite systems, particularly a finite-length wire. – garyp Aug 27 '16 at 21:34
  • The only in your question is puzzling. For example: The electric field of a sphere of uniform charge density and total charge charge Q can be obtained by applying Gauss' law – Alfred Centauri Aug 27 '16 at 22:23
  • @garyp Is there a glaring case where the infinite case doesn't approximate well any finite system? One example I can think of but isn't quite satisfying is the field inside an extremely large charged sphere of radius $R \lt \infty$ . If it is only extremely large, Gauss law yields that $|\vec E| \propto r$. However, for infinitely large spheres, $|\vec E| = 0$, by symmetry. The worst case error from the "infinitely large" approximation is thus catastrophically $\propto R$. – Real Aug 28 '16 at 03:53
  • @garyp just to nitpick on phrasing, the Maxwell-Ampere law is always true, but when it's applied incorrectly it can give incorrect results. –  Aug 28 '16 at 04:03
  • The basic answer, why we do not use gauss's law for finite distributions, is because of the lack of symmetry needed to use the flux equations efficiently. One can definitely use gauss law in this case too. But you would have to know what the field is at each point in space. You can see that it wont be very easy to do that. – Lelouch Aug 28 '16 at 04:34
  • What do you mean by lack of symmetry in the case of "finite" charged distribution?Please explain in detail. – Abu sayed Aug 28 '16 at 06:06

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Infinite regions are often easy to cope with owing to the high degree of symmetry an infinite region can allow. As such, they serve as excellent teaching examples because their analysis requires a keen understanding of the underlying physics without the need for too much mathematical wizardry which would distract the learning effort from physics. In short, they are simple, clear, instructive examples.

For your infinite charged sheet example, the infiniteness begets the symmetry of translation invariance in a plane. That is, suppose the sheet lies in the $x-y$ plane. Then, because the sheet never ends, the problem is exactly the same if shift the origin to anywhere within the plane. You can then begin from an assumption of the invariance to the solution to displacements in the $x-y$ plane and deduce rigorously that the electric field must be everywhere normal to the plane, but you can see this instantly intuitively too: if the field were not normal, there would need to be a preferred direction in the $x-y$ plane defined by the component of the field in the plane. Moreover, the field magnitude and direction cannot change as one moves from point to point in the plane. So you quickly deduce that the field is normal to the plane, and has a magnitude $\sigma/\epsilon$ on applying Gauss's law to the assumed normal, $x-y$ translation invariant field.

The same is not true if the sheet is finite. In that case, shifts in the co-ordinates in the $x-y$ direction must make a difference to the solution - imagine translating off the edge of the plane and you clearly have a different situation. This lack of symmetry means we cannot infer an everywhere normal field, and indeed near the edges the field fringing becomes significant.

Selene Routley
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    In fact, we cannot even assume a uniform charge distribution on a finite sheet, in a real world situation. At the edges of the sheet, a charged particle will be repelled by all the other charge on the sheet, with nothing to counterbalance that force. At this stage in your learning, there is no educational value in either ignoring such real world effects without comment, or in complicating the math (by orders of magnitude) to include them. For that reason, most undergraduate physics problems are "spherical cows" in some respects. – alephzero Aug 28 '16 at 04:48
  • So,in the case of a finite sheet ,the surface charge density is not uniformly distributed over the edges of the surface and thus we can't use Gauss's law to calculate the field near the sheet "easily"? Is my view right? Another question is that though the sheet is finite,can we take very small region and apply Gauss's law (because in that Case the surface charge will uniformly distributed over the small region of the surface)? – Abu sayed Aug 28 '16 at 05:52
  • I can't understand enough- the symmetry of translation invariance in a plane break down to the finite sheet in the answer of WetSavannaAnimal aka Rod Vance. – Abu sayed Aug 28 '16 at 06:03
  • @Abusayed Think about the electric field over the infinite charged plane. How could the field in a region have a certain value $c_1$, and in another a certain value $c_2$? Isn't the only mathemtical difference between $c_1$ and $c_2$ their coordinates? Now consider choosing a different coordinate origin (a translation operation). The problem stays exactly the same. This invariance of a problem to the translation operation is called a symmetry w.r.t. translation. – Real Aug 29 '16 at 02:03