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We know that the electric field is constant at any point irrespective of the radius due to a plane sheet of charge of infinite extent and can be proved by Gauss theorem however I don't understand why do we need an infinite length because we can apply the same gauss theorem to a finite length of plane sheet of charge and we get the same result. Also incase of capacitors, the electric field is taken constant even when the sheets are finite.

I have another question which is that the electric field due to an infinite linear charge distribution at any point is inversely proportional to the radius but we can apply the Gauss theorem there too and prove that electric field can be constant at any point irrespective of the radius by drawing a cuboid perpendicular to the linear charge.

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    $\begingroup$ If the sheet of charge is finite then the electric field magnitude (and vector direction) will depend upon location with respect to the finite sheet of charge. So no, the answer will NOT be the same. $\endgroup$
    – honeste_vivere
    Commented Jan 3, 2018 at 14:20
  • $\begingroup$ But for capacitors we consider the electric field to be constant even tho it's finite. And we can apply Gauss law on any finite charge and draw a cylinder across it and say that the electric field is constant $\endgroup$
    – user180358
    Commented Jan 3, 2018 at 14:21
  • $\begingroup$ That is an approximation used when you have two very closely spaced (with respect to the size of the parallel plates) sheets of charge held apart by mechanical forces and one uses a dielectric to contain the fields between. This is very different than an isolated, finite sheet of charge. $\endgroup$
    – honeste_vivere
    Commented Jan 3, 2018 at 14:24
  • $\begingroup$ But like I don't understand why we only need an infinite plane sheet of charge, because if we take a finite sheet of charge and apply Gauss theorem to it then we will also get the same result. Is it because the electric field lines will not be perpendicular to the surface of the plane if it's finitie? Apart from that I don't find any reason. And what about a infinitely long uniformly linear charged rod, it also produces electric field but the electric field gets reduced/increased proportional to the radius. But here too we can apply Gauss law and say electric field is constant ? $\endgroup$
    – user180358
    Commented Jan 3, 2018 at 14:40
  • $\begingroup$ If you are on the axis of symmetry for the finite sheet or not on the wire in the 2nd case, the electric field should be one dimensional. The problem with the finite sheet is that the electric field is not constant with distance. Eventually if you are far enough way, the finite sheet of charge can be approximated as a point charge. Does that make sense? $\endgroup$
    – honeste_vivere
    Commented Jan 3, 2018 at 15:05

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Gauss' law relates the total electric flux leaving a closed surface to the charge enclosed in the surface. As you say, the electric field due to an infinite sheet of charge must, by symmetry, be perpendicular to the sheet (if it had a component parallel to the sheet, how could it prefer any direction over any other?). That means that for a cylindrical Gauss volume intersecting the sheet and with axis perpendicular to the sheet, the flux is entirely through the ends, with none through the sides. The enclosed charge is the same regardless of the length of the cylinder, so the flux through the ends must be also. Therefore the electric field is the same at any distance from the sheet.

This argument fails for a finite sheet, because the electric field is in general not perpendicular to the sheet, and so the total flux through the side of the cylinder is not zero.

To apply Gauss' law to an infinite line charge, use a cylinder whose axis coincides with the line. Then, by symmetry, there is no flux through the ends of the cylinder, and the flux per unit area through the tubular part is uniform. As you increase the radius of the cylinder, the enclosed charge is unchanged, so the total flux is unchanged. But the area of the tubular surface is proportional to the radius, so the flux PER UNIT AREA (i.e. the electric field) must be inversely proportional to the radius.

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  • $\begingroup$ I get it now for the infinite plate. But can we say the same for an infinite line charge that it's electric field is also constant? Because since it's infinite in length, the electric field lines emerging out are also perpendicular to the line. If we take cylindrical Gauss volume instersecting it, will not it be the same as it is with the plate? $\endgroup$
    – user180358
    Commented Jan 3, 2018 at 17:42
  • $\begingroup$ No, not the same, because the surface with the flux is in that case the curved side of the cylinder, not the ends, and to move further from the line, you increase the radius of the cylinder, thus increasing the area of the surface with the flux without increasing the flux itself. As in the case of the sheet, the total flux stays the same as the cylinder expands. In contrast to that case, the area over which the flux is distributed is not constant, but increases. So the electric field decreases with radius. $\endgroup$
    – Ben51
    Commented Jan 3, 2018 at 18:07

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