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Obtain the expression of electric field by a plane of infinite size and with uniform charge distribution. |
Answer» Solution :Let a be the uniform surface charge density of an infinite plane sheet.  Take the x-axis normal to the given plane. By symmetry, the electric field will not depend on y and z COORDINATES and its direction at every point must be parallel to the x-direction. We can take the GAUSSIAN surface to be a rectangular parallel piped of cross sectional area A, as shown. Only the two FACES 1 and 2 will contribute to the flux, electric field lines are parallel to the other faces and they do not contribute to the total flux. The unit vector normal to surface 1 is in -x-direction while the unit vector normal to surface 2 is in the + x-direction. Therefore, flux `VECE. DeltavecS`through both the surfaces are equal and add up. Therefore the net flux through the Gaussian surface is 2 EA.The charge enclosed by the closed surface is `sigmaA`. Therefore by Gauss.s law, `2EA = (sigmaA)/epsilon_(0)` `therefore E = sigma/(2epsilon_(0))` `therefore vecE =sigma/(2epsilon_(0)).hatn`...........(1) where `hatn` is a unit vector normal to the plan and going away from it. E is directed away from the plate if `sigma`is positive and toward the plate if `sigma`is negative. For a finite large planar sheet equation (1) i approximately true in the middle regions of th planar sheet, away from the ends. |
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