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electric field due to thin infinite …

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electric field due to thin infinite charge sheet
  • 2 answers

Kislay Harsh 4 years, 9 months ago

Use Gauss theorem, consider a horizontal cylinder as the gaussian surface. Now, there are three parts of the cylinder that are two circular ends and a curve part. The angle between area vector of curve part and the electric field is 90°, so the curve part will not be considered. The electric field will be only due to the two circular ends.

Strange Men 4 years, 9 months ago

For an infinite charge sheet , the electric field will be perpendicular to the surface. Therefore only the ends of a cylindrical Gaussian surface will contribute to the electric flux. The resulting feild is half that of a conductor at equilibrium with this surface charge density.
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