Find the electric field at a distance L directly above an infinitely large plane carrying uniform surface charge density E.

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Chapter1: Units, Trigonometry. And Vectors
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Find the electric field at a distance L directly above an infinitely large plane carrying
uniform surface charge density E.
Solution: To find the electric field at a distance L above the plane using Gauss' law, we
construct a closed surface with a top surface at a distance L above the plane and a
bottom surface at a distance L below the plane. We close the sides off by another
surface through which we get zero electric flux because the electric field is vertical.
Therefore, electric flux is non-zero through the top and bottom surface which gives
2EA for the left-hand-side of Gauss' law with A the area of the top and bottom
surfaces. The charge inside is EA so Gauss' law reads
2EA = = EA/EO
and E= E/(280) in magnitude and E = En/(280) where n is a normal unit vector.
Transcribed Image Text:Find the electric field at a distance L directly above an infinitely large plane carrying uniform surface charge density E. Solution: To find the electric field at a distance L above the plane using Gauss' law, we construct a closed surface with a top surface at a distance L above the plane and a bottom surface at a distance L below the plane. We close the sides off by another surface through which we get zero electric flux because the electric field is vertical. Therefore, electric flux is non-zero through the top and bottom surface which gives 2EA for the left-hand-side of Gauss' law with A the area of the top and bottom surfaces. The charge inside is EA so Gauss' law reads 2EA = = EA/EO and E= E/(280) in magnitude and E = En/(280) where n is a normal unit vector.
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