The figure below shows a closed cylinder with cross-sectional area A 3.20 m². The constant electric field E has magnitude 1.85 x 10³ N/C and is directed vertically upward, perpendicular to the cylinder's top and bottom surfaces so that no field lines pass through the curved surface. Calculate the electric flux (in (N-m²)/C) through the cylinder's top and bottom surfaces. HINT (a) top surface (b) bottom surface N-m² N-m² (c) Determine the amount of charge (in C) inside the cylinder.

Physics for Scientists and Engineers: Foundations and Connections
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Author:Katz, Debora M.
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Chapter25: Gauss’s Law
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The figure below shows a closed cylinder with cross-sectional area A = 3.20 m².
The constant electric field E has magnitude 1.85 x 10 N/C and is directed vertically upward, perpendicular to the cylinder's top and bottom surfaces so that no field lines pass
through the curved surface. Calculate the electric flux (in (Nm²/C) through the cylinder's top and bottom surfaces.
WINTE
(a) top surface
(b) bottom surface
N-m²
N-m²
(c) Determine the amount of charge (in C) inside the cylinder.
Transcribed Image Text:The figure below shows a closed cylinder with cross-sectional area A = 3.20 m². The constant electric field E has magnitude 1.85 x 10 N/C and is directed vertically upward, perpendicular to the cylinder's top and bottom surfaces so that no field lines pass through the curved surface. Calculate the electric flux (in (Nm²/C) through the cylinder's top and bottom surfaces. WINTE (a) top surface (b) bottom surface N-m² N-m² (c) Determine the amount of charge (in C) inside the cylinder.
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