A closed surface is constructed in the space between two large parallel metal plates, where there is a constant electric field E. The magnitude of the field is El-1170 N/C. The numbers (1), (2), and (3) in the simulation label small increments of area of 1.5 cm² on the surface. The direction of each increment is defined to be the direction of a vector out of the surface and perpendicular to the increment's surface area. With that definition assume the direction of the first increment &A, is 65° from the direction of E, that the second increment AA, is oriented 90° from the direction of E, and that the third increment &A, is oriented 135 from the direction of E Find the flux through each of the three increments of area. For area A N-m²/C For area A N·m²/C For area &A Nm²/C

A closed surface is constructed in the space between two large parallel metal plates, where there is a constant electric field E. The magnitude of the field is El-1170 N/C. The numbers (1), (2), and (3) in the simulation label small increments of area of 1.5 cm² on the surface. The direction of each increment is defined to be the direction of a vector out of the surface and perpendicular to the increment's surface area. With that definition assume the direction of the first increment &A, is 65° from the direction of E, that the second increment AA, is oriented 90° from the direction of E, and that the third increment &A, is oriented 135 from the direction of E Find the flux through each of the three increments of area. For area A N-m²/C For area A N·m²/C For area &A Nm²/C

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter15: Electric Forces And Fields

Section: Chapter Questions

Problem 49P: Figure P15.49 shows a closed cylinder with cross-sectional area A = 2.00 m2. The constant electric...

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In which of the following contexts can Gausss law not be readily applied to find the electric field? (a) near a long, uniformly charged wire (b) above a large, uniformly charged plane (c) inside a uniformly charged ball (d) outside a uniformly charged sphere (e) Gausss law can be readily applied to find the electric field in all these contexts.
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Figure P15.49 shows a closed cylinder with cross-sectional area A = 2.00 m2. The constant electric field E has magnitude 3.50 103 N/C and is directed vertically upward, perpendicular to the cylinder's top and bottom surfaces so that no field lines paw through the curved surface. Calculate the electric flux through the cylinder's (a) lop and (b) bottom surface, (c) Determine the amount of charge inside the cylinder. Figure P15.49
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