
College Physics
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ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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A disk of radius 0.10 m is oriented with its normal vector n at 30 degrees to a unform electric field E vector of a magnitude of 2.0x10^3 N/C. What is the electric flux through the disk?
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Transcribed Image Text:A disk of radius 0.10 m is oriented with its normal vector n at 30 degrees to a uniform electric
field E vector of a magnitude of 2.0x103 N/C. What is the electric flux through the disk?
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Required
Solution =
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- could you solve (d) only please?arrow_forwardWe have calculated the electric field due to a uniformly charged disk of radius R, along its axis. Note that the final result does not contain the integration variable r: R. Q/A 2€0 Edisk (x² +R*)* Edisk perpendicular to the center of the disk Uniform Q over area A (A=RR²) Show that at a perpendicular distance R from the center of a uniformly negatively charged disk of CA and is directed toward the disk: Q/A radius R, the electric field is 0.3- 2€0 4.4.1barrow_forwardA spherical charge distribution of radius R is given by ρ = ρ0(1 – r/R), where r is the radial distance and ρ is the volume charge density. The electric field E is maximum at r equals: Zero 2.0R R/3 2R/3 0.745R Clear my choicearrow_forward
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