40-43. Electric Flux- Application

40-43 Electric - (b What is the electric flux E through the door opened at = 90(c At what angle 1 is the electric flux through the door zero(d At

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Electric Flux: Application (1) Consider a rectangular sheet oriented perpendicular to the yz plane as shown and positioned in a uniform electric field ~ E = (2 ˆ j ) N/C. y z 2m 3m 4m E A (a) Find the area A of the sheet. (b) Find the angle between ~ A and ~ E . (c) Find the electric flux through the sheet. tsl40 – p.1/4
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Electric Flux: Application (2) Consider a plane sheet of paper whose orientation in space is described by the area vector ~ A = (3 ˆ j + 4 ˆ k )m 2 positioned in a region of uniform electric field ~ E = (1 ˆ i + 5 ˆ j - 2 ˆ k ) N/C. y z A E (a) Find the area of the sheet. (b) Find the magnitude of the electric field. (c) Find the electric flux through the sheet. (d) Find the angle between ~ A and ~ E . tsl41 – p.2/4
Background image of page 2
Electric Flux: Application (3) The room shown below is positioned in an electric field ~ E = (3 ˆ i + 2 ˆ j + 5 ˆ k ) N/C. 2m 1m z y θ (a) What is the electric flux Φ E through the closed door?
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: (b) What is the electric flux E through the door opened at = 90 ? (c) At what angle 1 is the electric flux through the door zero? (d) At what angle 2 is the electric flux through the door a maximum? tsl42 p.3/4 Electric Flux: Application (4) Consider a positive point charge Q at the center of a spherical surface of radius R . Calculate the electric flux through the surface. ~ E is directed radially outward. Hence ~ E is parallel to d ~ A everywhere on the surface. ~ E has the same magnitude, E = kQ/R 2 , everywhere on the surface. The area of the spherical surface is A = 4 R 2 . Hence the electric flux is E = EA = 4 kQ . Note that E is independent of R . tsl43 p.4/4...
View Full Document

This note was uploaded on 04/30/2008 for the course PHYS 204 taught by Professor Andrevantonder during the Spring '07 term at Rhode Island.

Page1 / 4

40-43 Electric - (b What is the electric flux E through the door opened at = 90(c At what angle 1 is the electric flux through the door zero(d At

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online