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For an infinite conducting sheet (it has a top and bottom, but is infinitely long and wide; an example of this would be figure 6.33 from Vol 2), how many sides of the rectangular box will have a non-zero flux?  Choices: One, Two, Three

For an infinite conducting sheet (it has a top and bottom, but is infinitely long and wide; an example of this would be figure 6.33 from Vol 2), how many sides of the rectangular box will have a non-zero flux?  Choices: One, Two, Three

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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For an infinite conducting sheet (it has a top and bottom, but is infinitely long and wide; an example of this would be figure 6.33 from Vol 2), how many sides of the rectangular box will have a non-zero flux? 

Choices: One, Two, Three

Gaussian surface and flux calculation In the present case, a convenient Gaussian surface is a box, since the expected electric field points in one direction only. To keep the Gaussian box symmetrical about the plane of charges, we take it to straddle the plane of the charges, such that one face containing the field point P is taken parallel to the plane of the charges. In Figure 6.33, sides I and Il of the Gaussian surface (the box) that are parallel to the infinite plane have been shaded. They are the only surfaces that give rise to nonzero flux because the electric field and the area vectors of the other faces are perpendicular to each other. AÃ E Side I Ep P. ΔΑ y E A Side II Figure 6.33 A thin charged sheet and the Gaussian box for finding the electric field at the field point P. The normal to each face of the box is from inside the box to outside. On two faces of the box, the electric fields are parallel to the area vectors, and on the other four faces, the electric fields are perpendicular to the area vectors. Let A be the area of the shaded surface on each side of the plane and Ep be the magnitude of the electric field at point P. Since sides I and Il are at the same distance from the plane, the electric field has the same magnitude at points in these planes, although the directions of the electric field at these points in the two planes are opposite to each other. Magnitude at I or II: E(z) = Ep. If the charge on the plane is positive, then the direction of the electric field and the area vectors are as shown in Figure 6.33. Therefore, we find for the flux of electric field through the box
Transcribed Image Text:Gaussian surface and flux calculation In the present case, a convenient Gaussian surface is a box, since the expected electric field points in one direction only. To keep the Gaussian box symmetrical about the plane of charges, we take it to straddle the plane of the charges, such that one face containing the field point P is taken parallel to the plane of the charges. In Figure 6.33, sides I and Il of the Gaussian surface (the box) that are parallel to the infinite plane have been shaded. They are the only surfaces that give rise to nonzero flux because the electric field and the area vectors of the other faces are perpendicular to each other. AÃ E Side I Ep P. ΔΑ y E A Side II Figure 6.33 A thin charged sheet and the Gaussian box for finding the electric field at the field point P. The normal to each face of the box is from inside the box to outside. On two faces of the box, the electric fields are parallel to the area vectors, and on the other four faces, the electric fields are perpendicular to the area vectors. Let A be the area of the shaded surface on each side of the plane and Ep be the magnitude of the electric field at point P. Since sides I and Il are at the same distance from the plane, the electric field has the same magnitude at points in these planes, although the directions of the electric field at these points in the two planes are opposite to each other. Magnitude at I or II: E(z) = Ep. If the charge on the plane is positive, then the direction of the electric field and the area vectors are as shown in Figure 6.33. Therefore, we find for the flux of electric field through the box
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