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Unformatted text preview: PHYS-1200 PHYSICS II SPRING 2005 Class 2 Activity: Flux and Gauss' Law ANSWERS 1. Area Vectors On the figure, draw an area vector → A for the flat side facing you of the object shown below. On the figure, draw two different area vectors → A d on the curved side of the object. → A d → A d → A 2. Flux through an open surface . An open surface is a 2-dimentional area that does not enclose any volume, like the area enclosed by a ring. As shown in the figure, a ring has an area vector defined in the direction shown. There is a light source on the right and the ring casts a shadow on a wall on the left. What are the angles between the area vector and the direction of the light when there is maximum light coming through the ring? θ = zero and 180 __º How about when there is no light coming through? θ = ___ 90 or 270 ____º Electric flux, Φ E , is proportional to the number of electric field lines passing through a surface. For an open surface, like a sheet of paper or the area enclosed by a ring or a 1 doorway, the flux through that surface is positive for a field line that has a component parallel to the area vector of that surface. Using the ring and light idea, think about the light passing through the ring in terms of FLUX as a function of the angle θ . Deduce the value of the angle θ at each of the three points indicated on the graph. Point indicated on the graph....
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This note was uploaded on 09/17/2008 for the course PHYS 1200 taught by Professor Stoler during the Spring '06 term at Rensselaer Polytechnic Institute.
- Spring '06
- Gauss' Law