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# Chapter_3 - General Physics II Chapter 3 Gauss Law We now...

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General Physics II Chapter 3: Gauss’ Law We now want to quickly discuss one of the more useful tools for calculating the electric field, namely Gauss’ law. In order to understand Gauss’s law, it seems we need to know the concept of electric flux. We can image the flux as a volume rate of flow. We also can imagine the light come out to the room is the same as the light come out from the light bulk. Electric Flux Flux comes from the Latin word meaning “flow”. Field lines roughly describe the electric field strength. The strength of electric field is represented by the relative number of field lines passing through a unit area: Gaziantep University Faculty of Engineering Department of Engineering Physics 1 E K number of field lines/area θ φ cos i i i EA A E = = K K for whole surface: i A E K K = i A K Æ small Æ for continuous surface A d K Now we have EdA φ =⋅ K K v where is a whole surface integral. dA Example Find the flux that exits a sphere centered at the origin due to a point charge also at the origin. A small surface area on a sphere is, d G A = r 2 sin θ d θ d φ ˆ r . Using the field due to the point charge, G E = k q 2 r ˆ r , the flux can be calculated from its definition, Φ≡ G E d G A , k q 2 r ˆ r r 2 sin θ d θ d φ ˆ r Φ= k q sin θ 0 π d θ d φ 0 2 π = 4 π kq = q ε o Example Find the flux due to a point charge over the surface of a cube. Again start with the definition of flux, , and the field of a point charge, G E d G A G E = k q 2 r ˆ r . This time however, the electric field and the area element are not always parallel and the integral is very difficult. But, we know that the number of field lines that leave the charge is not affected by the shape of some imaginary surface that surrounds the charge . θ dA

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General Physics II That is, the same number of field lines exit the cube as exit the sphere in previous example so the flux must again equal Φ= q ε o . (Optional) In order to calculate flux through the cube due to point charge we can write the electric field in Cartesian coordinate system: ( ) 3/2 222 0 ˆˆ ˆ 4 qx x y y z z E xyz πε ++ = G Let us calculate the flux through the upper surface. The surface area is ˆ dA dxdyz = G , the the flux is () 0 4 upper q zdxdy πε ∫∫ I have evaluated this integral by using Mathematica( A computer program) and the result is 0 6 upper q ε . We have six surface the the total flux is 0 q ε .
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## This note was uploaded on 02/29/2012 for the course PHYS 227 taught by Professor Rabe during the Fall '08 term at Rutgers.

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Chapter_3 - General Physics II Chapter 3 Gauss Law We now...

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