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Unformatted text preview: Electric Flux and Gauss Law c Electric flux, definition cos E i i i i i E A = = E A r r surface lim i E i i A E E A d = = E A r r c Gauss law q in is the net charge inside the surface A E in E q d = = r r Applying Gauss Law c To use Gauss law, you need to choose a gaussian surface over which the surface integral can be simplified and the electric field determined c Take advantage of symmetry c Remember, the gaussian surface is a surface you choose, it does not have to coincide with a real surface Conditions for a Gaussian Surface c Try to choose a surface that satisfies one or more of these conditions: c The value of the electric field can be argued from symmetry to be constant over the surface c The dot product of can be expressed as a simple algebraic product EdA because and are parallel . c The dot product is 0 because and are perpendicular c The field is zero over the portion of the surface E r d E A r r E r d A r d A r Still no clue how to use Gauss Law? There are only three types of problems. See examples in the following pages. Problem type I: Field Due to a Spherically Symmetric Even Charge Distribution, including a point charge. c The field must be different inside ( r <a ) and outside ( r > a ) of the sphere. c For r > a, s elect a sphere as the gaussian surface, with radius r and Symmetric to the original sphere. Because of this symmetry, the electric field direction radially along r, and at a given r , the field magnitude is a constant. r E ) r r r 2 2 2 2 4 1 4 A E A E r Q k r Q k r q E q r E d d e e in in E = = = = = = = E is constant at a given r . Gauss Law As if the charge is a point charge Q Field inside the sphere c For r < a, s elect a sphere as the gaussian surface. c All the arguments are the same as for r > a. The only difference is here q in < Q c Find out that q in = Q( r/a ) 3 (How?) r r r r r 3 3 3 a a a Q k r Q k Q r r k r q k r q E q r E d d e e e in e in in E = = = = = = = = = E 3 2 2 2 2 1 4 1 4...
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 Spring '08
 Ye
 Charge, Magnetism

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