Physics 1 Problem Solutions 191

# Physics 1 Problem Solutions 191 - Chapt 24 Gauss’s Law...

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Unformatted text preview: Chapt. 24 Gauss’s Law 187 c) Coulomb’s law. d) Faraday’s law of induction. e) the Einstein equation. 024 qmult 00500 1 1 1 easy memory: Gauss law spherical 5. For a spherically symmetric charge distribution, Gauss’s law gives the magnitude of the electric field to be a) E = q encl 4 πε r 2 . b) E = λ encl 2 πε r , where λ encl is linear charge density. c) E = σ encl 2 ε , where σ encl is area charge density. d) E = q encl 4 πε r . e) V = q encl 4 πε r . 024 qmult 00510 2 5 1 moderate thinking memory: force on a spherically sym. distribution 6. From a Gauss’s law analysis, we know that a spherically symmetric charge distribution of total charge q creates an electric field outside of the distribution which is the same electric field that a point charge q located at the center of symmetry would create. Say we had a symmetrical charge distribution 1 and general charge distribution 2. The net force of distribution 1 on distribution 2 is vector F 12 . The net force of distribution 2 on distribution 1 is....
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## This note was uploaded on 11/16/2011 for the course PHY 2053 taught by Professor Buchler during the Fall '06 term at University of Florida.

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