Physics 1 Problem Solutions 194

Physics 1 Problem - 190 Chapt 24 Gauss’s Law planar slab with charge density dσ in a plate gives rise to a differential electric field d vector

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Unformatted text preview: 190 Chapt. 24 Gauss’s Law planar slab with charge density dσ in a plate gives rise to a differential electric field d vector E = dσ 2 ε ˆ n , where ˆ n points away from the slab on either side of the slab, ε is the vacuum permittivity, d vector E is constant outside of the differential slab, except that it direction changes from one side to the other of the slab as determined by ˆ n . This result follows from Gauss’s law. We can integrate up all these slabs immediately to get vector E = σ 2 ε ˆ n , where vector E is the electric field outside of the plate and ˆ n points away from the plate on either side of the plate. a) In the exact planar symmetry approximation, what is the electric field between the plates? Give both magnitude and direction. Only a symbolic answer is possible. b) At time zero, a proton (charge e ) is released from the positive plate and an electron (charge − e ) from the negative plate. Both particles are at rest at the time of release. They don’t) from the negative plate....
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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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