LECT4_159161

LECT4_159161 - Gauss Law E = Qencl 0 E through any closed...

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Gauss’ Law Φ E = Q encl / ε 0 Φ E through any closed surface is equal to the net charge enclosed, Q encl , div. by ε 0
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Sample Gaussian surfaces Hint: Choose surfaces such that E is or || to surface!
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Gauss’ Law: A sheet of charge Define σ = charge per unit area
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Gauss’ Law: Charged Spherical Shell At r < a: E = 0.
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Gauss’ Law: Charged Spherical Shell At r > b, E looks like that from a single point charge Q Divide both sides by area: 4 πε 0 r 2 E= Q encl 4 E π r 2
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Gauss’ Law: 2 planes with opposing charges
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Gauss’ Law: 2 planes with opposing charges + + + + + + + _ _ _ _ _ _ _ + + + + + + + _ _ _ _ _ _ _ E=– σ /(2 ε 0 ) E=+ σ /(2 ε 0 ) E=+ σ /(2 ε 0 ) E=– σ /(2 ε 0 )
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Gauss’ Law: 2 planes with opposing charges + + + + + + + _ _ _ _ _ _ _ Inside: E=+ σ /(2 ε 0 )+ σ /(2 ε 0 ) = σ / ε 0 E=0 outside E=0 outside
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Electrical potential energy corresponding to Coulomb force (e.g., assoc. with distributions of charges) Total Energy = K.E. + P.E. Electric Potential = P.E. per unit charge
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This note was uploaded on 09/30/2011 for the course PHYS 1B taught by Professor Briankeating during the Summer '07 term at UCSD.

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LECT4_159161 - Gauss Law E = Qencl 0 E through any closed...

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