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

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