lecture6 - L ecture 6 G ausss Law E lectric Potential...

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Lecture 6: Gauss’s Law Electric Potential Energy Electric Potential
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Physics Concept Components Learning Outcome Electric Force and Fields Charge and its analogy to mass Coulomb’s Law and its analogy to gravity Electric Fields Electric Flux Gauss’s Law Electric Potential Capacitance Material Properties Evaluate the forces of charged objects on one another. Evaluate value and magnitude of electric fields at arbitrary points in space. Evaluate the potential differences between objects (surfaces and points). Calculate flux with surface integrals. Deduce electric fields using Gauss’s Law. Discern between electric potential and electric potential energy Compute electric field and potential for point charges Compute electric field and potential from charge distributions August 12, 2013 2 Physics 2B -­‐ Summer Session 2 -­‐ C. Palmer
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We have a uniformly charged ball with radius, R, and total charge Q. August 12, 2013 3 Physics 2B -­‐ Summer Session 2 -­‐ C. Palmer R +Q
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What is the charge density ρ for this sphere? A) B) C) D) E) August 12, 2013 4 Physics 2B -­‐ Summer Session 2 -­‐ C. Palmer R +Q ρ = Q 4 3 π R 3 ρ = Q π R 2 ρ = Q 4 π R 2 ρ = Q 4 π R 3 ρ = Q 2 π R
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What is the charge density ρ for this sphere? A) B) C) D) E) August 12, 2013 5 Physics 2B -­‐ Summer Session 2 -­‐ C. Palmer R +Q ρ = Q 4 3 π R 3 ρ = Q π R 2 ρ = Q 4 π R 2 ρ = Q 4 π R 3 ρ = Q 2 π R
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Given the charge density is ρ for this sphere, what is the charge inside a spherical Gaussian surface with r<R? A) B) C) D) E) August 12, 2013 6 Physics 2B -­‐ Summer Session 2 -­‐ C. Palmer R +Q r q r ( ) = Q q r ( ) = Q r 3 R 3 q r ( ) = Q rR 2 R 3 q r ( ) = Q 3 r 2 4 R 3 q r ( ) = Q 3 r 2 R 3 ρ = Q 4 3 π R 3
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Given the charge density is ρ for this sphere, what is the charge inside a spherical Gaussian surface with r<R? A) B) C) D) E) August 12, 2013 7 Physics 2B -­‐ Summer Session 2 -­‐ C. Palmer R +Q r q r ( ) = Q q r ( ) = Q r 3 R 3 q r ( ) = Q rR 2 R 3 q r ( ) = Q 3 r 2 4 R 3 q r ( ) = Q 3 r 2 R 3 ρ = Q 4 3 π R 3
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We must write out the normal to the Gaussian surface (sphere with radius, r) and write out the direction of the electric field. August 12, 2013 8 Physics 2B -­‐ Summer Session 2 -­‐ C. Palmer R +Q r q r ( ) = Q r 3 R 3 ρ = Q 4 3 π R 3 E r ( ) ˆ r ˆ n = ˆ r
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We must write out the normal to the Gaussian surface (sphere with radius, r) and write out the direction of the electric field. Now we are ready to apply Gauss’s Law August 12, 2013 9 Physics 2B -­‐ Summer Session 2 -­‐ C. Palmer R +Q r ρ = Q 4 3 π R 3 E r ( ) ˆ r ˆ n = ˆ r E d A = Q Enclosed ε 0 q r ( ) = Q r 3 R 3
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Apply Gauss’s Law Left hand side August 12, 2013 10 Physics 2B -­‐ Summer Session 2 -­‐ C. Palmer R +Q r q r ( ) = Q r 3 R 3 ρ = Q 4 3 π R 3 E r ( ) ˆ r ˆ n = ˆ r E d A = Q Enclosed ε 0 E d A = E r ( ) ˆ r ˆ r dA = E r ( ) dA = E r ( ) 4 π r 2
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Apply Gauss’s Law Left hand side Right hand side August 12, 2013 11 Physics 2B -­‐ Summer Session 2 -­‐ C. Palmer
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