lecture4 - L ecture 4 E lectric Fields of Continuous Charge...

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Lecture 4: Electric Fields of Continuous Charge Distributions Electric Flux Gauss’s Law
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Tuesday – 12:30PM-1:30PM After class Wednesday – 9:30AM-10:30PM Before class Let me know if Tuesday’s OH are at a bad time August 7, 2013 2 Physics 2B -­‐ Summer Session 2 -­‐ C. Palmer
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Tuesday – 12:30PM-1:30PM After class Wednesday – 9:30AM-10:30PM Before class Let me know if Tuesday’s OH are at a bad time August 7, 2013 3 Physics 2B -­‐ Summer Session 2 -­‐ C. Palmer z y x Vector symbols for into and out of the page is visualized with x’s and dots LIKE ARROWS Into the page (back of arrow) Out of the page (front of arrow)
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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 7, 2013 4 Physics 2B -­‐ Summer Session 2 -­‐ C. Palmer
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Find dE Total from a single point in the distribution Use electric field from a single point Need to know what is your junk of charge You will need to know the charge density If the overall charge is infinite, charge density will be given. If the overall charge is finite, you may have to compute it. Recognize symmetry to cancel certain components of the dE Total Typically only one component will survive but not always August 7, 2013 5 Physics 2B -­‐ Summer Session 2 -­‐ C. Palmer
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August 7, 2013 6 Physics 2B -­‐ Summer Session 2 -­‐ C. Palmer Charge Density +λ (C/m) z y x d dq=λd l d l R y θ E Z = k E λ d cos θ ( ) d θ π 2 π 2 = 2 k E λ d
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What is the y-component of the electric field due to dq at the “x”? A) B) C) D) August 7, 2013 7 Physics 2B -­‐ Summer Session 2 -­‐ C. Palmer z y x dE y = k E λ dl R 2 dE y = sin θ ( ) k E λ dl R 2 dE y = cos θ ( ) k E λ dl R 2 Charge Density +λ (C/m) d dq=λd l d l R y θ dE y = sin θ ( ) k E λ dl R 2
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What is the y-component of the electric field due to dq at the “x”? A) B) C) D) August 7, 2013 8 Physics 2B -­‐ Summer Session 2 -­‐ C. Palmer z y x dE y = k E λ dl R 2 dE y = sin θ ( ) k E λ dl R 2 dE y = cos θ ( ) k E λ dl R 2 Charge Density +λ (C/m) d dq=λd l d l R y θ dE y = sin θ ( ) k E λ dl R 2
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From determining the differential electric field and integrating we find 0 From symmetry we could have also seen that the net field in the y-direction must be 0 August 7, 2013 9 Physics 2B -­‐ Summer Session 2 -­‐ C. Palmer Charge Density +λ (C/m) z y x d dq=λd l d l R y θ E y = k E λ d sin θ ( ) d θ π 2 π 2 = 0
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