Lecture_05_2097

Lecture_05_2097 - Physics 0175 Lecture 5 (June 29, 2009)...

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Physics 0175 Lecture 5 (June 29, 2009) Applications of Gauss’s Law Electric Potential Equipotential Surfaces Calculating the Potential From the Electric Field Examples: Point Charge, Group of Point Charges Calculating the Potential: Continuous Charge Distributions 1
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Where to Go for Help 1. Your Recitation Section. 2. Office Hours: Lecturer: Prof. Brian R. D’Urso E-Mail: dursobr@pitt.edu Office Hours: Monday and Wednesday, 3:00 – 4:00 pm 315 Allen Hall TAs: Shonali Dhingra, shd28@pitt.edu Office Hours: Tuesday and Thursday, 8:30 pm – 9:30 pm and Friday, 2:00 pm – 4:00 pm 514 Allen Hall (desk #16) Naufer Nusran, nmn6@pitt.edu Office Hours: Tuesday and Thursday, 3:00 pm – 5:00 pm 514 Allen Hall (desk #7) Sui Chi Woo, suw11@pitt.edu Office Hours: Tuesday and Thursday, 1:00 pm – 3:00 pm 419 Allen Hall 3. Appointment with Lecturer or any TA. 2
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Homework Assignment #2 Reading: Chapters: 24, 25 Problems: See WileyPLUS Homework due Tuesday night / Wednesday morning! 3
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Gauss’ Law ∫∫ = = = = Φ 0 enc q dA E dA cos E d ε θ A E r r q enc = net enclosed charge Gauss’s Law is stated in terms of a surface integral over a closed surface. 4
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Under electrostatic conditions: • The electric field within conducting material must be zero . •Any excess charge on a conductor resides on its surfaces. Application of Gauss’ Law to Conductors 5
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• Any charge outside of a conductor has no effect on the electric field inside a cavity within the conductor. • Any charge inside of a cavity within a conductor contributes to the charge on the outer surface of the conductor. The charge on the outer surface is equal to the net charge of the system (conductor plus any charges within cavities inside the conductor). Electrostatic Shielding 6
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Gauss’s Law is convenient for applications to charge distributions with sufficient symmetry . • Spherical • Planar • Cylindrical Applications of Gauss’ Law 7
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Gauss’ Law and Coulomb’s Law Single Point Charge: Choose a spherical Gaussian surface with the charge at its center. () 2 0 2 0 0 enc 0 r q 4 1 E q r 4 E q dA E q d πε π ε = = = = A E r r • Choose a spherical Gaussian surface with the point charge at the center. • Due to the spherical symmetry, the electric field has the same magnitude and direction (outward or inwards) for every point on the surface. Therefore, it can be pulled from the integral. • The surface integral than becomes the surface area of a sphere. • The result for the electric field is identical to Coulomb’s Law. 8
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Planar Symmetry: Four Cases Four useful cases with planar symmetry: • One charged conductive plate.
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Lecture_05_2097 - Physics 0175 Lecture 5 (June 29, 2009)...

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