Lecture07 - Physics 132 Introductory Physics Electricity...

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1 January 18, 2008 Physics 132-Winter 2008 Prof. Jim Beatty-Ohio State 1 Physics 132 Introductory Physics: Electricity and Magnetism Winter Quarter 2008 Lecture 7 January 18, 2008 Physics 132-Winter 2008 Prof. Jim Beatty-Ohio State 2
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2 January 18, 2008 Physics 132-Winter 2008 Prof. Jim Beatty-Ohio State 3 Electric Flux E A Gauss’ Law is a relationship between electric flux and the presence of charges. For uniform E to plane Φ = EA Flux is the number of field lines crossing A. Units N m 2 /C January 18, 2008 Physics 132-Winter 2008 Prof. Jim Beatty-Ohio State 4 Electric Flux Sum over the area to get flux. Larger θ smaller Φ E A n ˆ n is the unit vector to A d Φ =E dA cos θ ˆ For arbitrary E: d Φ =E n dA ˆ
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3 January 18, 2008 Physics 132-Winter 2008 Prof. Jim Beatty-Ohio State 5 Review of Electric Field Results Point Charge: E 1/r 2 Sphere: A r 2 Line Charge: E 1/r Cylinder: A r Infinite Charged Plane: E 1 Plane: A 1 (both independent of Z) + + + + + + + + In all cases, EA is a constant! January 18, 2008 Physics 132-Winter 2008 Prof. Jim Beatty-Ohio State 6 Gauss’ Law Gauss’ Law: The total electric flux through a closed surface is proportional to the total charge enclosed. The surface can be any shape. It does not have to be the surface of a real object! Q enclosed =0 Φ =0 Φ = r E d r A closed surface = Q enclosed ε 0 Q enclosed =Q Φ =Q/ ε 0
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4 January 18, 2008 Physics 132-Winter 2008 Prof. Jim Beatty-Ohio State 7 Gauss’ Law for a Point Charge r E q Use Gauss’ Law to calculate the E field a distance r from a point charge q. Use symmetry : draw an imaginary spherical surface of radius r centered on q E is constant and normal on the surface. Gauss' Law : Φ = r E d r A = q ε 0 Φ = EA = 4 π r 2 E 4 π r 2 E = q ε 0 E = q 4 πε 0 r 2 Coulomb's Law Gauss’ Law implies Coulomb’s Law!
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