Lecture 3-4 Ch23 - PH 222-3A Spring 2007 222 3A Gauss Law...

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H 222- A Spring 2007 PH 222 3A Spring 2007 Gauss’ Law Lectures 3-4 Chapter 23 (Halliday/Resnick/Walker, Fundamentals of Physics 8 th edition) 1
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Chapter 23 L Gauss’ Law In this chapter we will introduce the following new concepts: The flux (symbol Φ ) of the electric field Symmetry a ss’ la Gauss’ law We will then apply Gauss’ law and determine the electric field enerated by: generated by: An infinite, uniformly charged insulating plane n infinite, uniformly charged insulating rod An infinite, uniformly charged insulating rod A uniformly charged spherical shell A uniform spherical charge distribution We will also apply Gauss’ law to determine the electric field inside and outside charged conductors. 2
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Consider an airstream of velocity that is aimed at a loop of area . The velocity vector is at angle with respect to the ˆ op normal The product cos is know as v Av v A θ = Flux of a Vector. G G e In this lux loop normal . The product is know nas nv Φ= the . In this example the flux is equal to the volume flow rate through the loop (thus the name flux). flux depends on . It is maximum and equal to for 0 ( perpendicular to the loop vA v Note 1: G plane). It is minimum and equal to zero for 90 ( parallel t the loop plane). v G o e oop p a e). cos . The vector is parallel to the loop normal and h as magnitude equal to . vA v A A A =⋅ Note 2 : GG G ˆ n 3
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Consider the closed surface shown in the figure. Flux of the Electric Field. In the vicinity of the surface assume that we have a known electric field . The flux of the electric field thro h ug E Φ G the surface is defined as follows: 1. Divide the surface into small "elements" of area . 2. For each element calculate the term cos . A EA E A θ Δ ⋅Δ = G G G ˆ n ˆ 3. Form the sum . 4. Take the limit of the sum a E A Φ =⋅ Δ G s the area 0. he limit of the sum becomes the integra A Δ→ n 2 Flux SI unit: N m / C The limit of the sum becomes the integral: The circle on the integral sign indicates that EdA Φ= Note 1: G G v ˆ n the surface is closed. When we apply Gauss' law the surface is known as "Gaussian." is proportional to the net number of ote 2 : is proportional to the net number of electric field lines that pass through the surface. Φ Note 2: E dA G G v 4
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he flux of through any closed surface net charge enclosed by the surface Gauss' law can be formulated as follows: q = Gauss' Law G 0 enc 0e n c The flux of enclosed by the surface. In equ ation form: Equivalently: Eq q ε × Φ= n c ε E dA q ⋅= G G v Gauss' law holds for closed surface. Usually one particular surface makes the problem of determining the electric field very simple. Note 1: any ˆ n ˆ When calculating the net charge inside a c Note 2 : losed surface we take into account the algebraic sign of each charge.
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This note was uploaded on 09/26/2008 for the course PH 222 taught by Professor Mirov during the Spring '08 term at University of Alabama at Birmingham.

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Lecture 3-4 Ch23 - PH 222-3A Spring 2007 222 3A Gauss Law...

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