Chapter_23

Chapter_23 - Chapter 23: Gauss law In this chapter we will...

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11 In this chapter we will introduce the following new concept: flux of the electric field ( symbol Φ ) . We will then discuss Gauss’ law. We will apply Gauss’ law to determine the electric field generated by: Ø A point charge Ø An infinite, uniformly charged insulating plane Ø An infinite, uniformly charged insulating rod Ø A uniformly charged spherical shell Ø A uniform spherical charge distribution We will also apply Gauss’ law to determine the electric field inside and outside charged conductors. Chapter 23: Gauss’ law

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22 ˆ n ˆ n Flux Consider an airstream of velocity which is aimed at a loop of area A . The velocity vector is at angle I with respect to the loop normal. Then the product v A cos & is know as the volume flux I . It represents rate at which air volume flows through the loop . Flux = v A cos » depends on I : Ø It is maximum m max= vA when I =00 . Ø It is minimum min= 0 when m =900 . Since we can write that: Reminder: is an area vector. Area vector is directed in the direction of the loops normal and has magnitude equal to the loop’s area. v v A v vAcosθ = A v Φ = A
33 ˆ n ˆ n ˆ n Φ= E d A Flux of an Electric Field Let us consider the closed surface shown in the figure. In the vicinity of the surface we have a known electric filed . We can define flux I of the electric field through the surface as follows: Ø divide the surface into small “elements” of area I A . Ø For each such element calculate the flux Ø Calculate the sum Ø Let element I A become infinitely small so that the sum becomes an integral: E E ⋅∆ A = E ⋅∆ A cos θ Φ= E ⋅∆ A

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44 ˆ n ˆ n ˆ n Flux of an Electric Field (cont’d) Electric field flux I through a closed surface is given by Circle on the integral indicates that integration surface is closed. Such closed surface is called “ Gaussian surface ”. Since magnitude of electric field is proportional to the density of the field lines, we can conclude that: = A d E Φ The electric field flux I through a Gaussian surface is proportional to the net number of electric field lines passing through that surface.
55 ˆ n ˆ n ˆ n enc 0 q Φ ε = enc 0 q A d E ε = Gauss’ Law Gauss’s law can be formulated as follows: The flux of an electric filed though any closed surface 0 = net charge enclosed by the surface. In equation form: or equivalently Gauss’s law holds for any closed surface. But usually there is one surface which make calculation of the integral easiest. When calculating charge inside a closed surface one should take into account the algebraic signs of all charges enclosed. If enclosed charge

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This note was uploaded on 10/04/2010 for the course PHY 340410 taught by Professor Shi,z during the Fall '10 term at SUNY Buffalo.

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Chapter_23 - Chapter 23: Gauss law In this chapter we will...

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