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

Info iconThis preview shows pages 1–6. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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
Background image of page 2
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
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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.
Background image of page 4
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
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 6
This is the end of the preview. Sign up to access the rest of the document.

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.

Page1 / 25

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

This preview shows document pages 1 - 6. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online