Lecture 3 Gauss’s Law Chp. 24FluxGauss’s LawElectric lines of flux and Gauss’s LawPowerPoint PresentationFind the electric flux through a cylindrical surface in a uniform electric field EElectric lines of flux and Derivation of Gauss’ Law using Coulombs lawSlide 8Applications of Gauss’s LawElectric field in and around conductorsSlide 11Slide 12Slide 13Slide 14Slide 15Warm up set 3Summer July 2004 1Lecture 3 Gauss’s Law Chp. 24•Cartoon - Electric field is analogous to gravitational field•Opening Demo -•Warm-up problem •Physlet /webphysics.davidson.edu/physletprob •Topics•Flux•Electric Flux and Example•Gauss’ Law•Coulombs Law from Gauss’ Law•Isolated conductor and Electric field outside conductor•Application of Gauss’ Law•Charged wire or rod•Plane of charge•Conducting Plates•Spherical shell of charge•List of Demos–Faraday Ice pail: metal cup, charge ball, teflon rod, silk,electroscopeSummer July 2004 2Flux € Φ =v cosθ AΦ =r v . r A r A = Aˆ n € θ =0Φ = vA€ θ =45Φ = 0.707vA€ Φ =(normal component) x Area€ =ΦSummer July 2004 3Gauss’s Law•Gauss’s law makes it possible to find the electric field easily in highly symmetric situations.•Drawing electric field lines around charges leads us to Gauss’ Law•The idea is to draw a closed surface like a balloon around any charge distribution, then some field line will exit through the surface and some will enter or renter. If we count those that leave as positive and those that enter as negative, then the net number leaving will give a measure of the net positive charge inside.Summer July 2004 4Electric lines of flux and Gauss’s Law• The flux  through a plane surface of area A due to a uniform field E is a simple product:  = E A where E is normal to the area A . € ˆ n € r E € ˆ n € r E € ˆ n € r E •  = En A =E cos •  = En A = 0 x A = 0 because the normal component of E is 0Summer July 2004 5 € Φ =r E ∑• Δr A € Φ =r E ∫• dr A Approximate FluxExact FluxCircle means you integrate over a closed surface. € dr A =ˆ n dASummer July 2004 6Find the electric flux through a cylindrical surface in a uniform electric field E € Φ =r E ∫• dr A € dr A =ˆ n dA€ = E∫cosθdA€ Φ = E∫cos180dA = − EdA∫= −EπR2a.b.€ Φ = E∫cos180dA = EdA∫= EπR2€ Φ = E∫cos90dA = 0c.Flux from a. + b. + c. What is the flux if the cylinder were vertical ?Suppose it were any shape?Summer July 2004 7Electric lines of flux and Derivation of Gauss’ Law using Coulombs law•Consider a sphere drawn around a positive point charge. Evaluate the net flux through the closed surface. € Φ =r E ∫• dr A dANet Flux = € = E∫cosθdA = EdA∫ E II nCos 0 = 1For a Point charge E=kq/r2€ Φ = EdA = kq /r2∫∫dA€ Φ =kq /r2dA∫= kq/r2(4πr2)€ Φ =4πkq€ 4πk =1/ε0 where ε0= 8.85x10−12C2Nm2€ Φnet=qencε0Gauss’ Law € dr A =ˆ n dA€ ˆ nSummer July 2004 8€ Φnet= qenc/ε0This result can be extended to any shape surfacewith any number of point charges inside and outside the surface as long as we evaluate thenet flux through it.Gauss’ LawSummer July 2004 9Applications of Gauss’s Law•Find electric filed of an infinite long uniformly charged wire of negligible radius.•Find electric field of a large thin flat plane or sheet of charge•Find electric field around two parallel flat planes•Find E inside and outside of a long solid cylinder of charge density and radius r.•Find E for a thin cylindrical shell of surface charge density •Find E inside and outside a solid charged sphere of charge density Summer July 2004 10Electric field in and around conductors•Inside a conductor in electrostaticequilibrium the electric field is zero( averaged over many atomic volumes).The electrons in a conductor move around so that they cancel out any electric field inside the conductor resulting from free charges anywhere including outside theconductor. This results in a net force ofF = eE = 0 inside the conductor.Summer July 2004 11Electric field in and around conductors•Any net electric charge resides on the surface of the conductor within a few angstroms (10-10 m).Draw a gaussian surface just inside the conductor. We know E = 0 everywhere on this surface. Hence , the net flux is zero. Hence, the net charge inside is zero.Show Faraday ice pail demo.Summer July 2004 12Electric field in and around conductors•The electric field just outside a conductor has magnitude  /0 and is directed perpendicular to the surface.–Draw a small pill box that extends into the conductor. Since there isno field inside, all the flux comesout through the top.– EA=q/0= A/ 0,– so E=  / 0Summer July 2004 13Summer July 2004 14Summer July 2004 15Summer July 2004 16Warm up set 3 1. [15388] Halliday /Resnick/ Walker Chapter 24 Question 2What is for (a) a square of edge length a, (b) a circle of radius r, and (c) the curved surface of a cylinder of length h and radius r.2. [153811] Halliday /Resnick/ Walker Chapter 24 Question 3.€