Gauss’ Law

1 Fall 2008 Physics 231 Lecture 2-1 Gauss LawFall 2008 Physics 231 Lecture 2-2 Electric Field LinesThe number of field lines, also known as lines of force, are related to strength of the electric fieldMore appropriately it is the number of field lines crossing through a given surface that is related to the electric fieldFall 2008 Physics 231 Lecture 2-3 FluxHow much of something passes through some surfaceNumber of particles passing through a given surfaceTwo ways to defineNumber per unit area ( , 10 particles/cm2)Number passing through an area of interestFall 2008 Physics 231 Lecture 2-4 Electric FluxThe electric flux is defined to beAEE= Where E is the electric field and A is the areaFall 2008 Physics 231 Lecture 2-5 Electric FluxIf surface area is not perpendicular to the electric field we have to slightly change our definition of the flux cosAEE= Where is the angle between the field and the unit vector that is perpendicular to the surfaceFall 2008 Physics 231 Lecture 2-6 Electric FluxWe can see that the relationship between the flux and the electric field and the area vector is just the dot product of two vectorsnAEAEEE = = rrrn is a unit vector perpendicular to the surfaceFall 2008

2 Physics 231 Lecture 2-7A ConventionThe direction of a unit vector for an open surface isambiguousFor a closed surface, the unit vector is taken as being pointed outwardFall 2008 Physics 231 Lecture 2-8 Electric FluxWhere flux lines enter the surface, the surface normal and the electric field lines are anti-parallelWhere the flux lines exit the surface they are parallelFall 2008 Physics 231 Lecture 2-9 Electric FluxIs there a difference in the net flux through the cube between the two situations?No!It is important to remember to properly take into account the various dot productsFall 2008 Physics 231 Lecture 2-10 Electric FluxThe equation we have for flux is fine for simple situationsthe electric field is uniform and the surface area is planeWhat happens when either one or the other or both is not trueFall 2008 Physics 231 Lecture 2-11 Electric Flux We proceed as we did in the transition from discrete charges to a continuous distribution of chargesWe break the surface area into small pieces and then calculate the flux through each piece and then sum themIn the limit of infinitesimal areas this just becomes an integral = AdEErrFall 2008 Physics 231 Lecture 2-12 Electric Flux of

3 A Point ChargeWhat is electric flux that comes from a point charge?We start from = AdEErr2041rqE =The problem has spherical symmetry, we therefore use a sphere as the Gaussian surfaceThe electric field is given bySince E is radial, its dot product with the differential area vector, which is also radial, is always oneAlso E is the same at every point on the surface of the sphereFall 2008 Physics 231 Lecture 2-13 Electric Flux of a Point ChargeFor these reasons, E can be pulled out from the integral and what remains is = dAEEThe integral over the surface area of the sphere yields24rA =Pulling all this together then yields0220441; qrrqEAEEE= = = Notice that this is independentof the radius of the sphereFall 2008 Physics 231 Lecture 2-14dS1dS2 Example 1A positive charge is contained inside a spherical shell.

4 How does the differential electric flux, dФE, through the surface element dSchange when the charge is moved from position 1 to position 2?dФEa) increasesb)decreasesc)doesn t changeFall 2008 Physics 231 Lecture 2-15dS1dS2 Example 1 -continueddФEa) increasesb)decreasesc)doesn t changeThe total flux of a charge is constant, with the density of fluxlines being higher the closer you are to the chargeTherefore as you move the charge closer to the surface element, the density of flux lines increasesMultiplying this higher density by the same value for the area of dSgives us that the incremental flux also increasesFall 2008 Physics 231 Lecture 2-16dS1dS2 Example 2A positive charge is contained inside a spherical shell.

5 How does the total flux, ФE,through the entire surface change when the charge is moved from position 1 to position 2?a) ФE increasesb)ФE decreasesc)ФE doesn t changeAs we previously calculated, the total flux from a point charge depends only upon the chargeFall 2008 Physics 231 Lecture 2-17 Gauss LawThe result for a single charge can be extended to systems consisting of more than one charge = iiEq01 One repeats the calculation for each of the charges enclosed by the surface and then sum the individual fluxesGauss Law relates the flux through a closed surface to charge within that surfaceFall 2008 Physics 231 Lecture 2-18 Gauss LawGauss Law states thatThe net flux through any closed surface equals the net (total)

6 Charge insidethat surface divided by 00 netEQAdE= = rrNote that the integral is over a closedsurfaceFall 2008 Physics 231 Lecture 2-19 Example 3A blue sphere A is contained within a red spherical shell B. There is a chargeQAon the blue sphere and chargeQBon the red spherical electric field in the region between the spheres is completely independent ofQBthe charge on the red spherical 2008 Physics 231 Lecture 2-20 SurfacesChoose surface appropriate to problemIt doesnothave to be a sphereExploit symmetries, if anyFall 2008 Physics 231 Lecture 2-21 Example 4 Thin Infinite Sheet of ChargeA given sheet has a charge density given by C/m2By symmetry, E is perpendicular to the sheetEUse a surface that exploits this factA cylinderA Gaussian pillboxFall 2008 Physics 231 Lecture 2-22 Thin Infinite Sheet of Charge00 AAEEAAEAAdErightcurvedleft=++= rrBut E and Acurvedare perpendicular to each other so their dot product is zero and the middle term on the left disappears0022 == EAAEFall 2008 Physics 231 Lecture 2-23 Example 5 Infinite Line having a Charge Density Apply Gauss Law.

7 H+ + + +y+ + + + +ErEr+ + + + + + + ++ + + + + +By SymmetryTherefore, choose the Gaussian surface to be a cylinder of radius rand length haligned with the x-axisE-field must be to line of charge and can only depend on distance from the lineEquating these and rearranging yieldsOn the ends, 0= SdErrsince E//is zeroandOn the barrel,rhESdE 2= rrhq =rE02 =This is the same result as using the integral formulationFall 2008 Physics 231 Lecture 2-24 Example 6 Solid Uniformly Charged SphereA charge Q is uniformly distributed throughout the volume of an insulating sphere of radius is the electric field for r < R?3/3R4Q :Density Charge =Calculate average charge densityNow select a Gaussian sphere of radius r within this larger sphereCharge within this sphere is given by33333434 RrQrRQVQ enclencl= == /Fall 2008 Physics 231 Lecture 2-25 Example 6 Solid Uniformly Charged SphereElectric Field is everywhere perpendicular to surface, parallel to surface normalGauss Law then gives303302044 RrQERrQrEQAdEencl === rrField increases linearly within sphereOutside of sphere, electric field is given by that of a point charge of value QFall 2008 Physics 231 Lecture 2-26 Charges on ConductorsGiven a solid conductor.

8 On which is placed an excess chargethen in the staticlimitThe excess charge will reside on the surface of the conductorandEverywhere the electric field due to this excess charge will be perpendicular to the surfaceandThe electric field within the conductor will everywhere be zeroFall 2008 Physics 231 Lecture 2-27 Example 7A solid conductingsphere is concentric with a thin conductingshell, as shownThe inner sphere carries a charge Q1, and the spherical shell carries a charge Q2, such that Q2 = -3 is the charge distributed on the sphere? is the charge distributed on the spherical shell? is the electric field at r < R1? Between R1and R2?At r > R2? happens when you connect the two spheres with a wire?

9 (What are the charges?)Fall 2008 Physics 231 Lecture is the charge distributed on the sphere?Remember that the electric field insidea conductor in a static situation Gauss s Law, there can be no net charge inside the conductorThe charge, Q1,must reside on the outside surface of the sphere++++++++Fall 2008 Physics 231 Lecture is the charge distributed on the spherical shell?The electric field insidethe conducting shell is can be no net charge inside the conductorUsing Gauss Law it can be shown that the inner surface of the shell must carry a net charge of -Q1 The outer surface must carry the charge +Q1 + Q2, so that the net charge on the shell equals Q2 The charges are distributed uniformly over the inner and outer surfaces of the shell, hence2214 RQinner =2212212424 RQRQQ outer =+=andFall 2008 Physics 231 Lecture 2-30r< R1:This is inside the conducting sphere, thereforeThe electric field insidea conductor is is the Electric Field at r< R1?

10 Between R1and R2? At r> R2?0=ErBetween R1and R2 : R1< r< R2 Charge enclosed within a Gaussian sphere = Q1rrQkE 21=rr> R2 Charge enclosed within a Gaussian sphere = Q1 + Q2rrQkrrQQkrrQQkE 2 3 21211221 = =+=rFall 2008 Physics 231 Lecture happens when you connect the two spheres with a wire? (What are the charges?)After electrostatic equilibrium is reached,there is no charge on the inner sphere,and none on the inner surface of the shellThe charge Q1+ Q2resides on the outer surface0=ErAlso, for r< R2rrQkE 221 =rand for r > R2 Fall 2008 Physics 231 Lecture 2-32 Example 8-qAn uncharged spherical conductor has a weirdly shaped cavity carved out of it. Inside the cavity is a charge -q.(a) Less than q(b) Exactly q(c) More than qi) How much charge is on the cavity wall?