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Click to edit Master title style,Click to edit Master text styles,Second level,Third level,Fourth level,Fifth level,Fall 2008,Physics 231,Lecture 2-,*,Gauss Law,Electric Field Lines,The number of field lines,also known as lines of force,are related to strength of the electric field,More appropriately it is the number of field lines crossing through a given surface that is related to the electric field,Flux,How much of something passes through some surface,Number of particles passing through a given surface,Two ways to define,Number per unit area(e.g.,10 particles/cm,2,),Number passing through an area of interest,Electric Flux,The electric flux is defined to be,Where E is the electric field and A is the area,Electric Flux,If surface area is not perpendicular to the electric field we have to slightly change our definition of the flux,Where,f,is the angle between the field and the unit vector that is perpendicular to the surface,Electric Flux,We can see that the relationship between the flux and the electric field and the area vector is just the dot product of two vectors,is a unit vector perpendicular to the surface,A Convention,The direction of a unit vector for an open surface is,ambiguous,For a closed surface,the unit vector is taken as being pointed outward,Electric Flux,Where flux lines enter the surface,the surface normal and the electric field lines are anti-parallel,Where the flux lines exit the surface they are parallel,Electric Flux,Is 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 products,Electric Flux,The equation we have for flux is fine for simple situations,the electric field is uniform and,the surface area is plane,What happens when either one or the other or both is not true,Electric Flux,We proceed as we did in the transition from discrete charges to a continuous distribution of charges,We break the surface area into small pieces and then calculate the flux through each piece and then sum them,In the limit of infinitesimal areas this just becomes an integral,Electric Flux of a Point Charge,What is electric flux that comes from a point charge?,We start from,Since E is radial,its dot product with the differential area vector,which is also radial,is always one,The electric field is given by,Also E is the same at every point on the surface of the sphere,The problem has spherical symmetry,we therefore use a sphere as the Gaussian surface,Electric Flux of a Point Charge,The integral over the surface area of the sphere yields,For these reasons,E can be pulled out from the integral and what remains is,Pulling all this together then yields,Notice that this is,independent,of the radius of the sphere,A positive charge is contained inside a spherical shell.How does the differential electric flux,d,E,through the surface element,dS,change when the charge is moved from position 1 to position 2?,d,E,a)increasesb),decreasesc),doesnt change,dS,1,dS,2,Example 1,dS,1,dS,2,Example 1-continued,The total flux of a charge is constant,with the density of flux lines being higher the closer,y,ou are to the charge,Therefore as you move the charge closer to the surface element,the density of flux lines,increases,Multiplying this higher density by the same value for the area of,dS,gives us that the incremental flux also increases,d,E,a)increasesb),decreasesc),doesnt change,dS,1,dS,2,A positive charge is contained inside a spherical shell.How does the total flux,E,through the entire surface change when the charge is moved from position 1 to position 2?,a),E,increasesb),E,decreasesc),E,doesnt change,Example 2,As we previously calculated,the total flux from a point charge depends only upon the charge,Gauss Law,The result for a single charge can be extended to systems consisting of more than one charge,One repeats the calculation for each of the charges enclosed by the surface and then sum the individual fluxes,Gauss Law relates the flux through a closed surface to charge within that surface,Gauss Law,Gauss Law states that,The net flux through any closed surface equals the net(total)charge,inside,that surface divided by,e,0,Note that the integral is over a,closed,surface,A,B,A blue sphere A is contained within a red spherical shell B.There is a charge,Q,A,on the blue sphere and charge,Q,B,on the red spherical shell.,The electric field in the region between the spheres is completely independent of,Q,B,the charge on the red spherical shell.,True,False,Example 3,Surfaces,Choose surface appropriate to problem,It does,not,have to be a sphere,Exploit,symmetries,if any,Example 4Thin Infinite Sheet of Charge,A given sheet has a charge density given by,s,C/m,2,By symmetry,E is perpendicular to the sheet,Use a surface that exploits this fact,A cylinder,A Gaussian pillbox,E,Thin Infinite Sheet of Charge,But E and A,curved,are perpendicular to each other so their dot product is zero and the middle term on the left disappears,E
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