Week 2 Day 2 Review Electric Field Lines Direction of eld at any point is tangent to eld line at that point Field lines point away from positive charges and terminate on negative charges Field lines never cross each other Dipole in Uniform Field E E p 2qa cos sin j Total Net Force F F F qE q E 0 Torque on dipole r x F p x E rF sin 2a qE sin pEsin p tends to align with the electric eld Continuous Sources Charge Density Volume V R L Q V Q V uniform Area A wL Q A Q A uniform Length L Q L Q L uniform Continuous Charge Distributions Break distributions into parts Q q dq E eld at P due to q E Superposition E E dE Concept Question Flux through sphere Correct answer 3 The total ux through the below spherical surface is zero because no charge is inside the sphere Gauss s Law The First Maxwell Equation A very useful computational technique to nd the electric eld when the source has enough symmetry Sources with Enough Symmetry Spherical Symmetry Cylindrical Symmetry Planar Symmetry Gauss s Law The idea Gauss s Law The Equation Electric Flux Open surface Concept Question Sign of Flux Open and Closed Surfaces Sign of Flux Closed Surface The total ux of electric eld lines penetrating any of these closed surfaces is the same and depends only on the amount of charge inside Electric ux the surface integral of E over closed surface S is proportional to charge enclosed by the volume enclosed by S Case One E is a uniform vector eld perpendicular to planar surface S of area A Our Goal Always reduce problem to nding a surface where we can reduce the integral to either E times A or zero Case Two Electric eld is uniform vector eld directed at angle to planar surface S of area A Correct answer 2 The electric ux through the planar surface below positive unit normal to the left is negative A rectangle is an open surface it does not contain a volume A sphere is a closed surface it does contain a volume For closed surface A is normal to the surface and point outward For electric eld that has constant magnitude on surface Case One if electric eld points out then EA Case Two if electric eld points out then EA Case Three E not uniform surface curved For closed surface A is normal to surface and points outward 0 if E points out 0 if E points in E A Group Problem Electric Flux Sphere Consider a closed surface of two parts Gauss s Law Generalization Qenclosed V Concept Question Gauss s Law Choosing a Gaussian Surface Consider a point like charged object Q located at the origin What is the electric ux on a spherical surface of radius r E A 0 closed surface If you take an arbitrary closed surface and have it surround a point charge the ux through the surface is Q The ux of E through any surface is simply equal to the charge enclosed by that surface you can apply the superposition principle divided by Electric ux through any arbitrary closed surface is proportional to the charge enclosed inside that closed surface Correct answer 3 The grass seeds gure shows the electric eld of three charges with charges 1 1 and 1 The Gaussian surface in the gure is a sphere containing two of the charges The electric ux through the spherical Gaussian surface is zero Gauss s Law holds for all closed surfaces However it is only useful to calculate electric eld for sources with enough of symmetry in which there exists closed surfaces such that The electric eld on some faces of the closed surface is both perpendicular to the face and has constant magnitude on that face then E A EA The electric eld on some faces of the closed surface is parallel to that face Then E A 0 Applying Gauss s Law Preliminary Steps Step 1 identify the symmetry properties of the charge distribution Step 2 determine the direction of the electric eld Step 3 Decide how many different regions of space the charge distribution determines Step 4 For each region of space determined by the charge distribution separately determine both sides of the Gauss s Law Step 5 Equate the two sides of Gauss s Law in order to determine an expression for the magnitude of the electric eld Step 6 Electric ux through a closed surface is positive if the electric eld points in the outward normal direction Based on your result from Gauss s Law If the sign of the electric eld is positive then the electric eld points in the outward normal direction to the surface If the sign of the electric eld is negative then the electric eld points opposite the outward normal direction to the surface Applying Gauss s Law Based on the source identify regions in which to calculate electric eld Choose Gaussian surface S Symmetry Calculate E A Calculate Qenclosed charge enclosed by surface S Apply Gauss s Law to calculate electric eld E A Qenclosed Group Problem Gauss s Law Spherical Symmetry Q uniformly distributed throughout non conducting solid sphere of radius a Determine the direction and magnitude of the electric eld inside the sphere r a Concept Question Spherical Shell Correct answer 1 We just saw that in a solid sphere of charge the electric eld grows linearly with distance Inside the charged spherical shell at right r a what does the electric eld do Zero Worked Example Planar Symmetry Consider an in nite thin slab with uniform positive charge density Find a vector expression for the direction and magnitude of the electric eld outside the slab Make sure you know your Gaussian closed surface Symmetry is planar E E Note A is arbitrary its size and shape and should divide out Total charge enclosed Qenclosed V wA A No ux through sides of cylinder only ux through endcaps Group Problem Cylindrical Symmetry An in nitely long rod has a uniform positive linear charge density Find the direction and magnitude of the electric eld outside the rod STOPPED HERE IN CLASS Concept Question Superposition Correct answer Three in nite sheets of charge are shown above The sheet in the middle is negatively charged with charge per unit area 2 and the other two sheets are positively charged with charge per unit area Which set of arrows and zeros best describes the electric eld Electric Field for Charged In nite Plane Dipole E falls off like 1 r Spherical charge E falls off like 1 r Line of charge in nite E falls off like 1 r Plane of charge in nite E uniform on either side of plane
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