Physics 212 Lecture 4, Slide 1Physics 212Lecture 4Today's Concepts:Conductors +Using Gauss’ LawPhysics 212 Lecture 4MusicWho is the Artist?A) Marc RibotB) Bill FrisellC) Richard ThompsonD) Pat MethenyE) John ScofieldWhy??I think he is my favorite jazz guitarist !!He also is coming to Ellnora (as are two of the others) !!Physics 212 Lecture 4, Slide 3Your Comments“ABSOLUTELY POSITIVELY 100% CONFUSED!! I don't understand the conductors and how that affects the E field and i REALLY do not understand the solid cylindrical conductor part AT ALL! Help!!“04“I don't understand Gaussian surfaces”Most students are having difficulties with this topic: The Checkpoints show this clearly.This whole way of thinking (Gauss’ Law) is very unfamiliar to you: calculate a field means, first, pick a surface??The solution? DON’T PANIC… We’re confident you will master these concepts but it will take a little work.“Wow.....just wow”Our ResponseTODAY’S PLAN:• Do Checkpoints again! Try to understand the reasoning• Do a calculation using Gauss’ Law“If my charge was related to my understanding of this material, I would be negative.”“Gauss' Law is very confusing. I do not fully understand how to find the electic field using this law. I follow the examples but the problems don't seem to work for me.”Physics 212 Lecture 4, Slide 4Conductors = charges free to moveClaim: E = 0 inside any conductor at equilibriumCharges in conductor move to make E field zero inside. (Induced charge distribution). If E <>0, then charge feels force and moves!Claim: Excess charge on conductor only on surface at equilibriumWhy?• Apply Gauss’ Law• Take Gaussian surface to be just inside conductor surface E = 0SIMULATION 206• E = 0 everywhere inside conductor∫=⋅oencQAdEε• Gauss’ Law:0encQ=∫=⋅oencQAdEεPhysics 212 Lecture 4, Slide 5Gauss’ Law + Conductors + Induced ChargesALWAYS TRUE!If choose a Gaussian surface that is entirely in metal, then E=0 so Qenclosedmust also be zero!0ε=encQEA09How Does This Work??Charges in conductor move to surfaces to make Qenclosed= 0. We say charge is induced on the surfaces of conductors∫=⋅oencQAdEεPhysics 212 Lecture 4, Slide 6Charge in Cavity of ConductorA particle with charge +Q is placed in the center of an uncharged conducting hollow sphere. How much charge will be induced on the inner and outer surfaces of the sphere?A) inner = –Q, outer = +Q B) inner = –Q/2 , outer = +Q/2 C) inner = 0, outer = 0D) inner = +Q/2, outer = -Q/2E) inner = +Q, outer = -Q 10QQinnerQouter• Gauss’ Law:0encQ=Since E=0 in conductor∫=⋅oencQAdEεPhysics 212 Lecture 4, Slide 7Infinite CylindersA long thin wire has a uniform positive charge density of 2.5 C/m. Concentric with the wire is a long thick conducting cylinder, with inner radius 3 cm, and outer radius 5 cm. The conducting cylinder has a net linear charge density of -4 C/m.What is the linear charge density of the induced charge on the inner surface of the conducting cylinder (λi) and on the outer surface (λo)?λi: +2.5 C/m -4 C/m -2.5 C/m -2.5 C/m 0λo: -6.5 C/m 0 +2.5 C/m -1.5 C/m -4 C/mA) B) C) D) E)14λιλοE = 0 in material of conducting shellEnclosed charge = 0λi= -2.5 C/mλi + λo = -4 C/mλo= -1.5 C/mPhysics 212 Lecture 4, Slide 8Gauss’ Law ALWAYS TRUE!In cases with symmetry can pull E outside and get0ε=encQEA19In General, integral to calculate flux is difficult…. and not useful!To use Gauss’ Law to calculate E, need to choose surface carefully!1) Want E to be constant and equal to value at location of interestOR2) Want E dot A = 0 so doesn’t add to integral∫=⋅oencQAdEεPhysics 212 Lecture 4, Slide 9Gauss’ Law SymmetriesALWAYS TRUE!In cases with symmetry can pull E outside and get0ε=encQEASphericalCylindricalPlanar24π=A r204π ε=encQEr2π=A rL02λπ ε=Er22π=A r02σε=E21∫=⋅oencQAdEεPhysics 212 Lecture 4, Slide 10“We can treat the cube as a point charge and the all points on the sphere's surface are equidistant to this point charge.”“A cube would be best suited because the field lines produced by the charge would be perpendicular to all sides of a cubic Gaussian surface.”“Since the field lines are not always perpendicular and symmetric to the surface, we cannot calculate the electric field using Gauss' Law here.“010203040506070Checkpoint 1 23(D) The field cannot be calculated using Gauss’ Law(E) None of the aboveTHE CUBE HAS NO GLOBAL SYMMETRY !THE FIELD AT THE FACE OF THE CUBE IS NOT PERPENDICULAR OR PARALLEL3D POINT Ø SPHERICAL2D LINE Ø CYLINDRICAL1D PLANE Ø PLANARPhysics 212 Lecture 4, Slide 11Checkpoint 3.1260102030405060out in zero“the electric field lines point outward from the positive charge.““the positive charges are moving to the negative charges which is inward““E inside conductor is zero “Careful: what does inside mean? This is always true for a solid conductor (within the material of the conductor)Here we have a charge “inside”Physics 212 Lecture 4, Slide 12Checkpoint 3.3290102030405060out in zero“the electric field always points outward““Since the outer shell is a negative charge, the field points inward.”“If we create a gaussian surface of a sphere, we can see that the enclosed charge, q+ (-q) = 0, and therefore the field must also be zero.. “What is direction of field OUTSIDE the red sphere?Physics 212 Lecture 4, Slide 1301020304050Checkpoint 231“the formula is E = (p/3E)*r”“For all r<a the spherical shell one would use to apply gauss's law would encapsulate no charge, resulting in no electric field.““I figured that you guys wouldn't give us all of the information about a and b without there being a reason for it.“Physics 212 Lecture 4, Slide 14Checkpoint 42701020304050In which case is E at point P the biggest?A) A B) B C) the same“Case A it is bigger because in case B the charges cancel and it is zero““The two extra charged planes in Case B increased the magnitude of electric field at P.““They are the same because the distance from the sheet does not matter. The (+) and (-) in caseB cancel each other out, but there is another (+) plane whose distance from the point does not matter.. “Physics 212 Lecture 4, Slide 15+-+Case ACase B+Superposition:34NETPhysics 212 Lecture