Sep. 27, 2007Physics 208 Lecture 81 From last time… This Friday’s honors lecture: Biological Electric Fields Dr. J. Meisel, Dept. of Zoology, UW Continuous charge distributions y Motion of charged particles

Sep. 27, 2007Physics 208 Lecture 82 Exam 1 Wed. Oct. 3, 5:30-7 pm, 2103 Ch (here) Covers Chap. 21.6, lecture, lab, discussion, HW Sample exams on website. Solutions on website. HW4 (assigned today) covers exam material.

Sep. 27, 2007Physics 208 Lecture 83 Electric Flux Electric flux  E through a surface: (component of E-field  surface) X (surface area) Proportional to # E- field lines penetrating surface

Sep. 27, 2007Physics 208 Lecture 84 Why perpendicular component? Suppose surface make angle  surface normal  E = EA cos   E =0 if E parallel A  E = EA (max) if E  A Flux SI units are N·m 2 /C Component || surface Component  surface Only  component ‘goes through’ surface

Sep. 27, 2007Physics 208 Lecture 85 Total flux E not constant add up small areas where it is constant Surface not flat add up small areas where it is ~ flat Add them all up:

Sep. 27, 2007Physics 208 Lecture 86 q Quick quiz Two spheres of radius R and 2R are centered on the positive charges of the same value q. The flux through these spheres compare as A)Flux(R)=Flux(2R) B)Flux(R)=2Flux(2R) C)Flux(R)=(1/2)Flux(2R) D)Flux(R)=4Flux(2R) E)Flux(R)=(1/4)Flux(2R) q R 2R

Sep. 27, 2007Physics 208 Lecture 87 Flux through spherical surface Closed spherical surface, positive point charge at center E-field  surface, directed outward E-field magnitude on sphere: Net flux through surface: E constant, everywhere  surface R

Sep. 27, 2007Physics 208 Lecture 88 q Quick quiz Two spheres of the same size enclose different positive charges, q and 2q. The flux through these spheres compare as A)Flux(A)=Flux(B) B)Flux(A)=2Flux(B) C)Flux(A)=(1/2)Flux(B) D)Flux(A)=4Flux(B) E)Flux(A)=(1/4)Flux(B) 2q A B

Sep. 27, 2007Physics 208 Lecture 89 Gauss’ law net electric flux through closed surface = charge enclosed /  

Sep. 27, 2007Physics 208 Lecture 810 Using Gauss’ law Use to find E-field from known charge distribution. Step 1: Determine direction of E-field Step 2: Make up an imaginary closed ‘Gaussian’ surface Make sure E-field at all points on surface either perpendicular to surface or parallel to surface Make sure E-field has same value whenever perpendicular to surface Step 3: find E-field on Gaussian surface from charge enclosed. Use Gauss’ law: electric flux thru surface = charge enclosed /  o

Sep. 27, 2007Physics 208 Lecture 811 Field outside uniformly-charged sphere Field direction: radially out from charge Gaussian surface: Sphere of radius r Surface area where Value of on this area: Flux thru Gaussian surface: Charge enclosed: Gauss’ law:

Sep. 27, 2007Physics 208 Lecture 812 Quick Quiz Which Gaussian surface could be used to calculate E-field from infinitely long line of charge? None of these A.B.C.D.

Sep. 27, 2007Physics 208 Lecture 813 E-field from line of charge Field direction: radially out from line charge Gaussian surface: Cylinder of radius r Area where Value of on this area: Flux thru Gaussian surface: Charge enclosed: Gauss’ law: Linear charge density C/m

Sep. 27, 2007Physics 208 Lecture 814 Quick Quiz Which Gaussian surface could be used to calculate E-field from infinite sheet of charge? Any of these A.B.C.D.

Sep. 27, 2007Physics 208 Lecture 815 Field from infinite plane of charge Field direction: perpendicular to plane Gaussian surface: Cylinder of radius r Area where Value of on this area: Flux thru Gaussian surface: Charge enclosed: Gauss’ law: Surface charge  C/m 2

Sep. 27, 2007Physics 208 Lecture 816 Properties of Conductor in Electrostatic Equilibrium In a conductor in electrostatic equilibrium there is no net motion of charge Property 1: E=0 everywhere inside the conductor Conductor slab in an external field E: if E  0 free electrons would be accelerated These electrons would not be in equilibrium When the external field is applied, the electrons redistribute until they generate a field in the conductor that exactly cancels the applied field. E tot = E+E in = 0 E in E tot =0

Sep. 27, 2007Physics 208 Lecture 817 Induced dipole What charge is induced on sphere to make zero electric field? +

Sep. 27, 2007Physics 208 Lecture 818 Conductors: charge on surface only Choose a gaussian surface inside (as close to the surface as desired) There is no net flux through the gaussian surface (since E=0) Any net charge must reside on the surface (cannot be inside!) E=0

Sep. 27, 2007Physics 208 Lecture 819 Field’s Magnitude and Direction E-field always  surface: Parallel component of E would put force on charges Charges would accelerate This is not equilibrium Apply Gauss’s law Field lines