1. 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

2. Sep. 27, 2007Physics 208 Lecture 82 Exam 1 Wed. Oct. 3, 5:30-7 pm, 2103 Ch (here) Covers Chap. 21.6, 22-27 + lecture, lab, discussion, HW Sample exams on website. Solutions on website. HW4 (assigned today) covers exam material.

3. 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

4. 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

5. 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:

6. 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

7. 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

8. 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

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

10. 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

11. 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:

12. 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.

13. 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

14. 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.

15. 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

16. 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

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

18. 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

19. 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