From last time…

1. y + + + + + + + + + + + + + + + + + + + + From last time… Continuous charge distributions Motion of charged particles No Discussion Wed. Oct 1 — Exam grading Physics 208 Lecture 9

2. The story so far… • Charges • Forces • Electric fields • Electric field lines Now something a little different Electric flux : ‘flow’ of electric field through a surface Physics 208 Lecture 9

3. Electric Flux • Electric flux E through a surface: (component of E-field  surface) X (surface area) • Proportional to # E- field lines penetrating surface Physics 208 Lecture 9

4. Why perpendicular component? • Suppose surface make angle  surface normal Component || surface Component  surface Only  component‘goes through’ surface • E = EA cos  • E =0 if E parallel A • E = EA (max) if E  A • Flux SI units are N·m2/C Physics 208 Lecture 9

5. 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: Physics 208 Lecture 9

6. q q Quick quiz Two spheres of radius R and 2R are centered on the positive charges of the same value q. The electric flux through the spheres compare as R Flux(R)=Flux(2R) Flux(R)=2*Flux(2R) Flux(R)=(1/2)*Flux(2R) Flux(R)=4*Flux(2R) Flux(R)=(1/4)*Flux(2R) 2R Physics 208 Lecture 9

7. Flux through spherical surface R • 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 Physics 208 Lecture 9

8. q 2q Quick quiz Two spheres of the same size enclose different positive charges, q and 2q. The flux through these spheres compare as Flux(A)=Flux(B) Flux(A)=2Flux(B) Flux(A)=(1/2)Flux(B) Flux(A)=4Flux(B) Flux(A)=(1/4)Flux(B) A B Physics 208 Lecture 9

9. q Gauss’ law • net electric flux through closed surface = charge enclosed /  • Works for any closed surface, not only sphere • Works for any charge distribution, not only point charge Physics 208 Lecture 9

10. What is E here? + + + + + + Why Gauss’ law? Is a relation between charge and electric field. Can be used to calculate electric field from charge distribution. Uniform volume density of charge Physics 208 Lecture 9

11. 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 Physics 208 Lecture 9

12. 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: Physics 208 Lecture 9

13. Quick Quiz Which Gaussian surface could be used to calculate E-field from infinitely long line of charge? None ofthese A. B. C. D. Physics 208 Lecture 9

14. 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: Linear charge density  C/m • Gauss’ law: Physics 208 Lecture 9

15. Quick Quiz Which Gaussian surface could be used to calculate E-field from infinite sheet of charge? Any ofthese A. B. C. D. Physics 208 Lecture 9

16. 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: Surface charge  C/m2 • Gauss’ law: Physics 208 Lecture 9

17. Conductor in Electrostatic Equilibrium In a conductor in electrostatic equilibrium there is no net motion of charge • E=0 everywhere inside the conductor Ein • 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. Etot =0 Etot = E+Ein= 0 Physics 208 Lecture 9

18. Charge distribution on conductor • What charge is induced on sphere to make zero electric field? + Physics 208 Lecture 9

19. 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 Physics 208 Lecture 9

20. this surface this surface this surface E-Field 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 at surface Physics 208 Lecture 9

21. - - + + - + - + - - + + Summary of conductors • everywhere inside a conductor • Charge in conductor is only on the surface • surface of conductor Physics 208 Lecture 9