1. From last time… Motion of charged particlesy +
Continuous charge distributions
Motion of charged particles
Arrange exam 1 time.
Lab this week is electrostatics.
No Discussion Wed. Oct 1 — Exam grading
Sep. 30, 2008
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
Sep. 30, 2008
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
Sep. 30, 2008
Physics 208 Lecture 9

4. Why perpendicular component?
Suppose surface make angle  surface normal
Component || surface
Component  surface
Only  component ‘goes through’ surface
Other component is parallel to the surface
E = EA cos 
E =0 if E parallel A
E = EA (max) if E  A
Flux SI units are N·m2/C
Sep. 30, 2008
Physics 208 Lecture 9

5. Total flux E not constant Surface not flat
add up small areas where it is constant
Surface not flat
add up small areas where it is ~ flat
Add them all up:
Sep. 30, 2008
Physics 208 Lecture 9

6. 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) q 2R q
Sep. 30, 2008
Physics 208 Lecture 9

7. 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. 30, 2008
Physics 208 Lecture 9

8. 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 q 2q
Sep. 30, 2008
Physics 208 Lecture 9

9. Gauss’ law Works for any closed surface, not only sphere
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 q
Sep. 30, 2008
Physics 208 Lecture 9

10. Why Gauss’ law? Is a relation between charge and electric field.
Can be used to calculate electric field from charge distribution.
What is E here? +
Ask first what is its direction. Then ask how can use Gauss’ law - there is no surface here.
Uniform volume density of charge
Sep. 30, 2008
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
Need to add some slides after this giving examples of gaussian surfaces that work for particular field distributions (e.g. point charge, plane of charge, line of charge) before doing any calculations of electric flux. Could do it in form of questions. (sort of already do this later on).
Sep. 30, 2008
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:
Go through this one myself - students help with next ones.
Gauss’ law:
Sep. 30, 2008
Physics 208 Lecture 9

13. 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. 30, 2008
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:
Ask students for area, flux, and charge enclosed.
Linear charge density  C/m
Gauss’ law:
Sep. 30, 2008
Physics 208 Lecture 9

15. Quick Quiz
Which Gaussian surface could be used to calculate E-field from infinite sheet of charge?
Any of these
Say that reason why will be evident when doing problem. A. B. C. D.
Sep. 30, 2008
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:
Ask students for area, flux through surface, and charge enclosed.
Surface charge  C/m2
Gauss’ law:
Sep. 30, 2008
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
Sep. 30, 2008
Physics 208 Lecture 9

18. Charge distribution on conductor
What charge is induced on sphere to make zero electric field? +
Show applet
Sep. 30, 2008
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
Sep. 30, 2008
Physics 208 Lecture 9

20. 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
this surface
this surface
this surface
Sep. 30, 2008
Physics 208 Lecture 9

21. Summary of conductors everywhere inside a conductor
Charge in conductor is only on the surface
surface of conductor - +
Sep. 30, 2008
Physics 208 Lecture 9