2. R R Rr VQ 4   r Q 4   R

3. Announcements No lecture September 26 th. Everything else goes on as usual –3rd workshops September 26-30 th –Preflight #7 due October 1st, 7am Review Session: October 5 th 3-5pm –you chose it over office hours! Room TBA –This time is purposely long to allow students to attend and leave to accommodate workshops –I will try to find an alternate time and home for the 3-5pm workshop that Friday (Brian’s group, B&L 269) Homework #2 due September 26 th at 9am 1 st exam is Tuesday, October 9 th, 8-9:30am Hubbell Aud. –Covers material of Lecture 1-6, Workshop 1-3, Homework 1-3 –Question: “Practice exams”? I could post an old exam I’ve given, BUT the problem is I don’t have one that overlaps exactly this material. Let me see what I can do… –Question: “can we bring in a ‘cheat sheet’ with formulas?” That’s a reasonable request. One (1) 8.5”x11” sheet of paper with whatever you like written on both sides. Don’t abuse it, please.

4. Your Questions and Comments… “Lolrus http://i19.tinypic.com/35dbvxj.jpg” –No, this is not what Prof. McFarland goes to Japan to study –Neutrinos are much smaller in size, less massive, perhaps less cute –Shouldn’t that be “I am the lolrus? Goo goo ga-joob” anyway?

5. Your Questions and Comments… “At the end of a lecture, could you have a slide with all the equations we learned? It would be a helpful reference.” –Mmm, is learning in really done terms of equations –There is a relatively accurate concepts summary at the end Prof. McFarland wlil try to make sure he doesn’t miss the “math equivalent” when relevant “what is a partial derivative??? ” –Hopefuly the “ ” means you have encountered this in 143/162? –Short answer: a derivative with respect to one variable while holding other variables constant –We will make occasional use of this in a few places in the class If it’s causing problems for many of you, please let Prof. McF know You have some examples of calculating electric field from potential in this lecture, workshop, homework. Practice these. Ask a TA, Prof. “[on preflight] [Detailed description of your frustrating concern]” –Sometimes it’s hard to help without a one on one conversation –Please try e-mail (all you have this week  ) or face-to-face –Prof. McFarland says, “I want to try to help, and it’s OK to talk to me about concerns over occurrences in workshop, lecture, etc.”

6. Reading Assignment: Chapter 23.5-8 examples: 23.10,12 The electrostatic force is the essence of right-thinking conservatism –conservation of energy makes electrical potential a useful concept If we know the electric field E, Last time… Potential due to point charges Physically,  V is what counts The place where V =0 is “arbitrary” (at infinity)

7. Lecture 5, Concept 2 A single charge ( Q = -1  C) is fixed at the origin. Define point A at x = + 5m and point B at x = +2m. –What is the sign of the potential difference between A and B? ( V AB  V B - V A ) x -1  C   A B (1) V AB <  (2) V AB =  (3) V AB >  Maybe it’s a hard problem! Or, maybe one or two of you may have been tempted to not answer carefully on your way out the door… What did you say on Wednesday?

8. Lecture 5, Concept 2 A single charge ( Q = -1  C) is fixed at the origin. Define point A at x = + 5m and point B at x = +2m. –What is the sign of the potential difference between A and B? ( V AB  V B - V A ) x -1  C   A B (1) V AB <  (2) V AB =  (3) V AB >  Since, V AB <0 !! The simplest way to get the sign of the potential difference is to imagine placing a positive charge at point A and determining whether positive or negative work would be done in moving the charge to point B. A positive charge at A would be attracted to the -1  C charge; therefore NEGATIVE work would be done to move the charge from A to B. You can also determine the sign directly from the definition:

9. Equipotentials and conductors E from V Calculate electric field of dipole from potential Electric Potential Energy –of charge in external electric field –later: stored in the electric field itself Appendices: example calculations of a spherical charge configuration Today’s Lecture… Text Reference: Chapter 23.5,8 examples: 23.10,12

10. Equipotentials of a charged sphere The electric field of the charged sphere has spherical symmetry. The potential depends only on the distance from the center of the sphere, as is expected from spherical symmetry. Therefore, the potential is constant along a sphere which is concentric with the point charge. These surfaces are called equipotentials. Notice that the electric field is perpendicular to the equipotential surface at all points. ErEr Equipotential

11. Equipotentials GENERAL PROPERTY : –The electric field is always perpendicular to an equipotential surface. Defined as: The locus of points with the same potential. Example: as we saw, for a spherically symmetric charge distribution, the equipotentials are spheres centered on the charge. Along an equipotential surface, there is no change in V. If is zero, then this condition is satisfied. Why?? In words, if the dot product of the field vector and the displacement vector is zero, then these two vectors are perpendicular and there is no potential difference along this path. If the electric field is perpendicular to a surface everywhere, the surface is an equipotential.

12. Conductors Claim The surface of a conductor is always an equipotential surface (in fact, the entire conductor is an equipotential) Why?? If surface were not equipotential, there would be an electric field component parallel to the surface and the charges would move!! + + + + + + + + + + + + + +

13. Charge on Conductors? How is charge distributed on the surface of a conductor? –KEY: Must produce E =0 inside the conductor and E perependicular (normal) to the surface. Spherical example (with little off-center charge): - - - - - - - - - - - - - - - + + + + + + + + + + + + + + + + +q+q E =0 inside conducting shell. charge density induced on outer surface uniform E outside has spherical symmetry centered on spherical conducting shell. charge density induced on inner surface non-uniform.

14. Two spherical conductors are separated by a large distance. They each carry the same positive charge Q. Conductor A has a larger radius than conductor B. Compare the potential at the surface of conductor A with the potential at the surface of conductor B. A B a) V A > V B b) V A = V B c) V A < V B PF06 Question #2:

15. A B The two conductors are now connected by a wire. How do the potentials at the conductor surfaces compare now ? a) V A > V B b) V A = V B c) V A < V B What happens to the charge on conductor A after it is connected to conductor B ? a) Q A increases b) Q A decreases c) Q A doesn’t change PF06: Question #3-4

16. What you said … B Right: V=kQ/r, so the sphere with the small radius will have the greater potential Because potential is proportional to charge, the potential of A must increase so that it is equal to the potential of B. Since the potential increases, the charge must also increase. A B The greater surface area of A allows the charge to be distributed more sparesly across the sphere's surface area. Therefore, the potential at the surface of A would be less. [True because it would take less work to add a positive test charge from infinity to the larger sphere] VB>VA since took more work to get closer to center when starting from infinity. Now that they are connected the charge would disperse again until electrostatic equilibrium and the potential would be equal + + + + + + + + + + -∫ E dl

17. when they are connected by a wire, the charges spread out and distribute themselves evenly [conductors need only be at same potential, not same charge] throughout the two spheres and the wire. the potential is the same all over the place now. because they are connected, some charge from A goes into the wire, so it decreases [good point! but remember the wire is a very small thing, so at the sphere’s potential, it would hold very little charge. the bigger effect is the charge flowing from B to A, so the net effect is an increase for A] Adding a wire between the charged spheres would only decrease the overall charge on both spheres because they would evenly distribute their charge onto the wire. The larger end of this system, A, would still have greater potential than the smaller side, B [charge need not be even] When the spheres are connected, the potential equalizes. However, the charge doesnt change. The charge continues to be equal [but charge must be changing to equalize potential] The charge is equal. However, the potential on the first is larger because the radius is larger. Wrong: dl is larger in sphere A than sphere B. [yes -E*dl>0, but path comes from infinity] Because V is proportional to Q/r, with Q the same for both, and r, the distance away from the surface the same for both, Q/r is the same, thus V is the same [in “kQ/r” (potential of point charge) r is the distance from charge. here potential at surface is related to radius of sphere.]

18. A point charge Q is fixed at the center of an uncharged conducting spherical shell of inner radius a and outer radius b. –What is the value of the potential V a at the inner surface of the spherical shell? (3) (2) (1) a b Q Lecture 6, Concept 1 Hint: remember how potential is defined!

19. A point charge Q is fixed at the center of an uncharged conducting spherical shell of inner radius a and outer radius b. –What is the value of the potential V a at the inner surface of the spherical shell? a b Q (3) (2) (1) E out How to start?? The only thing we know about the potential is its definition: To calculate V a, we need to know the electric field E Outside the spherical shell: Apply Gauss’ Law to sphere: Inside the spherical shell: E = 0 Lecture 6, Concept 1

20. Charge on Conductor Example 2 spheres at same potential (wired). How are charge, potential, surface field related? Which one has biggest charge? Density: r S r L  Smaller sphere has the larger surface charge density and local field! Potentials equal, so… Is surface field proportional to charge or charge density?

21. Electric Potential Energy: Coming Full Circle The Coulomb force is a CONSERVATIVE force (i.e. the work done by it on a particle which moves around a closed path returning to its initial position is ZERO.) The total energy (kinetic + electric potential) is then conserved for a charged particle moving under the influence of the Coulomb force. Therefore, a particle moving under the influence of the Coulomb force is said to have an electric potential energy defined by: this “ q ” is the ‘test charge” in other examples...

22. E A B C If a negative charge is moved from point A to point C, its electric potential energy a) increases b) decreases c) doesn’t change PF06: Question #6

23. What you said … E A B C Right: Work must be done to get the charge to C from A because it is opposing the electric field. (A positive charge at A would flow freely towards C) a neg charge has its greatest potential energy at pt c which is closer to a negative plate [it’s OK to think of the charge distribution causing the field, as long as you know what it is] Potential is lower at C from the last preflight. U=q x potential so U is bigger at C Wrong: Since E points right, electric potential energy decreases towards the right [The electric potential decreases as you move to the right. But the potential energy of a negative charge increases.] The electric field is the same at both spots, so there is no change according to the formula for potential. [Integrate electric field to get a change in potential, constant E field <> constant V] My answer… and which is negative since…

24. Potential Confusion: Potential energy, potential and potential energy Potential energy stored in a static charge distribution –work we do to assemble the charges. More on this later… Electrostatic potential at any point in space –sources give rise to V ( r ) Electric potential energy of a charge in the presence of a set of source charges –potential energy of the test charge equals the potential from the sources times the test charge

25. Energy Units MKS: U = QV 1 Coulomb-Volt = 1 Joule elementary particles (e.g. e, p) good unit is “electron-Volt” q e =1.6x10 -19 Coulomb, so 1 e V = 1.6x10 -19 J Accelerators Electrostatic: Van de Graaff protons  1 Me V ( 10 6 e V ) Electromagnetic: Fermilab protons  1Te V ( 10 12 e V ) 1 mile

26. V V +d V Expressed as a vector, E is the negative gradient of V Cartesian coordinates: Spherical coordinates: E from V ? We can obtain the electric field E from the potential V by inverting our previous relation between E and V :

27. This graph shows the electric potential at various points along the x-axis. At which point(s) is the electric field zero? A B C D PF06: Question #8

28. What you said … Right: the electric field is zero at point B because the electric potential is not changing delta V = -Int(E*dl), implies E is the derivative of V E is the [negative] derivative of V, so when the slope of V(x) is zero, E is zero Wrong: if there is negative potential there is a field of 0. [Potential is not zero to left and right of B] Since the potential is the integral of the E field, the point where the net area is zero means the field is zero [potential is graphed here and question asks about field, not the reverse.] THe potntial is 0 when E is 0, and visa versa the book, it is given that E=V/d. Since d is never zero, when V is zero, E will also be zero [but does this formula apply here? this is a special case of a uniform field and the potential difference across a distance d along the field lines.] my answer: the gradient is a derivative w/r to position. -dV/dx is 0 only at B.

29. Consider the following electric potential: What electric field does this describe?... expressing this as a vector: E from V : an Example Cartesian coordinates:

30. allows us to calculate the potential function V everywhere (keep in mind, we often define V A = 0 at some convenient place)  If we know the electric field E everywhere, allows us to calculate the electric field E everywhere If we know the potential function V everywhere, Units for Potential! 1 Joule/Coul = 1 VOLT The Bottom Line

31. The electric potential in a region of space is given by The x -component of the electric field E x at x = 2 is (1)E x =0 (2)E x >0(3)E x <0 Lecture 6, Concept 2

32. We know V (x) “everywhere”. To obtain E x, use The electric potential in a region of space is given by The x -component of the electric field E x at x = 2 is (1)E x =0 (2)E x >0(3)E x <0 Lecture 6, Concept 2

33. Electric Dipole Now we can use this potential to calculate the E field of a dipole ! (remember how messy the direct calculation was?) Rewrite this for special case r >> a :  The potential is much easier to calculate than the field since it is an algebraic sum of 2 scalar terms. z a a  +q -q r r 1 r 2 r2-r1r2-r1 the dipole moment

34. Electric Dipole Calculate E in spherical coordinates:  the dipole moment z a a  +q+q -q-q r r 1 r 2

35. Dipole Field z a a  +q+q -q-q r

36. Summary Conductors are equipotentials Potential Energy Can get from electric field to potential and potential to field from: Reading Assignment: 23.9, 24.1-4 Examples: 24.1,3,7,8

37. Appendix a b c uncharged conductor solid sphere with total charge Q r I II III IV Calculate the potential V ( r ) at the point shown ( r < a )

38. Calculating Electric Potentials Calculate the potential V ( r ) at the point shown ( r < a ) Where do we know the potential, and where do we need to know it? Determine E ( r ) for all regions in between these two points Determine  V for each region by integration Check the sign of each potential difference  V... and so on... (from the point of view of a positive charge) V =0 at r = ... we need r < a... a b c uncharged conductor solid sphere with total charge Q r I II III IV  V > 0 means we went “uphill”  V < 0 means we went “downhill”

39. Calculating Electric Potentials a b c r I II III IV Calculate the potential V ( r ) at the point shown ( r < a ) Look at first term: Line integral from infinity to “c” has to be positive, pushing against a force: What’s left? Line integral is going “in” which is just the opposite of what usually is done - controlled by limits

40. Calculating Electric Potentials a b c r I II III IV Calculate the potential V(r) at the point shown ( r < a ) Look at third term : Line integral from “ b ” to “ a ”, again has to be positive, pushing against a force: Line integral is going “in” which is just the opposite of what usually is done - controlled by limits What’s left? Previous slide we have calculated this already

41. Calculating Electric Potentials Calculate the potential V ( r ) at the point shown ( r < a ) a b c r I II III IV Look at last term: Line integral from “ a ” to “ r ”, again has to be positive, pushing against a force. But this time the force doesn’t vary the same way, since “ r ’ ” determines the amount of source charge What’s left to do? ADD THEM ALL UP! Sum the potentials This is the charge that is inside “ r ” and sources field

42. Calculating Electric Potentials Calculate the potential V(r) at the point shown ( r < a ) a b c r I II III IV Add up the terms from I, III and IV: I III IV The potential difference from infinity to “ a ” if the conducting shell weren’t there An adjustment to account for the fact that the conductor is an equipotential,  V = 0 from c -> b Potential increase from moving into the sphere

43. Calculating Electric Potentials Summary a b c r I II III IV The potential as a function of r for all 4 regions is: I r > c : II b < r < c: III a < r < b: IV r < a: