GAUSS’ LAW

Class Activities: Gauss’ Law Discussion Gauss vs Coulomb Discussion re "which is more fundamental, Gauss or Coulomb" (and, why) Let them discuss. (Pointed out the Coulomb came first, historically. And that from one, you can show the other, in statics. But also pointed out Coulomb is *wrong*, but Gauss is always true, in non-static cases. Also pointed out Gauss is always true but not always *helpful* to solve for E in a given problem…) Whiteboard Charge distribution from E field Whiteboard to compute the charge distribution from E=c r(vector) (which is also a clicker question) and then to compute Q(enclosed) of the resulting rho. Tutorials Divergence Paul van Kampen – Dublin University (Tutorials 1-8, page 32) In document “Tutorials 1-8” Two students try to calculate the charge density, one uses Cartesian and one uses cylindrical. They calculate the divergence of the E field in both coordinate systems. They derive the divergence formula in cylindrical coordinates. Very nice tutorial. Boundary condition activity Oregon State University – Boundary E Groups investigate how components of electric field change as you cross a boundary. Gauss’ Law Oregon State University Students working in small groups practice using Gauss' Law to determine the electric field due to several charge distributions. Students practice using the symmetry arguments necessary to use Gauss' Law. Discussion Gauss vs Coulomb Discussion re "which is more fundamental, Gauss or Coulomb" (and, why) Let them discuss. (Pointed out the Coulomb came first, historically. And that from one, you can show the other, in statics. But also pointed out Coulomb is *wrong*, but Gauss is always true, in non-static cases. Also pointed out Gauss is always true but not always *helpful* to solve for E in a given problem…) Whiteboard Charge distribution from E field Whiteboard to compute the charge distribution from E=c r(vector) (which is also a clicker question) and then to compute Q(enclosed) of the resulting rho. Tutorials Divergence Paul van Kampen – Dublin University (Tutorials 1-8, page 32) In document “Tutorials 1-8” Two students try to calculate the charge density, one uses Cartesian and one uses cylindrical. They calculate the divergence of the E field in both coordinate systems. They derive the divergence formula in cylindrical coordinates. Very nice tutorial. Boundary condition activity Oregon State University – Boundary E Groups investigate how components of electric field change as you cross a boundary. Gauss’ Law Oregon State University Students working in small groups practice using Gauss' Law to determine the electric field due to several charge distributions. Students practice using the symmetry arguments necessary to use Gauss' Law.

A) (I) only B) (I) and (II) only C) (I) and (III) only 2.19 Which of the following are vectors? (I) Electric field (II) Electric flux (III) Electric charge A) (I) only B) (I) and (II) only C) (I) and (III) only D) (II) and (III) only E) (I), (II), and (III) CORRECT ANSWER: A USED IN: Spring 2008 (Pollock) and ‘13 (Pollock) LECTURE NUMBER: 4 STUDENT RESPONSES: [[67]], 23, 7, 2, 2 (Sp ‘13) INSTRUCTOR NOTES: From singh’s conceptual test. Did this as “pre-class” warm-up to Gauss’ law lecture. Fascinatingly, 2 people voted c, and 3 voted B. Answer is A, I only. (In ‘13, even more went for “Electric flux” as also a vector) CREDIT: "Student understanding of symmetry and Gauss's law of electricity",  Chandralekha Singh Am. J. Phys. 74, 923 (2006) 

We don't know (r), so can't answer. 2.22 The space in and around a cubical box (edge length L) is filled with a constant uniform electric field, . What is the TOTAL electric flux through this closed surface? Zero EL2 2EL2 6EL2 We don't know (r), so can't answer. Answer is A. CORRECT ANSWER: USED IN: Spring 2008 (Pollock) and SP ‘13 LECTURE NUMBER:4 STUDENT RESPONSES: [[78%]] 13% 9% 0% 0% [[80%]] ,5, 14, 2, 0 INSTRUCTOR NOTES: This is from Dubson’s pre test, got 3 votes for C and 2 for D! It's really a freshman level question, but seems to be worth revisiting –SJP I had a question about the direction of dA, (they thought it might be arbitrary, so the answer COULD be 2EL^2, for instance) WRITTEN BY: M. Dubson, adapted by Steve Pollock (CU-Boulder) 4

(A) positive (B) negative (C) zero A positive point charge +q is placed outside a closed cylindrical surface as shown. The closed surface consists of the flat end caps (labeled A and B) and the curved side surface (C). What is the sign of the electric flux through surface C? (A) positive (B) negative (C) zero (D) not enough information given to decide CORRECT ANSWER: B. USED IN: Fall 2008 (Dubson), Spring 2008 and 13 (Pollock), Spring 2009 (Kinney), Fall 2009 (Schibli) LECTURE NUMBER: Dubson (Week 2, lecture 4). Pollock (Lecture 4), Kinney (Week 2, Lecture 4), Schibli (Week 2, Lecture 5) STUDENT RESPONSES: 2% [[90%]] 4% 4% 0% (FALL 2008) 16% [[68%]] 13% 3% 0% (SPRING 2009) 17% [[42%]] 42% 0% 0% (FALL 2009) 16, [[39%]], 46%, 09, 0, (Sp ‘13) INSTRUCTOR NOTES: The question is slightly subtle, I added the next slide (showing field lines) to explain it more clearly. We now have several “tools”, including Gauss’ law (flux through the CLOSED surface is zero) but must couple that with thinking about flux through individual bits. WRITTEN BY: Mike Dubson, adapted by Steve Pollock (CU-Boulder) A q C q B (Side View)

(A) positive (B) negative (C) zero A positive point charge +q is placed outside a closed cylindrical surface as shown. The closed surface consists of the flat end caps (labeled A and B) and the curved side surface (C). What is the sign of the electric flux through surface C? (A) positive (B) negative (C) zero (D) not enough information given to decide Helpful slide to explain the previous one. (Note that the red arrows generate negative flux for C, and positive for A and B, while the black ones are not contributing to the net flux through C. q A C q (Side View) B

Which of the following two fields has zero divergence? 1.4 Which of the following two fields has zero divergence? II I Both do B) Only I is zero C) Only II is zero D) Neither is zero E) ?? CORRECT ANSWER: B USED IN: Spring 2008 (Pollock) LECTURE NUMBER: 5 in SP ‘08, 4 in Sp ‘13 STUDENT RESPONSES: 4% [[83%]] 9% 4% 0% 16%, [74%]], 4, 7, 0 (Sp ‘13) INSTRUCTOR NOTES: 83% correct, some interesting discussion about why/how. (One student loved my using the math, d/dx(Fx)+d/dy(Fy) etc to “see” it). My answer is B, I has no divergence (no dEx/dx) (But again, I’m making an assumption about uniformity in z. For example, II is a good graph of the e-field if you have a long line charge offscreen to the left, and then *in the region shown,* there is no divergence!!). You could make this explicit, but I valued leaving that ambiguity in and letting it be part of the discussion – SJP In Sp ‘13 we talked more about the basic definition of div as “outward flux from a small local volume” per unit volume. With this in mind, Fig I has equal flux in and out for small boxes located anywhere in the picture, but Fig 2 has “more exiting than entering”, thus posit flux. I then asked what if the arrows all flipped sign – would that change the sign of div(E), or not? (Answer is no, still more lines exiting than entering!) Written by:?? (We got it from somewhere!)

What is the divergence of this vector field in the boxed region? 1.5 What is the divergence of this vector field in the boxed region? Zero Not zero ??? CORRECT ANSWER: A USED IN: Spring 2008 (Pollock) LECTURE NUMBER: 5 in SP ‘08, 4 in Sp ‘13 STUDENT RESPONSES:[[95%]] 5% 0% 0% 0% [[60%]], 38, 2, 0, 0 in Sp ‘13 INSTRUCTOR NOTES:95% correct, (only 60 the next time) but very good source of conversation. In ‘08 I made them DO it (using spherical coordinates, find the divergence of the Coulomb field from the front flyleaf). We also discussed “lines in = lines out” argument. I also went through the argument that “Ex/dx is negative, but dEy/dy is positive”, and those can cancel if the radial dependence is “just so”. This leads in to the delta function… Answer is A, (although again, there’s implicit 3-D assumptions being made here, I was viewing this as “E field lines from a point charge” . Fact is that the “box” is 3D and you aren’t SEEING information about the field in the 3rd dimension, you’re just assuming... But I think this was not important for the discussions at hand here) -SJP WRITTEN BY: Steven Pollock (CU-Boulder)

A Gaussian surface which is not a sphere has a single charge (q) inside it, not at the center. There are more charges outside. What can we say about total electric flux through this surface ? It is q/0 We know what it is, but it is NOT q/0 Need more info/details to figure it out. (Did not get to this one in SP 13) CORRECT ANSWER: A USED IN: Spring 2009 (Kinney), Fall 2009 (Schibli) LECTURE NUMBER: Pollock (Skipped), Kinney (Week 2, Lecture 4) STUDENT RESPONSES: [[97%]] 3% 0% 0% 0% (SPRING 2009) INSTRUCTOR NOTES: Answer is A. I think they’ve got this by now... (?) -SJP WRITTEN BY: Steven Pollock (CU-Boulder)

2.17 An infinite rod has uniform charge density . What is the direction of the E field at the point P shown? A) P B)  C) CORRECT ANSWER: C USED IN: Spring 2008 (Pollock), Fall 2009 (Schibli) Skipped in ‘13 LECTURE NUMBER: 6 STUDENT RESPONSES: 9% 0% [[91%]] 0% 0% (Sp ‘08) INSTRUCTOR NOTES: Used in lecture 6, as lead-ins to the role of symmetry in Gauss’ law. (91% correct, this one is not hard for them. Might need to think about how to make it a little more challenging...). -SJP WRITTEN BY: Steven Pollock (CU-Boulder) Origin 10

Deep questions to ponder Is Coulomb’s force law valid for all separation distances? (How about r=0?) What is the physics origin of the r2 dependence of Coulomb’s force law?   What is the physics origin of the 1/e0 dependence of Coulomb’s force law? What is the physics origin of the 1/4π factor in Coulomb’s force law? What really is electric charge? Why is electric charge quantized (in units of e)? What really is negative vs. positive electric charge (i.e. –e vs. +e)? Why does the Coulomb force vary as the product of charges q1q2? What really is the E-field associated with e.g. a point electric charge, e? Are electric field lines real? Do they really exist in space and time? Just fun! Did this in ‘08, no time in ‘13, too bad. (I posted it for them to look at, but nobody ever asked me about it)

An electric dipole (+q and –q, small distance d apart) sits centered in a Gaussian sphere. What can you say about the flux of E through the sphere, and |E| on the sphere? q -q Flux=0, E=0 everywhere on sphere surface Flux =0, E need not be zero everywhere on sphere Flux is not zero, E=0 everywhere on sphere Flux is not zero, E need not be zero… CORRECT ANSWER: B USED IN: Spring 2013 (Pollock) LECTURE NUMBER: 5 (Gauss’ Law) STUDENT RESPONSES: 12, [[84]], 2,2, 0 Sp ’13 (preclass question) INSTRUCTOR NOTES This was the “review’ question to start class. Most remembered Gauss law (which was on the board) to vote for A or B, but the debate was about whether the fact that the integral vanishes, does that mean E vanishes EVERYWHERE. It’s so important to understand that knowing the integral doesn’t by itself tell you about the integrand. (There was a lot of discussion, even though the end vote was solid) -SJP WRITTEN BY: Steven Pollock (CU-Boulder) NEW in 2013, combining two older questions.

Flux=0, E=0 everywhere on sphere surface An electric dipole (+q and –q, small distance d apart) sits centered in a Gaussian sphere. What can you say about the flux of E through the sphere, and |E| on the sphere? q -q Flux=0, E=0 everywhere on sphere surface Flux =0, E need not be zero on sphere Flux is not zero, E=0 everywhere on sphere Flux is not zero, E need not be zero… Just to “expllain” the previous slide, Gauss law pops up.

1.1 In spherical coordinates, what would be the correct description of the position vector “r” of the point P shown at (x,y,z) = (0, 2 m, 0) z y P r Origin x CORRECT ANSWER: A USED IN: Spring 2008 (Pollock) and 2013 LECTURE NUMBER: 5 (Gauss’ Law) STUDENT RESPONSES: [[10%]] 5% 14% 5% 67% (but specific options were different in version given in class) [[17]], 16, 14, 41, 12 (silent), then discussion, then [[25]], 4, 2, 7, 63 Sp ‘13 INSTRUCTOR NOTES : Only tiny number really voted correctly. The position vector points in the “r” direction! There was a lot of discussion about this, it was clearly confusing to them. We spent a good 10 minutes on this. Answer D has nonsensical units. And it says that r vector has a “down” component over there. I picked a coordinate system in the room and had everyone “point” in the rhat, and then phi-hat, directions, to show them that these are not constants, they are different for everyone in the room. “r_hat” has a “hidden notation”, it’s really “r_hat(r,theta,phi)”. Students wanted to know when a vector COULD have a thetahat or phihat component (how about VELOCITY of a point passing DOWN at P?) Students wanted to know if we could “fix it up” by putting meters in with the Pi/2’s… (Not in this case, A is correct). Student was bugged that answer A is ALSO the answer for OTHER points on that sphere, that’s right, R=2 m r_hat is by itself NOT sufficient. I point out that writing r = (r, theta, phi) is fine, describing a point in spherical components, those three “bits” of information are what you need, but that that notation is NOT the same as the cartesian, it is not saying “r r_hat + theta theta_hat + phi phi_hat”) (Note: I gave a slightly different version, in which option “D” above above was not available, so that’s why most of my students voted “None of these”, because they WANTED it to be what is now D above ) -SJP WRITTEN BY: Steven Pollock (CU-Boulder) This comes from research by Brant Hinrichs, it’s a known student difficulty. E) None of these

Divergence of the E-field, work and energy -- WEEK 3 -- Divergence of the E-field, work and energy

Consider the vector field where c = constant . MD-1 Consider the vector field where c = constant . The divergence of this vector field is: Zero everywhere except at the origin Zero everywhere including the origin Non-zero everywhere, including the origin. Non-zero everywhere, except at origin (zero at origin) (No fair computing the answer. Get answer from your brain.) CORRECT ANSWER: C USED IN: Fall 2008 (Dubson), Fall 2009 (Schibli), Spring 2013 (Pollock) LECTURE NUMBER: Dubson (Week 2, Lecture 5, and again in Week 3 as a 3D vector field), Schibli, Pollock (Week 2, Lecture 5) STUDENT RESPONSES: 26% 12% [[48%]] 14% 0% (FALL 2008) - Week 2 15% 13% [[61%]] 11% 0% (FALL 2008) - Week 3 (3D Vector Field) 11% 11% [[50%]] 28% 0% (FALL 2009) 64%, 13, 9, 15, 0 (silent), then I provided a hint (next slide AND the formula from the front flyleaf!!), still only 46, 0, [[53]],2, 0 (SP 2013) INSTRUCTOR NOTES: Good question. I like BOTH conceptual AND computational aspects. Students quickly decided that div(V) is nonzero at the origin, you can “see” the outward flux from a tiny sphere. But most did not realize that it would be nonzero ELSEWHERE. Last class we had done the “coulomb’s law” version of this (which IS zero away from the origin), so they were I think using that as some sort of “rule” now. See the next slide, which is helpful for the discussion of how we see that this field diverges everywhere (in fact, at the origin, the field is pathololgical at the origin, it JUMPS from 0 to c in any irection. When you compute 1/r^2 d/dr (r^2 c) you get 2c/r, the divergence is in fact INFINITE right at the origin!) This took a long time to talk through! WRITTEN BY: Mike Dubson (CU-Boulder)

Consider the vector field where c = constant . MD-1 Consider the vector field where c = constant . The divergence of this vector field is: Zero everywhere except at the origin Zero everywhere including the origin Non-zero everywhere, including the origin. Non-zero everywhere, except at origin (zero at origin) (No fair computing the answer. Get answer from your brain.) Sp ‘13: This is to help with the previous one, after they all got it wrong. (It didn’t help as much as I thought, but that formula was on the board, not on the ppt, when I showed it)

The divergence of this vector field is: Consider the 3D vector field in spherical coordinates, where c = constant . The divergence of this vector field is: Zero everywhere except at the origin Zero everywhere including the origin Non-zero everywhere, including the origin. Non-zero everywhere, except at origin (zero at origin) (No fair computing the answer. Get answer from your brain.) CORRECT ANSWER: A USED IN: Fall 2008 (Dubson), Fall 2009 (Schibli) LECTURE NUMBER: Dubson (Week 3, Lecture 6). STUDENT RESPONSES: [[63%]] 4% 19% 13% 0%(FALL 2008) [[61%]] 0% 23% 16% 0% (FALL 2009) INSTRUCTOR NOTES: WRITTEN BY: Mike Dubson (CU-Boulder)

What is the divergence of this vector field in the boxed region? 1.5 What is the divergence of this vector field in the boxed region? Zero Not zero ??? Sp ‘13. Good question for discussion. Didn’t click, just used as a prop to talk about fact that div(rhat/r^2) is mostly zero, but infinite at the origin – i.e. proportional to delta^3(r), this is my “lead-in” to talking about Delta functions next! This is subtle, it looks like the lines “diverge”, but that’s not what mathematical divergence means. (In that box, “lines in = lines out”. There’s no CREATION of field lines in the box! Take div using the front flyleaf in spherical coordinates if that helps!

What is the value of CORRECT ANSWER: C USED IN: Spring 2013 (Pollock) LECTURE NUMBER: 5 (Gauss’ Law) STUDENT RESPONSES: Didn’t bother clicking INSTRUCTOR NOTES: I had already really gone over the delta function, so didn’t take the time to let them click. I suspect it would have been quite high.

E) None of these/more than one/??? A point charge (q) is located at position R, as shown. What is (r), the charge density in all space? E) None of these/more than one/??? CORRECT ANSWER: E USED IN: Spring 2008 and 2013 (Pollock) LECTURE NUMBER: 5 STUDENT RESPONSES:14% 9% 59% 9% [[9%]] (Sp 2008) 14, 2, 67, 9, [[9]] INSTRUCTOR NOTES. E is really correct (since both C and D are good), although that’s a *little* mean! 60% voted for C, we had votes for A and B as well, though. Quick but useful discussion, the delta function is +1 when it’s argument vanishes. It is never negative, and r-R=0 is the SAME as R-r=0, both simply say that this charge density “blows up” when r=R. Few voted for A again this time! Gotta nip this in the bud, A is not a function of r! Based on HW, delta function is hard and I didn’t treat it thoroughly enough. I think we could easily add even more CT’s about the delta function -SJP WRITTEN BY: Steven Pollock (CU-Boulder)

What are the units of (x) if x is measured in meters?  is dimension less (‘no units’) [m]: Unit of length [m2]: Unit of length squared [m-1]: 1 / (unit of length) [m-2]: 1 / (unit of length squared) CORRECT ANSWER: D USED IN: Fall 2009 (Schibli), Sp 2013 (Pollock) LECTURE NUMBER: Schibli (Week 2, Lecture 6) Pollock (Week 2, Lecture 5) STUDENT RESPONSES: 31% 16% 8% [[41%]] 4% (FALL 2009) 32%, 3% 5%, [[54%]], 5% (SP ‘2013) INSTRUCTOR NOTES: Added by trs in fall 09 based on the students difficulties in previous years.(Schibli) This was a very good question to have – good conversation, and it’s a known difficulty/issue for students. I wrote the integral of delta(x) dx = 1 formula on the board while they were clicking, as a reminder. WRITTEN BY: Thomas Schibli (CU-Boulder) 22

What are the units of 3( r ) if the components of r are measured in meters? [m]: Unit of length [m2]: Unit of length squared [m-1]: 1 / (unit of length) [m-2]: 1 / (unit of length squared) None of these. CORRECT ANSWER: E (should be m^-3!) USED IN: Fall 2009 (Schibli) LECTURE NUMBER: Schibli (Week 2, Lecture 6) Discussed (but, no time to click) in Pollock (Week 2, Lecture 5, Sp ‘13) STUDENT RESPONSES: 0% 0% 18% 8% [[74%]] (FALL 2009) INSTRUCTOR NOTES: Added by trs in fall 09 based on the students difficulties in previous years. (Schibli) Worthwhile to talk about. I had rho(r)=+q delta^3(r) on the board already from earlier in the class, so pointed out that the UNITS of that equation only make sense if this clicker answer is E. WRITTEN BY: Thomas Schibli (CU-Boulder) 23

What is the divergence of this vector field in the boxed region? 1.5 What is the divergence of this vector field in the boxed region? (It’s zero there.) Divergence is nonzero where “r” is! Reminder, after math discussion. Didn’t reclick, just used as a prop to talk about fact that div(rhat/r^2) is proportional to delta^3(r)

The four shapes are I: cylinder II: cube III: cylinder IV: sphere Consider these four closed gaussian surfaces, each of which straddles an infinite sheet of constant areal mass density. The four shapes are I: cylinder II: cube III: cylinder IV: sphere  For which of these surfaces does gauss's law, help us find E near the surface?? From Classical Mechanics: Phys 2210 Sp 12 SJP L16 4, [[80]], 4, 5, 7 Sp 11: 0, [85], 9, 2, 4 Although they do fine, I think after a few weeks this gets lost. WHEN do you use Gauss’ law, and when “direct integration”. Should revisit this, perhaps in class, definitely in homeworks! Pre-class question, they did fine. Good discussion points – symmetry is required to tell us g is everywhere “in”, then we need surfaces over which g is constant/can be pulled out of the Gaussian integral. One student was puzzling about the fact that g is everywhere in, he was looking at surface I, and said basically “if only the mass inside that cylinder is needed to compute g, then you only have a small disc of mass, and then g will NOT be uniform on the top surface” Interesting, it was good to get that “out there” and talk it through (It’s not that all the outside mass is not contributing, it is, it just isn’t needed to COMPUTE g once your symmetry argument has been made!) Notes from 11 (Classical mechanics, gravity version) SJP, Sp ‘11 Lecture #15 Good discussion here (despite the high success rate). I used this to lead in to the symmetry discussion (this case is different from the previous one, now we CAN construct surfaces where g dot dA is constant) Someone liked IV, because he thought we could figure out the dot product geometrically. I then asked how you would deal with the unknown g(height), and he said “but we know g is constant”. Ah, but you know this not from symmetry, but from having previously solved this problem( with Gauss’ law! ) At least, if you don’t KNOW that g is constant, then III and IV are useless surfaces, ( and even if you somehow do know it, the geometry is nasty! ) We used this to go ahead and DO the calculation of g. I only integrated over the top of can I, getting an answer wrong by a factor of 2, and let the class brainstorm what was missing (they didn’t know it was wrong, but I told them it was!) Needed a lot of prompting, almost nobody realized you MUST integrate over the entire closed surface. But once I pointed that out, they were quick to get the other endcap. Good discussion of why that 2nd cap doubles (and does not cancel) the left hand side. Question closely related to one from Chandralekha Singh _________ All B) I and II only C) I and IV only D) I, II and IV only E) Some other combo

Extras on delta functions These came from another course (Class Mech), and I did not bother with them in E&M (though, I posted them)

What is the value of Classical Mechanics SJP, Sp ‘12 Lecture #25 Just had them shout this out. (A,0), no problem

What is the value of Classical Mechanics SJP, Sp ‘12 Lecture #25 Had them shout this out. (B, 1) Lots of participation, no apparent issues (we discussed the fact that the limit doesn’t matter as long as the delta function’s zero is “spanned”)

What is the value of Classical Mechanics 2012: I had NOT discussed the issue of the limits this year, did some hand raising, and got about 1/5 of the class voting for 4. But as soon as I pointed out the limits, and pointed to the graphical representation on the board, I saw some head slaps. So I think this is also effective! Maybe just let them click. 2011: Shouted this one out. (A,0) Class is doing fine, far as I can tell.I heard one or two “4”’s, but mostly, lots of students noted the issue here, that we did not span x=-2.

What is the value of Classical Mechanics SJP, Sp ‘11 Lecture #25 Shouted this one out. (C, 4) (Just follows up/confirms the previous one, not worrying about the limit issue)

What is the value of Classical Mechanics SJP, Sp ‘11 Lecture #25 2011: 0,0,[69],28,3 No time to click in 2012. This one generated good discussion. (The one person voting “different” wondered if it vanished, but he couldn’t articulate why he thought that). Lots of argument for negative 4, due to the sign flip in the delta function. I talked about our pictorial “limit” definition of the delta function as a limit of even functions, so that delta(x)=delta(-x), and thus the 2-x could just as well be x-2. I also argued that the delta function and x^2 are both intrinsically positive functions. (Our discussion of “even vs odd” was good, and someone pointed out that delta(x)=-delta(x) is true, at least, for all x away from 0. But another pointed out that the integral of odd functions vanishes, whereas the integral of delta clearly does not, so it cannot be odd!)

Classical Mechanics SJP, Sp ‘11 Lecture #25 (quick this year, we had already “derived” this in an early CT, no time to delve in or vote) I animated this slide, and asked all students to write down the integral at the top in their notebook. Many did, and then I pointed out that this is just a change of dummy index from what we already had on the board. I also pointed out this notation is perhaps how they will see the delta function in the future...

What are the UNITS of (where t is seconds) Recall that What are the UNITS of (where t is seconds) Classical Mechanics SJP, Sp ‘11 Lecture #25 16, [49], 32, 0, 3 Then on to the question itself. They were definitely struggling with this, so I walked through the example on the board without any f(t), just integral delta(t)dt = 1, and got them to argue the answer. I then went back to my original drawing of the limiting process, where the height was 1/tau, and that made several students go “ahh”. (See next slide for an animation)

1/τ τ Classical Mechanics SJP, Sp ‘12 Lecture #25

1.1a In cylindrical (2D) coordinates, what would be the correct description of the position vector “r” of the point P shown at (x,y) = (1, 1) y P r Origin x CORRECT ANSWER: A USED IN: Spring 2013 (Pollock) LECTURE NUMBER: 6 (Gauss’ Law) STUDENT RESPONSES: [[84]], 16, 0, 0 0 Sp ’13 (preclass question) INSTRUCTOR NOTES This was the “review’ question to start class. We had done almost exactly the same question at the start of the previous class, so I thought it would be even higher. (The position vector is different from the coordinates! Students seem to be confused about this - the point P has r=Sqrt[2], phi=pi/4, but it is certainly not the case that, e.g. B would be the position vector. The units aren’t even correct!) -SJP WRITTEN BY: Steven Pollock (CU-Boulder) NEW in 2013 E) Something else entirely

A) I only B) I and II only C) I and III only 4 surfaces are coaxial with an infinitely long line of charge with uniform . Choose all surfaces through which CORRECT ANSWER: D USED IN: Spring 2013 (Pollock) LECTURE NUMBER: 6 (Gauss’ Law) STUDENT RESPONSES12, 5, 5, [[64]], 14, Sp ’13 (preclass question) INSTRUCTOR NOTES : Last time I taught I just showed this and didn’t click, but there were a surprisingly large number who had some issues. Some thought that the shape had to reflect the symmetry of the field, others were confused about that last surface (even though I very clearly described it in words before they voted) It let to a brief but useful discussion about Gauss’ law, that it is always true, even if not always “helpful” to find the E field. -SJP WRITTEN BY: "Student understanding of symmetry and Gauss's law of electricity",  Chandralekha Singh Am. J. Phys. 74, 923 (2006)  A) I only B) I and II only C) I and III only D) I, II, and III only E) All four.

You have an E field given by E = c r /eo, (Here c = constant, r = spherical radius vector) What is the charge density (r)? A) c B) c r C) 3 c D) 3 c r^2 E) None of these is correct CORRECT ANSWER: C USED IN: Spring 2008 and ‘13 (Pollock), Spring 2009 (Kinney), Fall 2009 (Schibli) LECTURE NUMBER: Pollock (Lecture 5 in ‘08, 6 in ‘13); Kinney (Week 3, Lecture 6) STUDENT RESPONSES: 17% 9% [[52%]] 13% 9% (SPRING 2008) 3% 0% [[97%]] 0% 0% (SPRING 2009) 35% 16% [[47%]] 2% 0% (FALL 2009) 14%, 14%, [[63%]], 8, 2 (Sp ’13) INSTRUCTOR NOTES: They really struggled. 52% got it, all other answers were represented in Spring 08. Answer is C, Del dot E = rho/epsilon0, so just need to take divergence of “r vector”. (I suggested they use the front flyleaf, many did not seem to know HOW. Or you can just take d/dx + d/dy + d/dz = 1+1+1!). -SJP In Spring 2009 this generated some good discussion, but wasn’t a real stumper. I gave this as a whiteboard activity and didn’t show them the answer choices until they had a chance to work through it a little. In ‘13 the class very strongly wanted to use the “spherical coord” formula for div. (which is fine) WRITTEN BY: Steven Pollock (CU-Boulder)

Hint: Can you find it two DIFFERENT ways? Given E = c r/eo, (c = constant, r = spherical radius vector) We just found (r) = 3c. What is the total charge Q enclosed by an imaginary sphere centered on the origin, of radius R? Hint: Can you find it two DIFFERENT ways? CORRECT ANSWER: D USED IN: Spring 2008 and ‘13 (Pollock) and Spring 2009 (Kinney), Fall 2009 (Schibli) LECTURE NUMBER: Pollock (Lecture 5 in ‘08, 6 in ‘13) and Kinney (Week 3, Lecture 6) Schibli (Week 2, Lecture 6) STUDENT RESPONSES: 0% 5% 10% [[81%]] 5% 6% 16% 24% [[53%]] 2% (FALL 2009) 0%, 2%, 7%, [[92%]],0 (Spring 2013) INSTRUCTOR NOTES: 81% got it, but very few thought to use Gauss’ law (and one who did ended up really effectively still doing the 3-D volume integral, not the surface integral) . Answer: D. The “other way” is to find the FLUX through that sphere, which is E times surface area (r is a constant on that surface and comes out, so the integral is trivial) -SJP WRITTEN BY: Steven Pollock (CU-Boulder) (4/3)  c B) 4  c C) (4/3)  c R^3 D) 4  c R^3 E) None of these is correct

An infinite rod has uniform charge density  An infinite rod has uniform charge density . What is the direction of the E field at the point P shown? A) P B)  C) CORRECT ANSWER: C USED IN: Spring 2008 (Pollock), Fall 2009 (Schibli) LECTURE NUMBER: 6 STUDENT RESPONSES: 9% 0% [[91%]] 0% 0% INSTRUCTOR NOTES: Used in lecture 6, as lead-ins to the role of symmetry in Gauss’ law. (91% correct, this one is not hard for them. Might need to think about how to make it a little more challenging...). -SJP In ‘13 I didn’t click on it, just used it as a conversation starter. WRITTEN BY: Steven Pollock (CU-Boulder) Origin D) Not enough info

The four shapes are I: cylinder II: cube III: cylinder IV: sphere Consider these four closed gaussian surfaces, each of which straddles an infinite sheet of constant areal charge density s. The four shapes are I: cylinder II: cube III: cylinder IV: sphere  For which of these surfaces does gauss's law, help us find E near the surface?? CORRECT ANSWER: B USED IN: Spring 2013 (Pollock) LECTURE NUMBER: 6 (Gauss’ Law) STUDENT RESPONSES: 20%, [[70%]], 0, 7, 3 Sp ’13 INSTRUCTOR NOTES : I have given this also in Phys 2210, (where students did better!) but I hadn’t gone THROUGH a Gauss’ law derivation yet, except for the point charge. Some students think “all”, so we discussed again the idea of how to use Gauss’ law to extract E from the integral. –SJP’ WRITTEN BY: This comes from Chandralekha Singh’s inventory, albeit slightly modified. See comments below for notes from Class Mechanics, where we gave the sae questions: Phys 2210 Sp 12 SJP L16 4, [[80]], 4, 5, 7 Sp 11: 0, [85], 9, 2, 4 Although they do fine, I think after a few weeks this gets lost. WHEN do you use Gauss’ law, and when “direct integration”. Should revisit this, perhaps in class, definitely in homeworks! Pre-class question, they did fine. Good discussion points – symmetry is required to tell us g is everywhere “in”, then we need surfaces over which g is constant/can be pulled out of the Gaussian integral. One student was puzzling about the fact that g is everywhere in, he was looking at surface I, and said basically “if only the mass inside that cylinder is needed to compute g, then you only have a small disc of mass, and then g will NOT be uniform on the top surface” Interesting, it was good to get that “out there” and talk it through (It’s not that all the outside mass is not contributing, it is, it just isn’t needed to COMPUTE g once your symmetry argument has been made!) Notes from 11: SJP, Sp ‘11 Lecture #15 Good discussion here (despite the high success rate). I used this to lead in to the symmetry discussion (this case is different from the previous one, now we CAN construct surfaces where g dot dA is constant) Someone liked IV, because he thought we could figure out the dot product geometrically. I then asked how you would deal with the unknown g(height), and he said “but we know g is constant”. Ah, but you know this not from symmetry, but from having previously solved this problem( with Gauss’ law! ) At least, if you don’t KNOW that g is constant, then III and IV are useless surfaces, ( and even if you somehow do know it, the geometry is nasty! ) We used this to go ahead and DO the calculation of g. I only integrated over the top of can I, getting an answer wrong by a factor of 2, and let the class brainstorm what was missing (they didn’t know it was wrong, but I told them it was!) Needed a lot of prompting, almost nobody realized you MUST integrate over the entire closed surface. But once I pointed that out, they were quick to get the other endcap. Good discussion of why that 2nd cap doubles (and does not cancel) the left hand side. Question closely related to one from Chandralekha Singh _________ All B) I and II only C) I and IV only D) I, II and IV only E) Some other combo!

What is the electric field inside the sphere? 2.26 A spherical shell has a uniform positive charge density on its surface. (There are no other charges around) What is the electric field inside the sphere? + A: E=0 everywhere inside B: E is non-zero everywhere in the sphere CORRECT ANSWER: A USED IN: Spring 2013 (Pollock) LECTURE NUMBER: 6 (Gauss’ Law) (I also used it in lecture 8 in ‘08 as a lead in to discontinuity boundary condition) STUDENT RESPONSES: [[98%]], 0, 2, 0, 0 (Sp ‘13) INSTRUCTOR NOTES: I clued them in by telling them that everyone “knows” the answer to this, so the real point is to figure out WHY, how do you CONVINCE me of the answer. I don’t want a vague handwavey argument, you canNOT just say “Gauss law gives it because Q(enc)=0”, because there’s still another couple of steps: Symmetry tells you NOT that E=0, but that E is uniform and radial at some inner radius, and thus you can pull it out of the integral, and THEN you can see that E=0. A student clearly and quickly articulated this. I pointed out that it’s really NOT so intuitive to me that it’s zero: if you’re near a wall, you’re near charges, why don’t they “dominate” making an inward pointing field? (The answer is that there are MORE charges on the other side, and the r^2 of the larger far-away area cancels the 1/r^2 of Coulomb, but we don’t need any of that complicated geometrical argument if we use Gauss) -SJP WRITTEN BY: SJP C: E=0 only at the very center, but non-zero elsewhere inside the sphere. D: Not enough info given

What is the electric field inside the sphere? 2.29alt We place a charge Q just outside an insulating, spherical shell (First fixing all surface charges uniformly around the sphere, and keeping them there) What is the electric field inside the sphere? + Q + A: 0 everywhere inside B: non-zero everywhere in the sphere C: Something else D: Not enough info given CORRECT ANSWER: B Note: I added extra wording (just did it now, AFTER the clicker question was given) to try to clarify the physics for next time I use it, see below. USED IN: Spring 2013 (Pollock) LECTURE NUMBER: 7 (review Gauss lecture) STUDENT RESPONSES:, 48, [[47]], 3, 2 Sp ’13 (preclass question) Then, SHOWED them the 50/50 split and told them it’s important, think it through – vote again! 35, [[64]], 0, 0, 2 (after 2 minutes) INSTRUCTOR NOTES : great question, gets at important ideas. It follows up the same question last time without the +Q, which they all know is 0. Many students do not understand the wording, even though I say it is insulating, and the charges are fixed, they WANT to believe the charges can and will readjust to give E=0 inside (i.e. they remember the “conducting sphere” story). Might want to make it even MORE clear in wording or setup. I followed this up with the next slide (which shows the field), and also discussed what the field looks like OUTSIDE too (I do that by considering first the field of two + charges side by side, sketch that, and then “erase” the hole inside the shell and replace it with the Coulomb field of the next slide. Also used this as an opportunity to point out that div(E) = rho/e0 means that you get “jumps” in field lines at surface charges. Final point to throw in here – surface charges may look “math-y” and artificial, but when we get to conductors, that IS very physical. So understanding the math of “sigma” is not just an exercise, it’s practically important. Last time I taught we just discussed this, no clicks. this is quite hard for people - they WANT to argue that E=0 because you’re inside a shell of charges, not realizing that those charges are fixed here,so they contribute nothing (and superposition gives you the field from the green Q) Note!! Sp08: I returned to this in week 10, after an exam question where they had forgotten it. Start of class, 77% got it (silent), then let them talk, it went up to 86% -SJP WRITTEN BY: Steven Pollock

What is the electric field inside the sphere? 2.29alt We place a charge Q just outside an insulating, spherical shell (fixing all surface charges uniformly around the sphere) What is the electric field inside the sphere? + Q + A: 0 everywhere inside B: non-zero everywhere in the sphere C: Something else D: Not enough info given Picture might help explain the answer to the previous question. (See comments on previous slide)

When you are done with “white sheet”, page 1, side 1, Click A When you are done with both sides, Click B If you are done with the YELLOW sheet, click C Used this for Tutorial #2. We had only ~12 minutes for the Tutorial. 63% finished page 1, 32% finished page 2, 2% finished the whole thing, in that short time. It was surprisingly hard for them to get dA! See Tutorial comments.

What is E(point A) and E(B)? 2.41 A point charge +Q is near a thin hollow insulating sphere (radius L) with charge +Q uniformly distributed on its surface. What is E(point A) and E(B)? +Q spread out uniformly on surface A + 2L L L L/2 +Q (point) B No time in ‘13 CORRECT ANSWER: A USED IN: Spring 2008 (Pollock) LECTURE NUMBER: 10 STUDENT RESPONSES: [[95%]] 0% 5% 0% 0% INSTRUCTOR NOTES: Started class. It was 50% correct, but I started talking about it, and then asked THEM to talk about it, and it swung to 95% correct. (Some where confused about whether it was sigma or rho, despite al the words, though I don't see why this would change the answer - and several thought maybe B was ON the surface, when I intended for it to be inside...) My answer is A, E(A)=0, but E(B) is E from the +Q point. -SJP CREDIT: "Student understanding of symmetry and Gauss's law of electricity",  Chandralekha Singh Am. J. Phys. 74, 923 (2006)  A) E(A)=0, E(B)≠0 B) E(A)≠0, E(B) =0 C) Both nonzero D) Both 0 45

SVC1 Given a pair of very large, flat, charged plates with surface charge densities +. Using the two Gaussian surfaces shown (A and B), what is the E field in the region OUTSIDE the plates? A B A)  B)  C) 2 D) 4 It depends on the choice of surface + + + + + + + + + + + + + + + + + Skippd in ‘13 CORRECT ANSWER: B USED IN: Spring 2009 (Kinney), Fall 2009 (Schibli) LECTURE NUMBER: Kinney (Week 3, Lecture 6, lecturer Chasteen), Schibli (Week 2, Lecture 5) STUDENT RESPONSES: 4% [[74%]] 15% 0% 7% (SPRING 2009) 9% [[77%]] 9% 0% 4% (FALL 2009) INSTRUCTOR NOTES: With the first surface, EA=sigma*A/epsilon, because E=0 between the plates. With the second surface, 2EA = 2*sigma*A/epsilon.. The trick is that you take into account the E field from all charges everywhere, not just the charges enclosed by the surface. This generated some good discussion about just why you could take into account the E field due to the other charged plate (whose sigma was not enclosed) in surface A. WRITTEN BY: Stephanie Chasteen and Mike Dubson (CU) + + + + + + + + + + + + + + + + + 46 46

Which is true about |E| at points on the imaginary dashed triangle? (II) |E| same everywhere (uniform) in (I) ONLY |E| uniform in (II) ONLY Uniform in both, but different in cases I & II D. Uniform in both, and same in cases I & II. E. |E| varies from point to point in both cases Singh’s conceptual evaluation

For which of these Gaussian surfaces will Gauss’ law help us to calculate E at point A due to the sheet of charge? Point A is at the top center of each Gaussian surface. From Singh’s conceptual evaluation. Original wording was: A - only the sphere is symmetric enough Only the cylinder because the side walls have zero flux and it has circular symmetry Only the cylinder and the cube because any shape with the side walls perpendicular to the sheet and end caps paralle ot the sheet will work ]Only the sphere and the cylinder because they have circular cross section All surfaces will work because they are symmetric A) Only the sphere B) Only the cylinder C) Only the cylinder and the cube D) Only the sphere and the cylinder E) All surfaces will work

A) I only B) I and II only C) I and III only 4 surfaces are coaxial with an infinitely long line of charge with uniform . Choose all surfaces which can be used to find E at point P using Gauss’ law From Singh’s conceptual evaulation A) I only B) I and II only C) I and III only D) I, II, and III only E) All four.

Click only when you are done with the yellow sheet: What is the answer to the last question: iv) Where is this E-field’s divergence non-zero? div(E) is nonzero everywhere div(E) is nonzero Nowhere (i.e. 0 everywhere) C) div(E) is nonzero everywhere OUTSIDE the pipe D) div(E)is nonzero on the surface of the pipe only E) Something else! Used this while Finishing up the “yellow sheet” (Tutorial 2B) (I changed the question and answer to make it “non-negative”, it’s easier to answer that way!) CORRECT ANSWER: D USED IN: Spring 2013 (Pollock) LECTURE NUMBER: 7 (review Gauss lecture) STUDENT RESPONSES:, 9, 5, 5, [[77]], 5 Sp ’13 (preclass question) INSTRUCTOR NOTES : The Tutorial and lectures have done a decent job on this, but it’s still far from “universal”! Gauss’ law is on the board starting at them, and by the time I closed this clicker question we had also written on the board that, for this question, rho = sigma delta(s-R) WRITTEN BY: Steven Pollock

Yes, and it would be pretty easy/elegant… Yes, but it’s not easy. I would like to find the E field directly above the center of a uniform (finite) charged disk. Can we use Gauss’ law with the Gaussian surface depicted below? Gaussian surface: Flat, massive, uniform disk CORRECT ANSWER: C USED IN: Spring 2013 (Pollock) LECTURE NUMBER: 7 (review Gauss) STUDENT RESPONSES:, 10, 39, [[51]], Sp ’13 INSTRUCTOR NOTES : I gave this in Phys 2210 also (in the gravitational context) – it was 41% preclass there, then 52% after discussion. Results are similar here, see comments below. It’s a good one. SJP WRITTEN BY: Steven Pollock 2210 comments: I love this question – it created a ton of buzz in the room, they didn’t want to stop talking about it. One student wanted to know how answers B and C are different. I had to emphasize that B says you COULD do it if you know enough math, C says you just can’t do it. (I claim the latter is correct, NOT the former! See below) One student gave such a clear and articulate reasoning for “C” that it shut down some conversation, but questions still came up. (E.g, “couldn’t a sufficiently powerful computer do it”, which another student articulately rebutted using my example from last time, knowing a 1D integral of f(x) dx = 17, does not let you figure out f(x), no matter how powerful your computer is!) Comments from Sp 11: A “classic” question, good discussions here. We had really just started on Gauss’ law. As a way to help them, I wrote the 1-D integral Integral(0 to 10) f(x) dx = 17 And asked them, can you tell me f(x), given this integral? Most said no. I asked, “what if you were REALLY good at calculus, or had a supercomputer at your disposal”? The point is, knowing the integral of an unknown function CANNOT tell you the function, not enough information! You MUST have some symmetry argument to simplify, to “pull the function out” (I asked, if you know apriori that f(x) was constant, THEN could you figure out what f(x) is?) Students were very clear in the discussion that g varies both in magnitude and angle across the surface. We had some discussion about “not apply” because we couldn’t evaluate the dot product without knowing g. Another student countered that physically, g IS physically defined everywhere, and thus the integral must be well defined, as is “M(enclosed)”, so really there is no problem, even if we can’t USE it. This was what I was after! I framed Gauss’ law as a deep, fundamental truth about nature (“even for black holes, even when relativity is involved”) S. Pollock, but I stole it from somewhere, perhaps Singh? Not sure! Yes, and it would be pretty easy/elegant… Yes, but it’s not easy. No. Gauss’ law applies, but it can not be used here to compute E directly. No, Gauss’ law would not even apply in this case

In the limit, z << R, the value of Ez approaches zero MD16-1 Consider the z-component of the electric field Ez at distance z above the center of a uniformly charged disk (charge per area = +s, radius = R). In the limit, z << R, the value of Ez approaches zero a positive constant a negative constant +infinity -infinity CORRECT ANSWER: B USED IN: Fall 2008 (Dubson) LECTURE: Dubson (Week 15, Lecture 44) STUDENT RESPONSES: 22% [[70%]] 0% 9% 0% (FALL 2008) INSTRUCTOR NOTES: Used as a review quesiton last day of class. WRITTEN BY: Mike Dubson (CU-Boulder) z R s

The E-field near an infinite plane of charge is constant.

A: E=0 everywhere inside B: E is non-zero everywhere inside 2.27 A cubical non-conducting shell has a uniform positive charge density on its surface. (There are no other charges around) What is the field inside the box? A: E=0 everywhere inside B: E is non-zero everywhere inside C: E=0 only at the very center, but non-zero elsewhere inside. D: Not enough info given No time in ‘13, though I talked about it with students after clas, but somewhat later in the term as review. It’s hard/subtle. CORRECT ANSWER: C USED IN: Fall 2008 (Dubson) and Spring 2008 (Pollock) LECTURE NUMBER: Dubson (Week 2, Lecture 5) and skipped in Spring 2008 (Pollock) STUDENT RESPONSES: 9% 4% [[87%]] 0% 0% (FALL 2008) INSTRUCTOR NOTES: Skipped - but it would surely be a good one! See note below: There is a great website where you can simulate this (and variations, e.g. keep the box neutral, and bring in a charge from the outside, or put it in the hole), go to http://www.falstad.com/emstatic/ Look for “setup: conductiong box”. It takes some advance prep (and a little practice), but make the mouse “adjust potential”, and then click on the box, (I can’t recall if you need to make it a floater first or not) and then adjust the potential to make ZERO field. (You can first increase the brightness, to REALLY make it vanish), Then you go back to make the mouse “add + draggable object”, and you’ve got a conducting neutral box with a charge outside. Answer is C, E=0 at center by symmetry, but NOT at other points. One argument I have for this is that if this was a conductor, the charges would distribute themselves to MAKE E=0 inside, but they would NOT distribute uniformly - we know that charges concentrate at “points”. SO this distribution cannot be consistent with E=0 throughout. -SJP Dubson: Lots of questions about why the field lines go out the corners (upon seeing sketch, next slide). How can they go past positive charges? If you drop a positive test charge, wouldn’t that charge move towards the corners? Isn’t that kind of weird? Lots of discussion and questions. Great question. WRITTEN BY: Steven Pollock (CU-Boulder) 54

E field inside cubical box (sketch) E-field inside a cubical box with a uniform surface charge. The E-field lines sneak out the corners! It would be helpful to show the field lines outside the box as well.

What is the curl of this vector field, in the region shown below? non-zero everywhere Non-zero at a limited set of points zero curl everywhere We need a formula to decide for sure CORRECT ANSWER: A USED IN: Spring 2013 (Pollock) LECTURE NUMBER: 8 (intro to voltage) STUDENT RESPONSES: [[77]], 16, 3, 5, 0 (preclass question, have not dicussed curl yet) INSTRUCTOR NOTES : This was used as a warmup, to see how many know the answer before we’ve talked about it. I didn’t discuss it much, I did have the FORMULA for curl in Cartesian coordinates up on the board, so I could point out that it has a nonzero “d/dy of Ex” bit, which gives a z-component of curl. But we came back later and really worked out the reasoning (by drawing a small loop and computing the circulation) WRITTEN BY: Steven Pollock 56

Common vortex USED IN: Fall 2009 (Schibli) Spring 2013 (Pollock) LECTURE NUMBER: 8 (intro to voltage) Discussion: We’re trying to develop intuition for Stoke’s theorem and curl. Here I argue that the fish “knows” there is a circulation. That’s ALL it knows. By Stoke’s theorem, however, the fish perhaps also intuits “don’t explore the central region”, because it already knows there is a nonzero curl (a vortex) SOMEWHERE in there! Nice comic relief here with the final animation. Credit: Thomas Schibli

Feline Vortex USED IN: Fall 2009 (Schibli) Spring 2013 (pollock0 Very cute, just gratuitous comic relief. Credit: Thomas Schibli 58

Pace Load Hours USED IN: Spring 2013 (Pollock) LECTURE NUMBER: 8 (intro to voltage) From an online survey, it all seems ok. Load is just a LITTLE on the high end to my taste. WRITTEN BY: Steven Pollock Hours

Added by: Thomas Schibli