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**ELECTRIC DISPLACEMENT**

WRITTEN BY: Steven Pollock (CU-Boulder)

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If you put a dielectric in an external field Eext, it polarizes, adding a new field, Einduced (from the bound charges). These superpose, making a total field Etot. What is the vector equation relating these three fields? CORRECT ANSWER: C USED IN: Spring 2013 (Pollock) LECTURE NUMBER: 24 STUDENT RESPONSES:0, 2, [[90]],6, 2 (‘13) INSTRUCTOR NOTES – I had worked through the field in the previous CT (a uniformly polarized sphere), so on the board we had an E_ext up the page, a P up the page, an E_induced DOWN the page inside the sphere, and then I asked this one. I was wondering if the “up and down arrows” in the image would make them want to introduce spurious minus signs, but it did not. It’s just superposition here! -SJP WRITTEN BY: Steven Pollock (CU-Boulder)

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**Eext B) Einduced C) Etot**

4.5 We define "Electric Displacement" or "D" field: D =0E + P. If you put a dielectric in an external field Eext, it polarizes, adding a new field, Einduced (from the bound charges). These superpose, making a total field Etot. Which of these three E fields is the "E" in the formula for D above? CORRECT ANSWER: C USED IN: Fall 2008 (Dubson), Spring 2008 and ‘13 (Pollock), Fall 2009 (Schibli) LECTURE NUMBER: Dubson (Week 8, Lecture 20), Pollock (24) STUDENT RESPONSES: % 0% [[74%]] 0% 0% (FALL 2008) 7% 7% [[87%]] 0% 0% (SPRING 2008) 2% 7% [[91%]] 0% 0% (FALL 2009) [[81]], 2??, 0 INSTRUCTOR NOTES: Start of class in earlier terms. 87% correct, then I had them discuss with each other, and it went to 100%! Answer is C: E(tot) They didn’t have too much trouble, the discussion was brief but interesting (“could you ever measure E induced”? My answer was Etot is physical, that’s what you measure) In 2013 this came halfway through the lecture on defining D, so I felt I had really just told them the answer. (But apparently that doesn’t completely help!) -SJP WRITTEN BY: Steven Pollock (CU-Boulder) Eext B) Einduced C) Etot

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**Linear Dielectric: is the “Electric Susceptibility**

(Usually small, always positive) Added Spring 2014, just to get the definition up there.

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**is the dielectric constant**

Linear Dielectric: is the “Electric Susceptibility” Added Spring 2014, just to get the definition up there. I introduced this a little earlier than last year, just want to emphasize that the VALUE of D arises mostly if you know the material is linear, so you can get D (from free charges only) and thus E… is the dielectric constant is the permittivity

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**C) q/(4 r2) for r<R, but q/(4 0 r2) for r>R **

4.7c-alt We define D =0E + P, with A point charge +q is placed at the center of a dielectric sphere (radius R). There are no other free charges anywhere. What is |D(r)|? R + +q A) q/(4 r2) everywhere B) q/(4 0 r2) everywhere C) q/(4 r2) for r<R, but q/(4 0 r2) for r>R D) None of the above, it's more complicated E) We need more info to answer! CORRECT ANSWER: A USED IN: Fall 2008 (Dubson), Spring 2008 and 13 (Pollock), Fall 2009 (Schibli) LECTURE NUMBER: Dubson (Week 8, Lecture 20), Pollock (24). Schibli STUDENT RESPONSES: [[89%]] 7% 2% 2% 0% (FALL 2008) [[ 70%]] 5% 20% 5% 0% (SPRING 2008) [[77%]] 11% 5% 7% 0% (FALL 2009) [[77]], 4, 8, 8, 4(Sp ‘13) INSTRUCTOR NOTES: Answer is A - comes directly from the Gauss’ law statement at the top (and symmetry) The presence of the dielectric is irrelevant. 70% correct. “C” was next most popular. Important to emphasize that E is complicated here by the existence of bound charges, but D is not. (We could not use D to FIND E unless we know it’s linear, though!) Spent a little time using this clicker question to clarify/emphasize: is that +q “free” (even though it’s stuck in the middle? Yes, free doesn’t mean free in that sense!) Also, I animated the slide and asked them to spend some time figuring out D(r) on their own first. Took some time to let them struggle, reminded them of the “symmetry” aspects of D field (which way could it point? What can it depend on?) -SJP WRITTEN BY: Steven Pollock (CU-Boulder) 6

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ERK-3.1 Expand the following expression out to order 1/r2 for the case that r >> a and r>>b (No longer sure what this was used for!) 7

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4.8 A solid non-conducting dielectric rod has been injected ("doped") with a fixed, known charge distribution (s). (The material responds, polarizing internally) When computing D in the rod, do you treat this (s) as the "free charges" or "bound charges" ? (s) "free charge" B) "bound charge" C) Neither of these - (s) is some combination of free and bound D) Something else. CORRECT ANSWER: A USED IN: Spring 2008 and 13 (Pollock) LECTURE NUMBER: Pollock (25) STUDENT RESPONSES: [[85%]] 5% 10% 0% 0% (Sp 08?) [[41]], 53, 6 (Sp ‘13) INSTRUCTOR NOTES: Started lecture with this. 85% correct. MUCH worse the second year. Answer is A, good (brief) discussion of the use of the word “free” (even when they are “stuck” like this!)Its’ really just lingo (and my students this year seemed bothered by the fact that free charges are not “free”) -SJP WRITTEN BY: Steven Pollock (CU-Boulder) 8

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4.6 A very large (effectively infinite) capacitor has charge Q. A neutral (homogeneous) dielectric is inserted into the gap (and of course, it will polarize) . We want to find D everywhere. Which equation would you head to first? +Q B B -Q CORRECT ANSWER: B USED IN: Fall 2008 (Dubson), Spring 2008 and 13 (Pollock) LECTURE NUMBER: Dubson (Week 8, Lecture 20), Pollock (24, 25 in ‘13) STUDENT RESPONSES: % [[84%]] 5% 9% 2% (FALL 2008) 15% [[35%]] 20% 0% 30% (SPRING 2008)\ 13, [[75]], 6, 6 (Sp ‘13) (We had just written this on the board, and boxed it, at start of class though!) INSTRUCTOR NOTES: In ‘08: This was towards the end of class, and it was all over the place. 35% voted B, nobody went for D, but all other answers well represented. Note that I did NOT say it was a linear dielectric, and I clearly pointed that out to them, I did NOT want them to *assume* it was a linear dielectric. My answer is B - that’s certainly what you do, you argue Qfree is known, and find D everywhere. (In the course of discussion, though, my students pointed out I could be wrong. If the material is nonlinear, then there’s e.g. no guarantee that it’s homogeneous, which means it could polarize more on the left than on the right. Which would rearrange the Q on the capacitor, which means you’ve lost symmetry.... I fixed this for ‘13 by adding the word “homogeneous” - I want to assume nonlinearity, which is nominally irrelevant IF you know sigma(free) is a uniform sheet, but still keep it symmetric in the x-y directions to ensure the latter... ) -SJP WRITTEN BY: Steven Pollock (CU-Boulder) i B) ii C) iii D) More than one of these would work OK. 9

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**Which equation would you go to first? i B) ii C) iii **

An ideal (large) capacitor has charge Q. A neutral dielectric is inserted into the gap (and of course, it will polarize) We want to find E everywhere (i) D = 0 E + P (ii) (iii) +Q B B -Q CORRECT ANSWER: E USED IN: Spring 2013 LECTURE NUMBER: 25 STUDENT RESPONSES: 6, 17, 7, 4, [[67]] INSTRUCTOR NOTES: Skipped. (We talked about it with the previous one). My answer: E If you don’t know it’s *linear*, then knowing D is not enough to determine E. (Even if it’s still x-y symmetric/infinite, you still don’t know how much it polarizes!) Asked it in 2013, tried to point them to answer E (Noted that there was a “new” answer that wasn’t on the previous question) It’s slightly mean, of course, to make E be the correct answer. It’s a decent question for discussion, though –merely pointing out that different materials could polarize totally differently, giving very different E fields, so clearly you can’t “find E everywhere” without more info! One student pointed out that you CAN find E outside the capacitor (it’s zero). Nice point… -SJP WRITTEN BY: Steven Pollock (CU-Boulder) Which equation would you go to first? i B) ii C) iii D) Your call: more than 1 of these would work! E) Can't solve, unless know the dielectric is linear!

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**is the dielectric constant**

Linear Dielectric: is the “Electric Susceptibility” Added Spring 2014, just to get the definition up there. (And review in L25) is the dielectric constant

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**A) B) B C) ( B)A D) ( B) E) Something else**

MD8-4 An ideal (large) capacitor has charge Q. A neutral linear dielectric is inserted into the gap. We want to find D in the dielectric. s B B +s For the Gaussian pillbox shown, what is Qfree,enclosed? A) B) B C) ( B)A D) ( B) E) Something else CORRECT ANSWER: A USED IN: Fall 2008 (Dubson), Spring 2008 and ‘13(Pollock), Fall 2009 (Schibli) LECTURE NUMBER: Dubson (Week 8, Lecture 20, 1st try, Week 9, Lecture 21, 2nd try), Pollock (25) STUDENT RESPONSES: [[80%]] 0 % 14 % 2 % 5 % (1st try) (FALL ST TRY) [[82%]] 5% 7% 0% 7% (2nd try) (FALL 2008, 2ND TRY) [[90%]] 0% 5% 5% 0% (SPRING 2008) [[97%]] 0 % 3 % 0 % 0% (FALL 2009) [78]], 4, 15, 0 (Sp ‘13) INSTRUCTOR NOTES: I added animation in 2013, let them figure it out first, THEN give answer options. 90% correct in SP08. At this point it was review, didn’t seem to have troubles with it. (It also came AFTER the homework on this stuff!) In ‘13 it’s before the homework, and still a few students falling for “Q_total” instead of “Q_free”. I turned it around, asked them what answer C IS correct for. Answer: same box, doing Gausses law for E (except for epsilon0 factor) -SJP WRITTEN BY: Steven Pollock (CU-Boulder) 12

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**A) f B) B C) (f B)A D) (f B) E) Something else**

MD8-4 An ideal (large) capacitor has charge Q. A neutral linear dielectric is inserted into the gap. We want to find D in the dielectric. sf B B +sf For the Gaussian pillbox shown, what is Qfree,enclosed? A) f B) B C) (f B)A D) (f B) E) Something else Just shifted notation to make it as clear as possible. (Did not click) 13

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**It is definitely NOT zero in there…**

MD8-4 An ideal (large) capacitor has charge Q. A neutral linear dielectric is inserted into the gap. We want to find D in the dielectric. sf B B +sf Is D zero INSIDE the metal? (i.e. on the top face of our cubical Gaussian surface) CORRECT ANSWER: A USED IN: Sp 2013 (Pollock), LECTURE NUMBER: Pollock (25) STUDENT RESPONSES: [[60]], 9, 30 INSTRUCTOR NOTES: I decided this was a subtle point, and one they really need to know. Glad I asked it. There was a lot of debate. Most quickly caught on to use e0E+P, and that E=0 up there. (Though a few seem to have forgotten that, or thought it didn’t have to be true anymore. So, e.g. some asked me during the question if sigma_free has to go to the “bottom of the top plate” like it used to. Yes!) But the debate was, is there a nonzero “P” inside the metal? One student was stressed about it, and came after class to pursue it further: clearly a piece of metal will polarize when placed in an external E field, so why would we NOT say it “has a nonzero P” if it is polarized! It’s a good question, the metal has an overall dipole moment in that case, but Polarization (P) and thus D are *local* quantities, intensive. P = dipole moment/unit volume LOCALLY, in a tiny volume. In the bulk of metal, no matter where you are, there are no “local dipoles”, no separation of charges. The charges on the surface are not bound, they don’t arise FROM any nearby local polarization. -SJP WRITTEN BY: Steven Pollock (CU-Boulder) It must be zero in there It depends It is definitely NOT zero in there… 14

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**D) sf + sb E) Something else**

MD8-4 An ideal (large) capacitor has charge Q. A neutral linear dielectric is inserted into the gap. We want to find D in the dielectric. sf B B +sf What is |D| in the dielectric? CORRECT ANSWER: A USED IN: Spring 2013 (SJP) LECTURE NUMBER: 25 STUDENT RESPONSES: [[67%]] 10 % 18% 3% 3% (FALL 2009) [[79]], 8, 11, 0, 0 (Sp 2013) INSTRUCTOR NOTES: I added animation in 2013, let them figure it out first, THEN give answer options This should be simple, I’ve been suggesting and implying it for much of the lecture, but given the chance to actually do it, still 20% of the class is not quite with me. This is just after a clear discussion (previous clicker question) that D=0 in the metal, so only one face contributes. I think that many still see these formulas as “given formulas”, not the direct outcome of a single quick calculation. Good review/practice. WRITTEN BY: Steven Pollock (CU-Boulder) A) f B) 2 f C) f / 2 D) sf + sb E) Something else 15

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4.6d An ideal (large) capacitor has charge Q. A neutral linear dielectric is inserted into the gap. Now that we have D in the dielectric, what is E inside the dielectric ? +Q -Q B B A) E = D 0 er B) E = D/e0er C) E = D D) E = D/ Not so simple! Need another method CORRECT ANSWER: B USED IN: Fall 2008 (Dubson), Spring 2008 and 13 (Pollock), Fall 2009 (Schibli) LECTURE NUMBER: Dubson (Week 9, Lecture 21), Pollock (25) STUDENT RESPONSES: % [[98%]] 0% 2% 0% (FALL 2008) 5% [[90%]] 0% 5% 0% (SPRING 2008) 0% [[100%]] 0% 0% 0% (FALL 2009) 9, [[81]], 2, 7 (Sp 13) INSTRUCTOR NOTES: 90% correct. Answer is B. Again, review, no problems. Discussion was OK, people were thinking about it, and in particular I poked on E a little (is it a problem that E creates P, but then P modifies E, so could there be some “feedback loop” that messes up the simple relationship). In 2013, someone in the front row asked “how could this question not get 100%”, since D = e0*er*E was *on the board*. Good question… -SJP WRITTEN BY: Steven Pollock (CU-Boulder) 16

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Linear Dielectric: is the dielectric constant

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4.6e An ideal (large) capacitor has charge Q. A neutral linear dielectric is inserted into the gap (with given dielectric constant) Now that we have D in the dielectric, what is E in that small gap above the dielectric ? +Q -Q B B A) E = D 0 er B) E = D/e0er C) E = D D) E = D/ Not so simple! Need another method CORRECT ANSWER: D USED IN: Fall 2008 (Dubson), Spring and 13(Pollock), Fall 2009 (Schibli) LECTURE NUMBER: Dubson (Week 9, Lecture 21), Pollock (25) STUDENT RESPONSES: % 9% 0% [[74%]] 17% 0% 10% 0% [[90%]] 0% 0% 5% 0% [[95%]] 17% (FALL 2009) 0, 14, 5, [77]], 4 (Sp ’13) INSTRUCTOR NOTES: Answer is D. Mostly no problems, but conversation was OK, it’s not a “waste of time” I think. They discussed how you could get this two ways: From treating that region as just another dielectric (via the D field calculation we just did), or straight from D = e0 E + P, with P=0. We discussed the physics - E is reduced IN the dielectric, but not outside of it. In ‘13 I also pointed out that in the small gap, you look down and see TWO sheets from the dielectric, one with –sigma_b, (making a uniform field at your location), the other with +sigma_b (canceling that out). The dielectric is itself like a capacitor, and OUTSIDE makes no field. -SJP WRITTEN BY: Steven Pollock (CU-Boulder) 18

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4.6f An ideal (large) capacitor has charge Q. A neutral linear dielectric is inserted into the gap (with given dielectric constant) Where is E discontinuous? near the free charges on the plates near the bound charges on the dielectric surface +Q -Q B B CORRECT ANSWER: C USED IN: Spring 2008 (Pollock), Fall 2009 (Schibli) LECTURE NUMBER: Pollock (25) STUDENT RESPONSES: 10% 10% [[80%]] 0% 0% (SPRING 2008) 3% 10% [[95%]] 3% 0% (FALL 2009) 16, 5, [77]], 2, 0 INSTRUCTOR NOTES: Ended class with this. 80% correct, with A and B getting a couple. Correct answer is C. In ‘13 I walked through Gauss law, and also what the field IS everywhere. Showed them the next question (where is P discontinuous) but didn’t click on it – next class. -SJP Dubson skipped for time. WRITTEN BY: Steven Pollock (CU-Boulder) i only B) ii only C) both i and ii (but nowhere else) D) both i and ii but also other places E) none of these/other/??? 19

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**Where is E discontinuous? near the free charges on the plates **

An ideal (large) capacitor has charge Q. A neutral linear dielectric is inserted into the gap (with given dielectric constant) Where is E discontinuous? near the free charges on the plates near the bound charges on the dielectric surface +Q B B -Q CORRECT ANSWER: C USED IN: Spring 2008 (Pollock), Fall 2009 (Schibli) LECTURE NUMBER: Pollock (25 and again 26) STUDENT RESPONSES: 10% 10% [[80%]] 0% 0% (SPRING 2008) 3% 10% [[95%]] 3% 0% (FALL 2009) 16, 5, [77]], 2, 0 (Sp 13) 2, 0, [95]], 2, 0(Sp 13, Repeated start of next class) INSTRUCTOR NOTES: Ended class with this. 80% correct, with A and B getting a couple. Correct answer is C. In ‘13 I walked through Gauss law, and also what the field IS everywhere. Showed them the next question (where is P discontinuous) but didn’t click on it – next class. -SJP Dubson skipped for time. WRITTEN BY: Steven Pollock (CU-Boulder) i only B) ii only C) both i and ii (but nowhere else) D) both i and ii but also other places E) none of these/other/???

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**Where is D discontinuous? near the free charges on the plates **

An ideal (large) capacitor has charge Q. A neutral linear dielectric is inserted into the gap (with given dielectric constant) Where is D discontinuous? near the free charges on the plates near the bound charges on the dielectric surface +Q B B -Q CORRECT ANSWER: A USED IN: Spring 2008 and 13 (Pollock) LECTURE NUMBER: 26 STUDENT RESPONSES: [[67%]] 13% 20% 0% 0% Sp 08 [[55]], 2, 43, 0, 0 (Sp 2013) INSTRUCTOR NOTES: Start of class (following up similar question about E which ended last class) 67% correct, with B and C splitting the rest. Answer is A, the D field is continuous throughout the middle region (and continuously 0 in the conductor) - we had explicitly *solved* for D everywhere in the last class! In 2013 I learned from the class why it was so split. On the board is D = e0 er E, so D is proportional to E. So they argued to themselves that if E is discontinuous at both free and bound charges, and D is proportional to E, then D must ALSO be discontinuous at both free and bound charges (so they voted C) (What they forgot is that er is ALSO discontinuous where the bound charges are, and this is precisely what “cancels out” the discontinuity in E to result in a smooth D. Also in 2013, we discussed the (important!) caveat that the answer to this question is ONLY correct for this (highly symmetric) situation given. Upcoming q’s are going to show that D can be discontinuous in OTHER places, e.g. if the dielectric was finite, at the end, P has a discontinuity, so therefore D does too (in the “parallel” component!) We’ll get to this shortly! -SJP WRITTEN BY: Steven Pollock (CU-Boulder) i only B) ii only C) both i and ii (but nowhere else) D) both i and ii but also other places E) none of these/other/???

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**Depends on other things not given**

MD8-5 A point charge +q is placed at the center of a neutral, linear, homogeneous, dielectric teflon shell. Can D be computed from its divergence? Yes No Depends on other things not given + q CORRECT ANSWER: A USED IN: Fall 2008 (Dubson) Spring 2013 (Pollock) LECTURE NUMBER: Dubson (Week 9, Lecture 21, 1st try, Week 9, Lecture 22, 2nd try) Lect 26 in 2013 STUDENT RESPONSES: [[73%]] 27% 0% 0% 0% (1st try) [[93%]] 7% 0% 0% 0% (2nd try) [87%]], 6, 6, 0 0 (Sp ‘13) INSTRUCTOR NOTES: Most people answered that D can be computed from divergence. But you need to know the curl too. Curl of D in spherical coordinates is zero. Shouldn’t this question be changed so they can’t get the right answer for the wrong reason? However, students do see the connection in the next nice series of questions. This question shouldn’t be asked without the other ones. SJP agrees with the above discussion, we’re getting towards the idea that finding D is simple from Gauss’ law if you have enough symmetry. But if there are “edge effects”, or broken symmetry, then the answer will be NO, Gauss law (Div(D)) alone is not enough, you also need to know Curl(D)=0 to get our simple results. WRITTEN BY: Mike Dubson (CU-Boulder) 22

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**Is the curl of the polarization P zero everywhere?**

MD8-6 A point charge +q is placed at the center of a neutral, linear, homogeneous dielectric teflon shell. The shell polarizes due to the point charge. Is the curl of the polarization P zero everywhere? + Yes No Depends on other things not given - + + - - + - + - + q CORRECT ANSWER: A USED IN: Fall 2008 (Dubson), Fall 2009 (Schibli) LECTURE NUMBER: Dubson (Week 9, Lecture 22) Pollock (Lect 26) STUDENT RESPONSES: [[45%]] 55% 0% 0% 0% (FALL 2008) [[87%]] 13% 0% 0% 0% (FALL 2009) [[57]], 43, 0, 0 (Sp 2013) INSTRUCTOR NOTES: Answer – you need polarized material to have a P. No P outside dielectric. If loop goes up inside dielectric and then outside dielectric then curl of P isn’t zero. But Mike drew a loop and showed that you would get negative on other side of loop so it’s zero. But I didn’t get a sense of whether students make the connection between zero curl and P dot dl = 0. SJP notes: I argued further that we just saw in previous question that D and E are radial. (By symmetry alone!) But then, since P is linear in E, P is also radial, and thus has no curl. WRITTEN BY: Mike Dubson (CU-Boulder) - - + + - + 23

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**Depends on the inner radius of the dielectric.**

MD8-7 A point charge +q is placed at the center of a neutral, linear, dielectric hemispherical shell. Can D be computed from its divergence? Yes No Depends on the inner radius of the dielectric. + q CORRECT ANSWER: B USED IN: Fall 2008 (Dubson), Fall 2009 (Schibli), Sping 2013 (SJP) LECTURE NUMBER: Dubson (Week 9, Lecture 22) Pollock (Lect 26) STUDENT RESPONSES: 19% [[81%]] 0% 0% 0% (FALL 2008) 72% [[28%]] 0% 0% 0% (FALL 2009) 51%, [49]], 0, 0, 0 (SP 13) INSTRUCTOR NOTES: (LA notes: Instructor modeled going through the calculation to get D = Qf/2*pi*r^2. But that assumes that D is the same around the entire Gaussian surface, but it’s not, because half the surface is in dielectric and half isn’t. This was a nice way to show how to find your mistakes. SJP comments: This is hard for them. In part, because up till now we HAVE had high symmetry, so they’re getting used to the idea that D JUST arises from free charges. We didn’t change the free charges here from the last problem, so it’s not crazy for a student to think that same rho_free means same D! I drew the picture on the board, with the polarization/bound charges shown, and then argued that the E field (and thus P, and thus D) must be very nasty, certainly NOT just pure rhat, pure function of r alone. If it’s not purely D(r) rhat, then you can’t pull it out of the integral, and can’t simply compute it. Period. A student pointed out that it’s an “edge effect”, P is NOT curl free at the ends of the sphere, so D is not curl free, so we don’t really know what’s going on. I agree. (I briefly discussed how I WOULD compute the E field here, in practice. Iteratively! Use the known q and our simple approximation D = Q/4 pi r^2, to get E = D/eps, then use THAT E (and linear dielectric) to get P and thus a “first pass estimate” of bound charges. Use those to figure out the correction to the E field fro those bound charges. Then, add that back in – we have a NEW E field, thus a NEW polarization estimate, thus a NEW bound charge estimate… thus a NEWER E field, and so on, till we converge. Students groaned, but I think the idea is important – not all problems can be solved analytically, but we DO have the tools now, in principle, to solve even nasty problems. WRITTEN BY: Mike Dubson (CU-Boulder) 24

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**Is the curl of the polarization P zero everywhere?**

MD8-8 A point charge +q is placed at the center of a neutral, linear, dielectric shell. The shell polarizes due to the point charge. Is the curl of the polarization P zero everywhere? Yes No Depends on the inner radius of the dielectric. + + + - - - + - - + + q CORRECT ANSWER: B USED IN: Fall 2008 (Dubson), Fall 2009 (Schibli) LECTURE NUMBER: Dubson (Week 9, Lecture 22) STUDENT RESPONSES: 2% [[96%]] 2% 0% 0% (FALL 2008) 23% [[77%]] 0% 0% 0% (FALL 2009) INSTRUCTOR NOTES: People seemed to get this by now. This is a very nice series of questions. But it does take a full 30 minutes to get to this point. It was very helpful for students to have whiteboards during them because they could sketch. One even showed the board to show their loop. SJP – did not click on this because we had already answered it in the discussion of the previous question. (But it was good to flash it up to wrap up the previous sequence) WRITTEN BY: Mike Dubson (CU-Boulder) 25

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**A) E = E0 B) E > E0 C) E < E0**

MD8-9 An infinite plane of charge with surface charge density sf is between two infinite slabs of neutral linear dielectric (of dielectric constant r), as shown. The "bare" E-field, due only to the plane of free charge, has magnitude Eo = sf / 2e0 X E = ? sf CORRECT ANSWER: A USED IN: Fall 2008 (Dubson), Spring 2013 (Pollock) LECTURE NUMBER: Dubson (Week 9, Lecture 22) (Lecture 26) STUDENT RESPONSES: [[78%]] 9% 13% 0% 0% [[47]], 15, 38 (Sp 13) INSTRUCTOR NOTES: LA notes: ““dielectric capacitor”. Gave this as a whiteboard activity. This seemed like a realy good activity. Took about 10 mins to get D. He stopped us here and checked our D, thinking about where to put pillbox. Made us think about where to put the pillbox. Then we continued and found E mins total for activity. SJP – Used as a clicker question, it was very hard. Maybe it does need the 10 minutes Mike gave it! It’s a good problem Students came up with various explanations, including the fact that “at the x”, the contribution from the dielectric amounts to a long sheet of +sigma_bound at the top, another long sheet of –sigma_bound at the bottom, but outside, those two cancel, they contribute nothing. Another student pointed out that you could figure out E directly by Gauss’ law, drawing a pillbox centered on the sigma_free and ending at the x on the top, and symmetric at the bottom, giving sigma_f/2 eps0 again (because the dielectric is overall neutral). Nice! My argument was to just COMPUTE D (it’s simple, just sigma_f/2 eps 0, do it on the board) and then you compute E by D/e0 er, which at the x is just D/e0, or E0. During the discussion, once we had worked it all out, I asked them to explain PHYSICALLY how it could be that E is “weak” inside the dielectric, and then suddenly “springs back to life” out there at the x, where did the increase come from? (It’s from passing the layer of + bound charges at the top of the dielectric!) WRITTEN BY: Mike Dubson (CU-Boulder) What is the magnitude of the E-field in the space above the top dielectric at the point x ? A) E = E0 B) E > E0 C) E < E0 26

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**The answer depends on what else is “held fixed” (V? Q?)**

4.11 We argued that C goes UP by a factor of r if you fill a capacitor with dielectric. What happens to the stored energy of a capacitor if it’s filled with a dielectric? It goes up It goes down It is unchanged The answer depends on what else is “held fixed” (V? Q?) CORRECT ANSWER: D USED IN: Spring 2008 and 13 (Pollock), Fall 2009 (Schibli) LECTURE NUMBER: 27 (28 in ‘13) STUDENT RESPONSES: 7% 53% 0% [[40%]] 0% (SPRING 2008) 17% 31% 3% [[49%]] 0% (FALL 2009) 29, 45, 10, [[17]] Sp 13 INSTRUCTOR NOTES: I was clear when presenting the problem that I was NOT telling them if it was isolated, or connected to a battery, or what. In ‘13 I drew pictures on the board. Still, many voted for B, about 40% for D (which is correct). It was more like 70% for B until I wrote (in the middle of the concept test, on the board) the THREE formulas for stored energy of a capacitor (1/2 C V^2, 1/2 Q^2/C, 1/2 QV) Good discussion followed - the point being that it could go up OR down depending on what you hold fixed. The *reason* for voting B that I heard seemed to be that the “E field is reduced by the dielectric, so the energy is lower” In ‘13 I had just stated that E goes down in a dielectric (which in retrospect was confusing to them, e.g. if connected to a battery, E is what it always was) -SJP WRITTEN BY: Steven Pollock (CU-Boulder) 27

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4.12b If we push this dielectric inside the isolated capacitor, will it be drawn into the capacitor or repelled? DIELECTRIC CORRECT ANSWER: A USED IN: Spring 2008 (Pollock), Fall 2009 (Schibli) LECTURE NUMBER: 27 (28 in ‘13) STUDENT RESPONSES: n/a [[78%]] 19% 3% 0% 0% (FALL 2009) [[95]], 5, 0, 0, (Sp 13) but I pretty much gave away the answer during the discusson INSTRUCTOR NOTES: Showed this, we talked about it, but did not click. SJP explanation: A) When the dielectric is inside (assuming it is also isolated from the plates!) the E field will be reduced through most of the volume, which means the stored energy in the volume has gone down. Where did it go? It must have done work ON the dielectric, pulling it in You can also draw the fringe fields and think about the forces on the bound charges -SJP (Interesting discussion point: We had asked a similar question where it was a conductor being brought in (but not in direct contact) Would the forcehere be larger or smaller than in that previous example? One student argued LARGER because now the field lines do NOT have to be perpendicular to the dielectric, so the induced charges on the top and bottom can contribute to the sideways force. But another argument was that overall energy “reduction” in the system is less here than for the conductor (there’s still some residual E field), so the F = -dW/dx argument implies WEAKER force here. I agree with the latter argument, but am not sure what the flaw is in the former (although the former is vague - there are more charges contributing to the force, but each is weaker, so who says what the net result is? I think “fringe field” arguments are intrinsically hard, and would go with the simpler energy arguments!) WRITTEN BY: Ward Handley (CU-Boulder) It gets sucked into the capacitor It gets pushed out from the capacitor I just don’t know. 28

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