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CONDUCTORS + CAPACITORS
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Class Activities: Conductors + Capacitors (slide 1)
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Class Activities: Conductors + Capacitors (slide 2)
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Class Activities: Conductors + Capacitors (slide 3)
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CORRECT ANSWER: USED IN: Fall 2009 (Schibli) LECTURE NUMBER: Schibli STUDENT RESPONSES: INSTRUCTOR NOTES: WRITTEN BY: Thomas Schibli (CU-Boulder) Did this on blackboard in Only new thing is Del^2 V formula, which is very quick to derive.
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+q r A A) |E| = kq/r2, to left B) kq/r2 > |E| > 0, to left
2.30 A point charge +q sits outside a solid neutral conducting copper sphere of radius A. The charge q is a distance r > A from the center, on the right side. What is the E-field at the center of the sphere? (Assume equilibrium situation). A) |E| = kq/r2, to left B) kq/r2 > |E| > 0, to left C) |E| > 0, to right D) E = 0 E)None of these +q r CORRECT ANSWER: D USED IN: Fall 2008 (Dubson) and Spring 2008 and 13 (Pollock), Fall 2009 (Schibli) LECTURE NUMBER: Dubson (Week 4, Lecture 9) and Pollock (SKIPPED in 08, Start of class Lecture 11 in ‘13), Schibli STUDENT RESPONSES: 4% 14% 14% [[69%]] 0% (FALL 2008) 0% 4% 7% [[89%]] 0% (FALL 2009) 7,9,18,. [[65]], 2 (Sp ‘13) pre class, then after discussion, 0, 0, 4, [[96]], 0 INSTRUCTOR NOTES: (Answer is D) level question, basically same whether there is or is not a charge on the sphere (see next slide) Answer B is some sort of “partial shielding” intuition. Answer C grabs a few votes, it is what the “induced field” is, so perhaps if you think the sphere polarizes, realizes this induces a rightward field at the center, but forget that the +q still contributes, you might go for this? -SJP WRITTEN BY: Steven Pollock (CU-Boulder) A 6
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E) None of these! / it’s hard to compute +q r
2.30 In the previous question, suppose the copper sphere is charged, total charge +Q. (We are still in static equilibrium.) What is now the magnitude of the E-field at the center of the sphere? A) |E| = kq/r2 B) |E| = kQ/A2 C) |E| = k(q-Q)/r2 D) |E| = 0 E) None of these! / it’s hard to compute +q r CORRECT ANSWER: D USED IN: Spring 2008 and ‘13 (Pollock) LECTURE NUMBER: 11 STUDENT RESPONSES: 13% 0% 0% [[83%]] 4% (2008) 0, 2, 2, [[96]], 0 (Sp ‘13) INSTRUCTOR NOTES: In ‘08: Started class with this as a review. 85% correct. Answer is D, natch. Pretty much an 1120 review question. In ‘13 this followed the simpler version and a blackboard discussion. Nice to talk about what happens to the extra charge. If there was no +q it would be uniform, but with +q, it need not be. It just is whatever it needs to be to make E=0 inside! -SJP WRITTEN BY: Steven Pollock? Michael Dubson? Someone else? Pretty generic problem (CU-Boulder) A
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A: E=0 everywhere inside B: E is non-zero everywhere in the cube
2.27 A cubical non-conducting shell has a uniform positive charge density on its surface. (There are no other charges around) What does Gauss’ law tell us about the E field inside the cube? A: E=0 everywhere inside B: E is non-zero everywhere in the cube Skipped C: E=0 only at the very center, but non-zero elsewhere inside the cube. D: Not enough info given
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We have a large copper plate with uniform surface charge density
2.34 We have a large copper plate with uniform surface charge density Imagine the Gaussian surface drawn below. Calculate the E-field a small distance s above the conductor surface. A) |E| = /e0 B) |E| = /2e0 C) |E| = /4e0 D) |E| = (1/4pe0)(/s2) E) |E| =0 CORRECT ANSWER: A USED IN: Spring 2008 and 13 (Pollock) LECTURE NUMBER: 10 STUDENT RESPONSES: [[61%]] 39% 0% 0% 0% (2008) [[20]], 59, 19, 2, 0 (Sp 2013) INSTRUCTOR NOTES: 61% correct. 20% the next term! Many fell for sigma/2 e0. I pointed out that there might be a GIANT charge +Q just down and left of this figure, as part of the discussion... ! This is a freshman level question (which we use in 1120), but it still clearly trips people up even at this point. In 2013 I had to really go through the derivation at the board in careful detail, comparing and contrasting the “conductor” version with the similar but different problem that is SIMPLY a sheet in space with sigma. (Thus, explaining the factor of 2, both mathematically through Gauss’ law, but also physically. In the case of THIS scenario, with a conductor, there MUST be other + charges on the “far side” of the conductor!) -SJP WRITTEN BY: Steven Pollock (CU-Boulder) (standard question, this one is from Jamie Nagle ) s
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Boundary conditions (2013, this just reminded me to lecture on Griffith 2.3.5, following up the previous question with the general rules for E at boundaries.
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The Periodic Table metal semiconductor or intermediate insulator
CORRECT ANSWER: USED IN: Spring 2008 (Pollock), Fall 2009 (Schibli) LECTURE: STUDENT RESPONSES: INSTRUCTOR NOTES: WRITTEN BY: Steve Pollock (CU-Boulder)
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Which situation has the larger electric field above the plane?
Consider two situations, both with very large (effectively infinite) planes of charge, with the same uniform charge per area s I. A plane of charge completely isolated in space: II. A plane of charge on the surface of metal in equilib: CORRECT ANSWER: B USED IN: Fall 2008 (Dubson) LECTURE: Dubson (Week 4, Lecture 11) STUDENT RESPONSES: 20% [[68%]] 12% 0% 0% (FALL 2008) INSTRUCTOR NOTES: WRITTEN BY: Mike Dubson (CU-Boulder) Which situation has the larger electric field above the plane? A) I B) II C) I and II have the same size E-field
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(Assume Electrostatic equilibrium.)
A neutral copper sphere has a spherical hollow in the center. A charge +q is placed in the center of the hollow. What is the total charge on the outside surface of the copper sphere? (Assume Electrostatic equilibrium.) qouter = ? Zero -q +q 0 < qoutter < +q -q < qouter < 0 +q CORRECT ANSWER: C USED IN: Fall 2008 (Dubson), Fall 2009 (Schibli), Sp 2013 LECTURE: Dubson (Week 4, Lecture 10) Pollock (Week 4, Lecture 11) STUDENT RESPONSES: 4% 0% [[91%]] 5% 0% (FALL 2008) 7% 7% [[78%]] 9% 0% (FALL 2009) 0, 2, [[92]], 4, 2 (Sp ‘13) INSTRUCTOR NOTES: This proves pretty straightforward for them at this point, but the addition this year (what about on the inside surface) also seemed to help. The argument is simple and important: draw the imaginary Gaussian spherical surface IN the bulk of the metal, E is manifestly 0, thus so is the integral of E, and thus we conclude q_enc=0. But it’s q in the figure… there MUST be exactly –q “polarized”. WRITTEN BY: Mike Dubson (CU-Boulder) To think about: What about on the inside surface?
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Click A as soon as you start page 2
Click A as soon as you start page 2! Click B as soon as you START page 3! When done, answer this: A long coax has total charge +Q on the OUTER conductor. The INNER conductor is neutral. s=0 s=c +Q What is the sign of the potential difference, DV = V(c)-V(0), between the center of the inner conductor (s=0) and the outside of the outer conductor? C) Positive D) Negative E) Zero USED IN: Spring 2013, during the Tutorial activity on Conductors. Not enough time, must follow up next class. (In 12 minutes, 47% were on page 2, 37% on page 3, and rest had answers (but the majority were wrong, for C) (To think about: how and where do charges distribute on surfaces?)
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What is true of 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 true of E(point A) and E(B)? + +Q (point) 2L A L B L/2 +Q spread out From Singh’s concept evaluation - she uses these questions in her own pre/post testing and so we should be sensitive to not putting these on the web. A) E(A)=0, E(B)<> B) E(A)<>0, E(B) =0 C) Both nonzero D) Both E) ??
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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 Skipped in 2013, although I discussed it after class. 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 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) 16
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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.
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A long coax has total charge +Q on the OUTER conductor
A long coax has total charge +Q on the OUTER conductor. The INNER conductor is neutral. s=0 s=c +Q What is the sign of the potential difference, DV = V(c)-V(0), between the center of the inner conductor (s=0) and the outside of the outer conductor? C) Positive D) Negative E) Zero CORRECT ANSWER: C (Note that answers A, B are missing, because this slide came from Tutorial last time) USED IN: Sp 2013 (Pollock) LECTURE: Pollock (Week 4, Lecture 12) STUDENT RESPONSES: 0,0, 40, 17, [[44]] (pre class). Then, class discussion at board of charge dist, it becomes 0, 0, 32, 8 [[60]] INSTRUCTOR NOTES: Although we later came to agreement that E=0 INSIDE (everywhere!) this is still a very difficult/unintuitive idea. No Delta V across regions where E=0. All that +Q sits on the very outside edge. WRITTEN BY: Steven Pollock (CU-Boulder) (To think about FIRST: how and where do charges distribute on all surfaces?)
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A point charge +q is near a neutral copper sphere with a hollow interior space. In equilibrium, the surface charge density s on the interior of the hollow space is.. s = ? Zero everywhere Non-zero, but with zero net total charge on interior surface C) Non-zero with non-zero net total charge on interior surface. +q CORRECT ANSWER: A USED IN: Fall 2008 (Dubson), Fall 2009 (Schibli), Spring 2013 (Pollock) LECTURE: Dubson (Week 4, Lecture 10) Schibli, Pollock (Week 4, Lecture 12) STUDENT RESPONSES: [[6%]] 89% 6% 0% 0% (FALL 2008) [33%]] 59% 9% 0% 0% (FALL 2009) [[25%], 57, 18, 0, 0 [Sp ‘13) INSTRUCTOR NOTES: Good question! You have to explain in words what B and C are saying (they both have nonzero distribution, but B is “polarized” with zero total, equal amounts of + and – at different spots.) Students struggle with this. They all quickly grasp that the outside of conductor will polarize, but there is a strong sense that the inside will “antipolarize” in some way. B is very popular. I first disprove C: suppose Q(t0t, inner) is nonzero. Then let someone in class come up with a Gaussian surface for which this causes a problem (one entirely in the metal, for which Q_enc must be 0!) I (and they) came up with a couple of explanations for why B is also no good, even though it DOES satisfy Gauss’ law (Q_enc=0) . One is to consider first a SOLID sphere, with +q outside. Then they all see it will polarize OUTSIDE, causing E=0 everywhere inside. But now I would argue that if you “chop out” some TOTALLY NEUTRAL METAL from the center, you have changed nothing at all. No charges have been budged, so E is STILL zero in there. And you end up with the situation here, so E=0 here too! But if they don’t like that, you can go to Griffiths’ argument: suppose answer B is correct. Draw it, perhaps some + q on the right side of the inner wall, and some – on the left side, so that Q_net=0. Well, if that were the case, there would be E field lines inside that hollow region (leaving the +’s and entering the –’s). There has to be, we proved last class that E “jumps” whenever you pass any local sigma. But in that case, you could draw a loop that follows an E field line in the hole, and wraps back through the bulk. Integral(E . Dl) is NOT zero for that loop, yet it MUST always be. Contradiction! Bottom line: the polarization happens on the OUTSIDE, the inside is neutral and field free. This is a faraday cage, and students all seem to know about THAT from freshman physics! WRITTEN BY: Mike Dubson (CU-Boulder)
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E) None of these! / it’s hard to compute +q
A HOLLOW copper sphere has total charge +Q. A point charge +q sits outside at distance a. A charge, q’, is in the hole, at the center. (We are in static equilibrium.) What is the magnitude of the E-field a distance r from q’, (but, still in the “hole” region) A) |E| = kq’/r2 B) |E| = k(q’-Q)/r2 C) |E| = 0 D) |E| = kq/(a-r)2 E) None of these! / it’s hard to compute +q +q’ r CORRECT ANSWER: A USED IN: Fall 2008 (Dubson) and Spring 2008 and 13 (Pollock), Fall 2009 (Schibli) LECTURE NUMBER: Dubson (Week 4, Lecture 10). Pollock (Lecture 11 in 08, 12 in ‘13). Schibli STUDENT RESPONSES: [[92%]] 2% 6% 0% 0% (FALL 2008) [[87%]] 0% 13% 0% 0% (SPRING 2008) [[89%]] 9% 2% 0% 0% (FALL 2009) [[83]], 3, 8, 3, 2 (Sp 13) INSTRUCTOR NOTES: . 87% correct (it started worse, got better after discussion) This follows from the logic of the previous one, students don’t seem to have too much difficulty accepting it. E can’t be 0 there (Gauss’ law!) I spend some time talking about how charge distributes, e.g. there MUST be –q’ on the “inner wall” (by Gauss law, Q_enc=0 if you draw a circle in the bulk) In this case, despite the q outside, that –q’ will be symmetric. Then the +Q moves OUTSIDE, along with an additional q’ (so that the charge of the outer shell is still +Q, even though we just moved –q’ to the inside wall). That distribution would be symmetric if +q wasn’t outside, but because of that +q, we get a little LESS + charge on the right, a little more on the left. -SJP WRITTEN BY: Steven Pollock (CU-Boulder) a +Q
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Simple Coulomb field (straight away from q’, right up to the wall)
A HOLLOW copper sphere has total charge +Q. A point charge +q sits outside. A charge, q’, is in the hole, SHIFTED right a bit. (We are in static equilibrium.) What does the E field look like in the “hole” region? Simple Coulomb field (straight away from q’, right up to the wall) B) Complicated/ it’s hard to compute +q CORRECT ANSWER: B USED IN: Fall 2008 (Dubson) and Spring 2008 and ‘13 (Pollock), Fall 2009 (Schibli) LECTURE NUMBER: Dubson (Week 4, Lecture 10). Pollock (Lecture 11, 12 in ‘13). Schibli STUDENT RESPONSES: 81% [[19%]] 0% 0% 0% (FALL 2008) 57% [[43%]] 0% 0% 0% (SPRING 2008) 76% [[24%]] 0% 0% 0% (FALL 2009) 5%, [[50%], 0,0,0 (Sp ’13) INSTRUCTOR NOTES: Only 43% got B. VERY Good class discussion starter! Lots to talk about. (One student came up with an energy argument that was very elegant, thinking about work to move a test charge from near q’ to either part of the surrounding wall needing to be the same, since it’s an equipotential, but that wouldn’t come out right if E was “simple coulomb”, it’d be less work on the short path... Cute, I hadn't thought like this!). Answer is B. (It CAN’T be straight towards the wall, because then it would hit the metal surface “obliquely” at many points). I also take advantage of this along with the previous questions to talk, again, about what else is going on. I threw away the +q outside (which is distracting for what I’m about to say). I also throw away the +Q on the shell. I just want to focus on a +q’, OFF-CENTER, in a hole in a conductor. (There is now –q’ on the inner wall, distributed asymmetrically. That’s the next clicker question if you want, but I just explained it here) That distribution with the +q’ are CANCELLING “outside” of the inner shell, so they contribute all together NOTHING to the E field there. So +q’ distributes itsef on the outer shell. How, symmetrically or asymmetrically? Remember, the +q’ and the –q’ (asymmetric) on the inner wall are conspiring to make ZERO E FIELD outside themselves… So there is no reason, no force, to make the +q’ distribute itself anything but SYMMETRICALLY on the outer wall. So outside, you have Q_net =q’, you can not “hide” a charge, but you HAVE lost all information about where it is. (Field outside is coulomb, centered on the center) This is tough stuff, lots for them to take in. I have several homework problems on it to follow this up. -SJP WRITTEN BY: Steven Pollock (CU-Boulder) +q’ +Q
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+q +Q A HOLLOW copper sphere has total charge +Q.
A point charge +q sits outside. A charge, +qc, is in the hole, SHIFTED right a bit. (Assume static equilibrium.) What does the charge distribution look like on the inner surface of the hole? All - charges, uniformly spread around B) - charges close to qc, + charges opposite qc C) All - but more close to qc and fewer opposite D) All + but more opposite qc and fewer close E) Not enough information +q CORRECT ANSWER: USED IN: Spring 2009 (Kinney) LECTURE NUMBER: STUDENT RESPONSES: INSTRUCTOR NOTES: Skipped in Sp 2013, because students had already asked about it in the previous question so we had discussed it. WRITTEN BY: Ed Kinney (CU-Boulder) +qc +Q 22
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CAPACITORS
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2.49 Given a pair of very large, flat, conducting capacitor plates with surface charge densities +/- , what is the E field in the region between the plates? +Q B) C) 2 D) 4 E) Something else -Q CORRECT ANSWER: B USED IN: Fall 2008 (Dubson) and Spring 2008 (Pollock), Fall 2009 (Schibli) LECTURE NUMBER: Dubson (Week 4, Lecture 11). Pollock (Lecture 10). STUDENT RESPONSES: 0% [[62%]] 38% 0% 0% (FALL 2008) 4% [[57%]] 26% 4% 9% (SPRING 2008) 0% [[53%]] 47% 0% 0% (FALL 2009) 4% [[78%]] 17% 0% 0% (FALL 2009) 5, [[16]], 79, 0, 0 (Sp 13) Yikes! INSTRUCTOR NOTES: 2/3 were voting for C, so I said “2/3 of you are wrong”, and the vote swung to 60% B. Good discussion, they REALLY wanted to use superposition of the previous question to get C… This led to a great conversation about superposition (and how charges on conductors can "adjust"!) In ‘13 I didn’t do the above trick, and just walked them through the explanation. They wanted to “superpose”! This is a reminder that our rule that E(above)-E(below) = sigma/epsilon0 is ALWAYS TRUE, no matter what! I explained it that way, I also superposed E = sigma/2 epsilon0 for ONE sheet with the same for the other sheet. I also drew a Gaussian surface, partly in one metal, partly in the middle. ‘There are lots of ways to get this. But none of them gives answer C! -SJP WRITTEN BY: Steven Pollock (CU-Boulder) 24
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2.49m Given a pair of very large, flat, conducting capacitor plates with total charges +Q and –Q. Ignoring edges, what is the equilibrium distribution of the charge? +Q -Q A) Throughout each plate B) Uniformly on both side of each plate C) Uniformly on top of + Q plate and bottom of –Q plate D) Uniformly on bottom of +Q plate and top of –Q plate E) Something else CORRECT ANSWER: D USED IN: Spring 2009 (Kinney), Fall 2009 (Schibli) LECTURE NUMBER . STUDENT RESPONSES: 0% 2% 9% [[89%]] 0% (FALL 2009) INSTRUCTOR NOTES: WRITTEN BY: Ed Kinney (CU-Boulder) 25
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Which of the following cases could actually occur above and below a sheet of surface charge?
3.X E1 E1 A B E2 E2 E1 E1 (Used later, we’re not really at BC’s here, though it might be a lead-in, could be more intuitive at this point…?) CORRECT ANSWER: B USED IN: Fall 2008 (Dubson) and Spring 2008 (Pollock) LECTURE NUMBER: Dubson (Week 5, Lecture 13). Pollock (Lecture 13). STUDENT RESPONSES: 4% [[88%]] 8% 0% 0% (FALL 2008) 61% [[35%]] 4% 0% 0% (SPRING 2008) INSTRUCTOR NOTES: Dubson: Some confusion on what Eperp and E parallel meant. WRITTEN BY: Steven Pollock (CU-Boulder) C D E2 E2 E: None of these
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#1 has twice the stored energy B) #1 has more than twice
2.50 You have two very large parallel plate capacitors, both with the same area and the same charge Q. Capacitor #1 has twice the gap of Capacitor #2. Which has more stored potential energy? #1 +Q #1 has twice the stored energy B) #1 has more than twice C) They both have the same D) #2 has twice the stored energy E) #2 has more than twice. -Q CORRECT ANSWER: A USED IN: Fall 2008 (Dubson) and Spring 2008 and ‘13 (Pollock), Fall 2009 (Schibli) LECTURE NUMBER: Dubson (Week 5, Lecture 12). Pollock (Lecture 12, lecture 13 in 2013). STUDENT RESPONSES: [[65%]] 2% 2% 27% 4% (FALL 2008) [[26%]] 9% 4% 48% 13% (SPRING 2008) [[56%]] 4% 0% 33% 6% (FALL 2009) [[35]], 8, 12, 43, 2 (Sp ‘13) INSTRUCTOR NOTES: Started class with this (as review). Excellent question! Only 25% got it, but I solicited arguments, and the re-voted, and it went to 90% correct. Their intuition is “#2 is a stronger capacitor, so it must store more energy”. Answer is A (Same field, same E^2, but twice the volume. ) -SJP WRITTEN BY: Steven Pollock (CU-Boulder) #2 +Q -Q
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E) Some other combination!
2.51 You have two parallel plate capacitors, both with the same area and the same gap size. Capacitor #1 has twice the charge of #2. Which has more capacitance? More stored energy? C1>C2, PE1>PE2 B) C1>C2, PE1=PE2 C) C1=C2, PE1=PE2 D) C1=C2, PE1>PE2 E) Some other combination! #1 +2Q -2Q CORRECT ANSWER: D USED IN: Spring 2008 (Pollock)and ‘13 LECTURE NUMBER: 12 (13 in 2013) STUDENT RESPONSES: 13% 0% 0% [[87%]] 0% 27, 9 2, [[60]],2 INSTRUCTOR NOTES: Follow-up. This was quite easy given the previous one’s discussion. Couple of students fell for C1>C2, all rest got it. (almost everyone sees PE1>PE2). Could consider ways to make this more challenging, though it was nice to address the old freshman issue of capacitance being independent of the charge on the plates -SJP WRITTEN BY: Steven Pollock (CU-Boulder) #2 +Q -Q
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V A parallel plate capacitor is attached to a battery which maintains a constant voltage difference V between the capacitor plates. While the battery is attached, the plates are pulled apart. The electrostatic energy stored in the capacitor A) increases B) decreases C) stays constant. CORRECT ANSWER: B USED IN: Fall 2008 (Dubson) LECTURE: Dubson (Week 4, Lecture 11) STUDENT RESPONSES: 50% [[30%]] 20% 0% 0% (FALL 2008) INSTRUCTOR NOTES: WRITTEN BY: Mike Dubson (CU-Boulder)
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Two very strong (big C) ideal capacitors are well separated.
3.4 What if they are connected by one thin conducting wire, is this electrostatic situation physically stable? -Q Q -Q + Q - Yes No ??? + - (This question belongs in next batch of questions, Griffiths Ch 3, “Uniqueness”) CORRECT ANSWER: A USED IN: Fall 2008 (Dubson) and Spring 2008 (Pollock) LECTURE NUMBER: Dubson (Week 5, Lecture 13). Pollock (Lecture 12). INSTRUCTOR NOTES: Not voted on, but discussed it in the context of the previous one. My answer is yes, the conducting wire changes nothing at all. This can be a little baffling for students, we talked a lot about the ideas here in the Tutorial of this week. WRITTEN BY: Steven Pollock (CU-Boulder)
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