# Electricity and Magnetism II

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Electricity and Magnetism II
Griffiths Chapter 1-6 Review Clicker Questions

Some other direction, or E = 0, or ambiguous
Two charges +q and -q are on the y-axis, symmetric about the origin. Point A is an empty point in space on the x-axis. The direction of the E field at A is… +q x y -q A Up Down Left Right Some other direction, or E = 0, or ambiguous Class: CONCEPTUAL CORRECT ANSWER: B ______________________________ Physics 3320, Fa11 (SJP) Lecture 1 0, [100], 0, 0, 0 Fall 2011 Comments In Lecture 1, 3320 Fa11, it was 100% correct (but I left it going while we discussed it, I think it was 86% before talking) Basically, no problem here, a warm up. I had one student explain their answer by discussing the general “dipole field pattern”, which sweeps around and points down at A, which is of a different nature than the more direct (Coulomb’s law) response I would expect in Phys 2. ============================== ALSO USED IN: Fall 2008 (Dubson) and Spring 2008 (Pollock), 3320 Sp11 (Kinney) LECTURE NUMBER: Dubson (Week 1, Lecture 2). Pollock (Lecture 2) STUDENT RESPONSES: 2% [[94%]] 0% 0% 4% (FALL 2008) 10% [[81%]] 0% 0% 10% (SPRING 2008) 0%, [[92%]],4%,4%,0% (3320) INSTRUCTOR NOTES: This is from Dubson’s pre-test, 81% got it, it's an 1120 question really. HOWEVER, looking at the readout, they started off very quiet, and it was 63% correct So I poked them to talk with their neighbor, and it became 81%… -SJP WRITTEN BY: Mike Dubson, adapted by S. Pollock (CU-Boulder) Note: E(dip) = p/(4 pi e0 r^3) (2 cos(theta) rhat + sin(theta) thetahat) 2

Some other direction, or E = 0, or ambiguous
Two charges +q and -q are on the y-axis, symmetric about the origin. Point A is an empty point in space on the x-axis. The direction of the E field at A is… +q x y -q A Up Down Left Right Some other direction, or E = 0, or ambiguous E- E+ Etot Fall 2011 – SJP: Didn’t bother with this. 3

Consider the vector field shown by this field line diagram.
Inside the dotted box, the divergence of the field is: zero everywhere non-zero everywhere zero some places and non-zero other places impossible to tell without more information Class: CONCEPTUAL Correct Answer: A _____________________________ Physics 3320, Sp12 (MD) Lecture 1 [75], 25, 0, 0 Spring 2012 Comments Some students continue to associate spreading field lines with a divergence. No sources or sinks in the field line diagram; just as many field lines entering the dotted region as leaving. ============================ Written by Mike Dubson, Sp12

Consider the vector field shown. Could this be a B-field? yes, sure!
no, impossible! not enough info to answer question The curl of this field is : zero everywhere non-zero everywhere zero some places and non-zero other places impossible to tell without more information Class: CONCEPTUAL, CONCEPTUAL Correct Answer(s): A, B _____________________________ Physics 3320, Sp12 (MD): Top Question: Lecture 1 end of class [63], 37, 0, 0 AND at beginning of class in Lecture 2 Bottom Question: Lecture 2 0, [100], 0, 0 Spring 2012 Comments Opportunity to discuss representations and the difference between field lines and a vector plot. Confusion could come from associating bunched up field lines with the closely spaced (but smaller magnitude) vectors. Each of the arrows represents the magnitude and direction of the field at that point – don’t have to plot so many vectors and still represents the same field. ============================ Written by Mike Dubson, Sp12

Same B-field as in last question. The current density j is ..
everywhere zero non-zero and into page non-zero and out of page non-zero and in plane of page Class: CONCEPTUAL Correct Answer: B _____________________________ Physics 3320, Sp12 (MD) Lecture 1 0,[88], 22, 0 Spring 2012 Comments Students seem comfortable with the “paddlewheel” explanation, and using the right-hand rule. ============================ Written by Mike Dubson, Sp12

R8 Can you use Ampere’s Law to compute the B-field at center of circular loop carrying current I? A) Yes B) No Class: CONCEPTUAL Correct Answer: B __________________________ Not used in Fall 2011 Spring 2012: Excellent question! This question was part of an Ampere’s law review tutorial (#0C) in SP12, and most students got this wrong – led to lots of good discussion about symmetry. Many students thought you could get the field at the center due to a differential element of current using Ampere’s law, then integrate around the loop to get the total field. Good follow-up question is to ask exactly how could you calculate the field? (Biot-Savart law) ===============================

R10 To find E at P from a negatively charged sphere (radius R, volume charge density ρ), P=(X,Y,Z) (x,y,z) z y R x Answer to previous concept test (R9) 10

R11 To find E at P from a thin ring (radius R, charge density λ), which is the correct formula for the x-component of ? P=(x,y,z) x-x’ (x-x’)/R (x-R cos ϕ’) (x – x’)/Sqrt[(x-x’)2+(y-y’)2+(z-z’)2] More than one of the above is correct! y dl' x R Class: MATH/ PHYSICS CORRECT ANSWER: E _______________________________ Physics 3320, Fa11 (SJP) Lecture 2 27, 0, 23, 18, [32] _______________________________ Fall 2011 Comments See notes below. Began pulling on fact that curly R = r-r’, and we need x component, so A is correct, Then discussed fact that you CAN choose phi’ if you like (and mix it up with x,y,z, even!) So C is good too. Pointed out unit issue of B and D, and discussed what D is (it’s the x-component of curly R hat) LA/TA NOTES: Most students chose a response that was correct – 32% voted that more than one was correct, 18% chose the expression for the unit vector. ====================================== USED IN: LECTURE NUMBER: SKIPPED STUDENT RESPONSES: INSTRUCTOR NOTES: (Skipped, but I think it would be fine) They struggled (all term!) with curly R! Used in 3320, Sp09, Lecture #1, (Kinney), 44, 0, 12, 4, [[40]] WRITTEN BY: Steven Pollock (CU-Boulder) It’s a little mean, because E is the correct answer. Although the notation is not specified in the statement, a reasonable interpretation of “phi’” allows both A and C to be the correct answer. (thus, E)

(A) positive (B) negative (C) zero

Both B) Only I C) Only II D) Neither E) ??
Which of the following could be a (physical) electrostic field in the region shown? Why? II I Both B) Only I C) Only II D) Neither E) ?? Class: CONCEPTUAL CORRECT ANSWER: C _____________________________ Physics 3320, Fa11 (SJP) Lecture 3 Silent, they were about 67% correct, then opened it up to discussion. Final result: 9,0,[91],0,0 Physics 3320, Sp12 (MD) Lecture 2 0, 0, [97], 3, 0 ______________________________ Fall 2011 Comments No problem here. All the points below got covered, long discussion of what charge distribution would create such a field. (I discussed a thick slab of constant, uniform + charge density, the LEFT HALF of which would have this field) LA/TA NOTES: The correct answer is like the kind given in the last question on the pretest, so maybe some connection with how they’re thinking then and here. What kind of charge distribution could create this: an infinite plane, a line of charge? Question of representation: 2-dimensional drawing, some students imagined it could be a slice of the field from a line of charge, when it’s meant to be uniform along the axis into/out of board. One student suggests there should be a charge distribution in the area where the field is diverging, which leads to the correct reasoning. ________________________________ USED IN: Spring 2008 (Pollock) LECTURE NUMBER: 5 STUDENT RESPONSES: 4% [[83%]] 9% 4% 0% 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!!). As in the previous slide, you could make this explicit, but I valued leaving that ambiguity in and letting it be part of the discussion - SJP Written by:

R14 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) Not quite sure how to get this (without computing from the front flyleaf of Griffiths!) Class: MATH/ PHYSICS CORRECT ANSWER: A __________________________________ Not used in Fall 2011 ================================= USED IN: Fall 2008 (Dubson) LECTURE NUMBER: Dubson (Week 3, Lecture 6). STUDENT RESPONSES: [[63%]] 4% 19% 13% 0%(FALL 2008) INSTRUCTOR NOTES: WRITTEN BY: Mike Dubson (CU-Boulder)

R15 Stokes’ Theorem says that for a surface S bounded by a perimeter L, any vector field B obeys S L Does Stokes’ Theorem apply for any surface S bounded by a perimeter L, even one such as this balloon-shaped surface S : L S Yes B) No C) Sometimes Class: MATH/PHYSICS CORRECT ANSWER: A _____________________________ Physics 3320, Sp12 (MD) Lecture 3 [89], 5, 5 ___________________________ Good time to talk about what Curl.B looks like as a vector field – students focus their attention on B as a vector field, and don’t think about the field lines representing Curl.B. (Not used in Fall 2011) ================================== USED IN:  Fall 2008 (Dubson) LECTURE NUMBER:  Dubson (Week 11, Lecture 27) STUDENT RESPONSES: [[82%]] 16% 2% 0% 0% INSTRUCTOR NOTES:  WRITTEN BY: Mike Dubson (CU-Boulder)

Why is in electrostatics?
Because Because E is a conservative field Because the potential (voltage) between two points is independent of the path All of the above NONE of the above - it's not always true! Class: MATH/ PHYSICS CORRECT ANSWER: D __________________________________ Physics 3320, Fa11 (SJP) Lecture 3 No problem here, silent or with discussion: 0,0,4,[[91]], 4 ___________________________________ Fall 2011 Comments Took time with the discussion – how would you DEFINE “Conservative” (I think of “conservation of energy”, so hooked that back in) I also went through the derivation of Stokes’ theorem, which they couldn’t quite remember. (Some thought answer A was “by definition of conservative”?) Used this as a chance to review Stokes’ (and frame it as just another “fundamental theorem of calculus” example) Used this to motivate our use of voltage, and Poisson’s equation... LA/TA NOTES: Students were slow to arrive at Stokes’ theorem as what makes the connection with a curl and a line integral. Not sure if I like Steve’s motivation – by the fundamental theorem of calculus, it’s an integral over one higher dimension with a derivative. OK, but why is it the curl, and not the divergence? I think it’s helpful to talk about something being “curly” and going around a loop. ======================================== USED IN: Fall 2008 (Dubson) and Spring 2008 (Pollock) LECTURE NUMBER: Dubson (Week 3, Lecture 6). Pollock (Lecture 7). STUDENT RESPONSES: 6% 6% 9% [[79%]] 0% (FALL 2008) 9% 9% 14% [[55%]] 14% (SPRING 2008) INSTRUCTOR NOTES: . I used it as a “start of class” clicker question, before covering this material, so it was sort of a “reading question” 57% chose d) I then came back to it later in the lecture (though we didn’t re-vote) to talk through points a, b, and c…-SJP WRITTEN BY: Steven Pollock (CU-Boulder) 16

B) Could be E(r), but can't be V(r) C) Can't be E(r), could be V(r)
Could this be a plot of |E|(r)? Or V(r)? (for SOME physical situation?) A) Could be E(r), or V(r) B) Could be E(r), but can't be V(r) C) Can't be E(r), could be V(r) D) Can't be either E) ??? Class: CONCEPTUAL CORRECT ANSWER: B ____________________________ Physics 3320, Fa11 (SJP) Lecture 3 About 3/4 got this when it was silent, then in the end after their peer discussion 22, [74], 4, 0 Physics 3320, Sp12 (MD) Lecture 2 9, [88], 3, 0, 0 ____________________________ Fall 2011 Comments Good discussions, I got them to come up with a real object (e.g. charged metal sphere) for E, and used this as a segue to formal discussion about boundary conditions on E (and V). (Also, discussed generality, even though we’re moving to time-dependence this term, of those BC’s) ============================================ USED IN: Fall 2008 (Dubson) and Spring 2008 (Pollock) LECTURE NUMBER: Dubson (Week 3, Lecture 7). Pollock (Lecture 8). STUDENT RESPONSES: 31% [[57%]] 12% 0% 0% (FALL 2008) 27% [[73%]] 0% 0% 0% (SPRING 2008) In Phys 3320, Sp 2011, (Lecture #2, Kinney): 23, [[69]], 8 INSTRUCTOR NOTES: 73% correct. (I preceded it by 2.26, which made the E part easier, since they had just, 10 minutes earlier, SEEN this E field!) Nice discussion possibilities, we had some good arguments brought out about WHY we know V(r) can not be discontinuous. My answer is B, E can jump (e.g. when moving past a sheet of charge, E is discontinuous, but V cannot (if V has a discontinuity, E would be infinite since it’s a derivative) in physical situations -SJP WRITTEN BY: Steven Pollock (CU-Boulder)

V = non-zero constant everywhere inside
Given a thin spherical shell with uniform surface charge density σ (and no other charges anywhere else) what can you say about the potential V inside this sphere? (Assume as usual, V(∞)=0) V=0 everywhere inside V = non-zero constant everywhere inside C) V must vary with position, but 0 at the center. D) None of these/something else/not sure. Class: CONCEPTUAL CORRECT ANSWER: B ___________________________________ Fall 2011 Comments Didn’t click on this one since someone brought it up spontaneously! ================================== USED IN: Spring 2008 (Pollock) LECTURE NUMBER: 9 STUDENT RESPONSES: 17% [[ 67%]] 17% 0% 0% In 3320, Sp ‘11 (Kinney) Lecture #2, 5, [[90]],5, 0 INSTRUCTOR NOTES: Used this as the “lead in” question start of class. 2/3 got it, lots argued for A (which I also heard during office hours) but several also argued for C. Not sure if the confusion was just misreading the problem (thinking of it as a solid sphere), but I got *asked* about that during the question, and was extremely clear that this was supposed to be JUST A SHELL of charges with nothing inside. (SO, I just now edited the problem to say “and no other charges anywhere else”, we’ll see if that helps?) Although as usual, leaving in ambiguities sometimes makes for more fruitful and active discussions/arguments!) -SJP WRITTEN BY: Steven Pollock (CU-Boulder)

R19 A point charge +q is near a neutral copper sphere with a hollow interior space. In equilibrium, the surface charge density σ on the interior of the hollow space is.. 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. σ = ? +q Class: CONCEPTUAL CORRECT ANSWER: A “Visual Hint” in next slide (R20) ______________________________ Physics 3320, Fa11 (SJP) Lecture 4 [[34]] 45, 21 _______________________________ Fall 2011 Comments This is quite challenging – lots to discuss. (We got to everything described in notes below) LA NOTES: 1) The students I was near had a short discussion on how charges move on a conductor. They then arrived at the proper answer and were fairly confident in their decision. It was a bit difficult to get much more out of them than a short response. 2) The student I was sitting next to said that negative charge should build up close to the positive charge, and that positive charge will have to build up on the interior surface of the shell in order to counteract that. It was difficult to get a discussion going around this question, and the students seemed reluctant to talk to each other. ==================================== USED IN: Fall 2008 (Dubson) LECTURE: Dubson (Week 4, Lecture 10) STUDENT RESPONSES: [[6%]] 89% 6% 0% 0% (FALL 2008) In 3320, Sp ‘11, (Kinney) Lecture #2: [[0]], 41, 59 INSTRUCTOR NOTES: WRITTEN BY: Mike Dubson (CU-Boulder) By Gauss’ law, (Using a closed surface in the bulk of the conductor), the total net charge on the interior surface must be 0 (eliminating answer C) But how do we eliminate B, how do we know it’s not “polarized” on that inner surface, yet summing up to zero? Probably the most direct argument would be using curl(E)=0: If there were charges on that inner surface, E field lines would have to emerge from the +’s, cross the empty space (we know there’s no E in the conductor!) and END on the –’s. If you draw a loop which follows one of those field lines and then “closes” itself in the bulk, the integral of E dot dl would be nonzero. No good! (A more physical argument might be starting with a SOLID sphere. Clearly E=0, no charges anywhere. But, removing the “hole” in the middle would not change ANY charges or fields anywhere. So, the “new” solution for charge distribution and fields when you hollow out the sphere should be the SAME as when it was solid – no charges anywhere in there!!) (The outside will polarize, of course, with more – close to the +q. Charges move to the outside edge of a conductor in e-statics)

R20 A point charge +q is near a neutral copper sphere with a hollow interior space. In equilibrium, the surface charge density σ on the interior wall of the hollow conductor is.. 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. σ = ? +q This is just a “visual hint” to help eliminate option C from the question! [see slide R19]

A HOLLOW copper sphere has total charge +Q.
A point charge +q sits outside. A charge qc is in the hole, right at the center. (As usual, assume static equilibrium.) What is the magnitude of the E-field a distance r from qc, (but, still inside the hole) . |E| = kqc/r2 B) |E| = k(qc-Q)/r2 C) |E| = 0 D) None of these! / it’s hard to compute +q r +qc Class: CONCEPTUAL, MATH/ PHYSICS CORRECT ANSWER: A __________________________________ Skipped in Fall 2011 ======================================= USED IN: Fall 2008 (Dubson) and Spring 2008 (Pollock) LECTURE NUMBER: Dubson (Week 4, Lecture 10). Pollock (Lecture 11). STUDENT RESPONSES: [[92%]] 2% 6% 0% 0% (FALL 2008) [[87%]] 0% 13% 0% 0% (SPRING 2008) In 3320 (Sp ‘11) (Kinney) Lecture 2, [[97]], 0,0,3,0 INSTRUCTOR NOTES: . 87% correct (it started worse, got better after discussion) -SJP WRITTEN BY: Steven Pollock (CU-Boulder) +Q 21

+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 E field look like in the hole? It’s zero in there Simple Coulomb field (straight away from qc, right up to the wall) Simple (but not B) Complicated/ it’s hard to compute +q +qc Class: CONCEPTUAL CORRECT ANSWER: D (changed from the data below, where it used to be “B”) __________________________________ Physics 3320 Fa11 (SJP) Lecture 4 15, 30, 11, [[44]], 0 _________________________________ Fall 2011 Comments Discussion of (non) role of +Q and q outside, of shielding. I invoked “Boundary conditions” to discuss E(parallel)=0 at edge, although my student brought up “E-parallel would drive currents if it was nonzero” LA Notes: 1) After some silence the students attempted to simplify the problem by only considering the conductor with its own charge. They debated for a while without arriving at any definite answer. 2) One student said that in order to be in static equilibrium, more of the + charge on the outside would migrate to the left, outermost part of the shell (because the +q charge is situated closer to the right). ============================================== USED IN: Fall 2008 (Dubson) and Spring 2008 (Pollock) LECTURE NUMBER: Dubson (Week 4, Lecture 10). Pollock (Lecture 11). STUDENT RESPONSES: 81% [[19%]] 0% 0% 0% (FALL 2008) 57% [[43%]] 0% 0% 0% (SPRING 2008) In 3320, Sp ‘11 (Kinney) Lecture #2, 85, [[15]], 0,0,0 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). -SJP WRITTEN BY: Steven Pollock (CU-Boulder) +Q 22

+q +Q +qc 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 -, uniformly spread out B) - close to qc, + 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 +qc Class: CONCEPTUAL CORRECT ANSWER: C ___________________________________ Skipped in Fall 2011 ================================== USED IN: Spring 2009 (Kinney) LECTURE NUMBER: STUDENT RESPONSES: 5, 8, [[82]], 3, 3 INSTRUCTOR NOTES: WRITTEN BY: Ed Kinney (CU-Boulder) The total charge on the inner surface must be –qc, so that Gauss’ law is satisfied for a surface in the bulk. But, how do you know it’s “all –”, and not a mix of – (near) and + (farther?) I think I can use the same argument as several slides ago – if there WERE some +’s, field lines would have to emanate from there, and then “end” on a minus. Follow that field line, then close it through the bulk, and you’ll get a nonzero integral of E dot dl. +Q 23

A) B) C) D) E) None of the above!
R25 Current I flows down a wire (length L) with a square cross section (side a) If it is uniformly distributed over the entire wire area, what is the magnitude of the volume current density? A) B) C) D) E) None of the above! Class: MATH/ PHYSICS CORRECT ANSWER: A _______________________________ Physics 3320, Fa11 (SJP) Lecture 4 [[67]]], 0, 33,0 _______________________________ Fall 2011 Comments (I changed the wrong answers, but didn’t lure anyone) The issue is “what does volume current density” mean, and I used this to get all the relevant formulas (for I, K, and J, in terms of current passing, also in terms of charge density* velocity, and finally UNITS, all on board) LA NOTES: Student saw the words “volume current density” and immediately thought that meant current /volume. It seemed that a few students did not remember or learn this from EM I. ======================================= USED IN: Spring 2008 (Pollock) LECTURE NUMBER: 29 STUDENT RESPONSES: [[87%]] 0% 0% 13% 0% In 3320, Sp ‘11 (Lecture #4) (Kinney) it was [[89]],0,0,11,0 INSTRUCTOR NOTES: 87% on this one. This is too easy! The only confusion was the 2 students who wanted D. (Correct answer is A, pretty trivial) I would skip this - but there WAS a very good conversation that came out of this one, which was about whether current WOULD be uniformly distributed over the area. (The student thought the like charges would repel). So this gave an opportunity to talk about the model of a wire as electrically neutral, but two “fluids” on top of each other (with the electrons mobile). So that in reality this IS a decent picture of charge flow in a wire, I think. -SJP WRITTEN BY: Steven Pollock (CU-Boulder) 25

A) B) C) D) E) None of the above
R26 Current I flows down a wire (length L) with a square cross section (side a) If it is uniformly distributed over the outer surfaces only, what is the magnitude of the surface current density K? A) B) C) D) E) None of the above Class: MATH/ PHYSICS CORRECT ANSWER: C _________________________________ Physics 3320, Fa11 (SJP) Lecture #4 7,4, [[89]], 0, 0 __________________________________ Fall 2011 Comments After previous one, this is quick .Finishes off the “Table” I was writing on the board. ========================================== USED IN: Fall 2008 (Dubson), Spring 2008 (Pollock) LECTURE NUMBER: Dubson (Week 10, Lecture 25), Pollock (29) STUDENT RESPONSES: % 2% [[78%]] 0% 0% (FALL 2008) 0% 7% [[93%]] 0% 0% (SPRING 2008) In 3320, Sp ‘11 (Lecture #4) (Kinney) it was2, 7, [[91]],0,0, INSTRUCTOR NOTES: , 93% correct. No problems (only mistake was person who went for B). Very easy for them (good!) We had some good physics discussion about whether charges WOULD distribute evenly over the outer surface (not on a conductor, I wouldn’t think, for a variety of reasons. But you could “paint” them on the wire and then drag the wire, I suppose). -SJP WRITTEN BY: Steven Pollock (CU-Boulder) To think about: does it seem physically correct to you that charges WOULD distribute evenly over the outer surface? 26

What is B at the point shown?