APPENDIX A: STUDENT PERFORMANCE ON FINAL CUE QUESTIONS

Transcription

1 APPENDIX A: STUDENT PERFORMANCE ON FINAL CUE QUESTIONS Q# Points Short name Student Performance Average Score (correctness + explanation) Correctness only A B C SD (%) Gain * Q1 5 V of theta Q2 5 Cube Q3 5 Cube far away Q4 5 Images Q5 5 Superposition Q6 5 Current Loop Q7 5 Gauss Q8 5 Delta function Q9 10 Reference for V Q10 10 Sketch E cylinder Q11 6 BC s on E and V Q12 22 Graph E and V of disk A B C D Q13 5 Cartesian BC s Q14 8 Dielectric Q15 5 Circle BC s Q16 5 Multipole Q17 7 Ampere

2 All problems on the CUE, including point allocations and student performance. This data is intended to enable comparison of rubric-reliability on individual items, and to allow instructors administering the CUE to compare student performance on individual items with our published results. Those items with a gain score also appear on the pre-test and are indicated with an asterix (*). Student performance is taken from the N=103 students in the three mostly-traditionally taught classes which used the final form of the exam: Classes A, B and C described in Methods: CUE Administration. Student averages are given as the overall student average (%) in courses A, B and C, sequentially. The next line reports the average student percent correct (in italics) when correctness was able to be coded separately from other considerations such as explanation. This is followed with the average standard deviation for that question across courses (standard deviation was similar for all courses, varying by no more than 6%). The Gain is calculated by subtracting the student s score (%) for that question on the pre-test from their score for that question (%) on the post-test, and then averaging for the N=25 students who took the final form of the pre-test (in Course A), as well as the post-test. Total exam is worth 118 points.

3 APPENDIX B: THE CUE ASSESSMENT --- START ASSESSMENT --- Junior-Level Electrostatics Content Quiz Please fill out the following exam to the best of your ability. This will not count towards your final grade in the course. Do your best to get to all the questions on the test. When we ask you to explain your answer, please keep it brief, but clear. In most cases, the right answer only gets half credit, and the rest of the credit is given for your explanation. When asked to do so, be sure to explain your reasoning. If you don t know an answer, write I don t know (instead of leaving it blank). Name:

4 Page 1 of 8 For each of the following, give a brief outline of the EASIEST method that you would use to solve the problem. Methods used in this class include but are not limited to: Direct Integration, Ampere s Law, Superposition, Gauss Law, Method of Images, Separation of Variables, and Multipole Expansion. DO NOT SOLVE the problem, we just want to know: The general strategy (half credit) Why you chose that method (half credit) EXAMPLE PROBLEM Find the electric field at point P outside a uniformly charged sphere, with total charge +Q. ANSWER Gauss Law with a spherical Gaussian surface centered around the origin. Because the E field is symmetric in theta and phi and so is constant on that surface. Q1. An insulating sphere with radius R, with a voltage on its surface V (θ) = k cos(3θ). Find E (or V) inside the sphere at point P. Q2. A solid, neutral non-conducting cube, centered on the origin, with side length a. It has a charge density that depends on the distance z from the origin, ρ(z) = kz, so that the top of the cube is strongly positive and the bottom is strongly negative, as in the figure. Find E (or V) outside, at point P, on the z-axis, at a distance a from the cube.

5 Page 2 of 8 Q3. The same, neutral non-conducting cube as above, with ρ(z) = kz, but where P is off-axis, at a distance 50a from the cube. Q4. A grounded conducting plane with a point charge Q at a distance a. Find E (or V) at point P. Q5. A charged insulating solid sphere of radius R with a uniform volume charge density ρ 0, with an off-center spherical cavity carved out of it (see Figure). Find E (or V) at point P, a distance 4R from the sphere. Q6. A current loop of radius a that carries a constant current I. Find B (or A) at point P, off-axis, at a distance r=100a. Q7. A solid non-conducting sphere, centered on the origin, with a non-uniform charge density that depends on the distance from the origin, ρ(r) = ρ 0 e r 2 / a 2 where a is a constant. Find E (or V) inside at point P.

6 Page 3 of 8 Q8. A mass density is given by ρ( r ) = m 1 δ 3 ( r r 1 )+ m 2 δ 3 ( r r 2 ), where m 1 and m2 are constants. What is the value of ρ( r)dτ? all space What physical situation does this mass density represent? Q9. You are given a problem involving a non-conducting sphere, centered at the origin. The sphere has a non-uniform, positive and finite volume charge density ρ(r). You notice that another student has set the reference point for V such that V=0 at the center of the sphere: V(r=0)=0. +ρ(r) What would V=0 at r=0 imply about the sign of the potential at r? (a) V (r ) is positive (+) (b) V (r ) is negative (-) (c) V (r ) is zero (d) It depends Briefly explain your reasoning:

7 Page 4 of 8 Q10. You are given an infinite solid conducting cylinder whose vertical axis runs along the y direction, that is placed in an external electric field, E 0ˆ z, as in the figure to the right. The cylinder extends infinitely in the +y and y directions. On the two-dimensional figure below: (a) Sketch the induced charge, σ. (b) Sketch the electric field everywhere. Q11. For the conducting cylinder shown above we want to use the method of separation of variables to solve for: (a) the potential everywhere and (b) the surface charge σ. List the boundary conditions on V and/or E at the surface needed to do this. Do not solve for V, just tell us the boundary conditions on V or E. Boundary conditions:

8 Page 5 of 8 Q12. The following set of problems refer to the uniform flat, infinitely thin disk of radius R carrying uniform positive surface charge density +σ 0 as in the figure. (A) What is the value of the z-component of the electric field (E z ) very near the origin (z<<r)? (B) How does E z behave as a function of z as you get very far from the disk (z>>r)? (C) Draw a qualitative graph of E z as you move away from the disk, along the z-axis. We are looking for the relative magnitude and sign of E z as a function of distance from the disk, not field lines. Include both z>0 and z<0 on your graph. (D) Draw a qualitative graph of V as you move away from the disk, along the z-axis. Include both z>0 and z<0.

9 Page 6 of 8 Q13. You are given a 2-D box with potentials specified on the boundary as indicated in the figure to the right. The general solution to Laplace s equation in Cartesian coordinates is V (x, y) = (Ae kx + Be kx ) (Csin ky + Dcosky) OR V (x, y) = (Ae ky + Be ky ) (Csin kx + Dcoskx). That is, you can choose to associate the sin and cos with either the x or y coordinate. To solve this by separation of variables, which form of the solution should you choose? (a) V (x, y) = (Ae ky + Be ky ) (Csin kx + Dcoskx) (b) V (x, y) = (Ae kx + Be kx ) (Csin ky + Dcosky) (c) it doesn t matter Briefly explain your reasoning (no credit for right answer without reasoning) : Q14. A dielectric is inserted into an isolated infinite parallel plate capacitor, as shown. The dielectric fills the space without quite touching the plates, which are fixed in position. Describe what happens to the dielectric (both in the bulk and at the surfaces) when it is inserted into the capacitor. If a sketch would help, use the space above. In the limit that the dielectric is infinitely polarizable (i.e. χ e ) what would be the limiting values of the charge(s) and the net electric field in the dielectric?

10 Page 7 of 8 Q15. Circle all of the following boundary conditions that are suitable for solving Laplace s equation for finding V(r,θ) everywhere due to a charge density σ on a spherical surface of radius R. (I) V in =V out at r=r (II) E in = E out at r=r (III) E in E out = σ /ε 0 at r=r (IV) E in E out = σ /ε 0 at r=r Q16. You are given the following charge distribution made of 4 point charges, each located a distance a from the x- and y-axis. The dipole moment of this distribution is: (a) Zero (b) Non-zero (c) Not sure Briefly explain your reasoning:

11 Page 8 of 8 Q17. Consider an infinite non-magnetizeable cylinder with a uniform volume current density J. Where is the B field maximum? Explain how to determine this. How seriously did you just take this diagnostic exam? (a) I pretty much blew it off, didn't think much about a lot of the answers. (b) I took it sort of seriously, but when I didn't know an answer I didn't think very hard about it. (c) I took it seriously, and thought about my answers. If you imagine getting a letter grade on the portion of this test that you were able to complete within the time limit, what do you think that grade would be? Any other comments?

12 APPENDIX C: CUE GRADING RUBRIC This document is not to be used for grading the CUE. In order for scores generated using this rubric to be compared to the results discussed above, graders must go through a calibration exercise. The full training packet can be accessed by contacting Steven Pollock at steven.pollock@colorado.edu. Using the rubric: Partial points are given as (+), starting from zero, up to the maximum points for the question. In most cases, (+) points are indicating partial credit assignments for different pieces of the answer. codes enable us to determine what the most common student errors are in a class. There is often a code for a common mistake, but if a student response does not fit one of these common mistakes, there is no need to use a code. When assigning partial points that are not indicated on the rubric, consider how far towards the correct answer the student response is. Out of 3 points, for example, 1 point would be mostly wrong but there is something that is correct,, 2 points would be mostly correct but there is something that is wrong. For all questions, subtract up to 1 point for wrong statements at your discretion. (Eg., for Q7 (pretest Q3) if a student writes Use Gauss Law with Q total = 4/3(πr 3 ρ) you may wish to take up to a point off for that incorrect statement of Q total. When the rubric says that a certain item is not graded (e.g., Q9 correct ) then that item is recorded but does not count towards the final CUE score. Only give credit for what is actually written on the paper! Don t fill in gaps for the students because you know the answer and think you know where they were going with it. Unlike an exam that is for a grade, this diagnostic is intended to determine student understanding, and so it is not necessary to give credit for a students intentions or possible understanding.

13 Q1 (V of theta). 5 points total. Correct Answer 3 points Correct answer is separation of variables. Full credit for Laplace s Equation +2.5 for Legendre polynomials +2 point if implies sep. of var. without being explicit Explanation 2 points Typically give full credit. A integrate or direct integration B Gauss C N/A D E = -grad V Q2 (cube). 5 points total. Pre-test Q1. Correct Answer 3 points Correct answer is direct integration or Coulomb s Law. +1 point if they say that it s a dipole/multipole Explanation 2 points Full credit requires some mention of what the integral would look like or a correct statement of why they chose this method for a poor explanation of how they would go about it (eg., writing down Coulomb s Law). Q3 (cube far away). 5 points total. Correct Answer A N/A B Gauss C Separation of variables D Multipole 3 points Correct answer is multipole expansion using the dipole component. +1 point if say direct integration for dipole only +0.5 for approximation or multipole + 2 for multipole only + 1 for dipole or + 1 for approximation Explanation 2 points Full credit for saying dipole dominates because the observation point is far away. 1.5 points for multipole because r>>a +1 point if said that it s a dipole but give no further explanation +1 point if mention higher order multipoles (but not a dipole) +0.5 for saying the integration is hard because P is off-axis. If they answered direct integration, full credit requires some mention of what the integral would look like or why they chose this method for a poor explanation of how they would go about it (eg., writing down Coulomb s Law). A Integrate or direct integration B Gauss C Separation of variables

14 Q4 (images). 5 points total. Correct 3 points Correct answer is method of images. Explanation 2 points +2 for must match V=0 at boundary or match boundary conditions +1 for opposite charge Q +1 for mentioning that Q must be on the other side of the wall. Look for marks on the illustration as well. A Integrate or direct integration B Gauss C Separation of variables Q5 (superposition). 5 points total. Pre-test Q2 Correct Answer 3 points Correct answer is superposition 0 points for only saying Gauss Law. +1 point for saying integration or dipole +1 point for superposition of charges but not fields (e.g. for 4/3 π (R 3 - r 3 )ρ ο ) 0 for total charge of sphere with cavity Explanation 2 points Full answer is superposition of two oppositely charged spheres and then Gauss Law to solve for E of each sphere. Need to indicate what is being superposed for full credit (e.g., an antisphere of negative charge density). +1 point for stating what is superposed two oppositely charged sphere (+0.5 point if they don t state the spheres are oppositely charged) +1 point for explaining how to solve using the two charged spheres Q6 (current loop). 5 points total. Correct Answer A Integrate or direct integration B Gauss C Separation of variables D Use total Q of sphere 3 points Correct answer is multipole expansion using the dipole approximation. +1 points for saying Biot-Savart or integration for mentioning dipole +0.5 for also mentioning approximation or multipole + 2 for mentioning multipole + 1 for also mentioning dipole or approximation Explanation 2 points Full credit for saying that it looks like a dipole because P is far away. 1.5 points for multipole because r>>a +1 point for only saying dipole (without mention of distance), or vice versa point for mentioning that the integration is hard because P is offaxis If they answered Biot-Savart, full credit requires some mention of what the integral would look like or why they chose this method for a poor explanation of how they would go about it (eg., writing Biot-Savart Law). A Integrate or direct integration or Biot-Savart B Gauss

15 C Separation of variables D - Ampere Q7 (gauss). 5 points total. Pretest Q3 Correct Answer 3 points Correct answer is Gauss Law +1 point for saying direct integration Explanation 2 points Full credit requires some explanation of why (not just how) Gauss Law is used. This would include some mention of the Gaussian surface used or the symmetry (such as charge distribution depends only on r, or E field is radial). +1 point if the correct Gaussian surface is drawn point for explaining how to solve by Gauss s Law. If answer direct integration must give explanation of how they would solve this integral +0.5 for a poor explanation of how they would go about it (eg., writing down Coulomb s Law). A Integrate or direct integration B N/A C Separation of variables Q8 (delta function). 5 points total. Value of integral 3 points Correct answer is m 1 +m points for 4πm or some variant. Physical situation represented 0 points for 4πmr or 4πmr 2, mδ 3 (r-r 1 ) or some variant (wrong units) 2 points Correct answer is two point masses of mass m 1 and m 2. Full credit given for just saying two point masses. 0 points for two spheres or two shells 0.5 for two of something (but not spheres, shells, or masses) 1 for a single point mass 1 point for two masses (without saying that they re points). A Two spheres B Three charges or masses C -- mr or mr 2 or some variant D A and C E single charge or mass

16 Q9 (reference for V). 10 points total. Pretest Q4 Correct answer Not graded. 0 for A, C, or D 1 for B (correct answer) Notes: Some common answers: +6 for V decreases as you move away or increases as you move towards +2 for explicitly mentioning positive charge +1 for also having the answer right There are three main approaches to this problem: Potential, work, and energy. We break the answers into those three approaches. In all cases subtract -1 for each completely wrong statement (up to 2) POTENTIAL APPROACH A) The physics of positive charges 4 points Since ρ is positive: ΔV is positive OR V(r=0) > V( ) +2 if don t explicitly state that this is because the charge is positive. B) Definition of potential difference 4 points ΔV is defined as: V(ref)-V(r) OR V(r=0) V( ) OR Vf-Vi OR V decreases as you move away OR V(r=0) > V( ) (nb. Statement gives points for A and B). +2 for noting that usually V( ) =0 and it is shifted. C) Logic 2 points And since V(r=0)=0 THEN V( ) = XXX. (Students get credit for logic if a sign error is introduced from A or B). +1 for having correct answer without explicitly stating logic. WORK/ENERGY/FORCE APPROACH A) The physics of positive charges B) Definition of potential difference and connection to work 4 points Since ρ is positive, it takes positive work to bring a positive charge from infinit OR that a test charge would be repelled. +2 if don t explicitly state that this is because the charge is positive. 4 points It takes positive work to bring positive charge from low to high V. If they state W=QΔV they must make it clear what ΔV is defined as, or what direction you re moving (infinity to origin) in going through ΔV. OR V decreases as you move away +2 for noting that usually V( ) =0 and it is shifted C) Logic 2 points Since V(r)=0=0 then positive work implies V( )<V(r=0) and negative numbers are less than zero. +1 for having correct answer without explicitly stating logic. ELECTRIC FIELD APPROACH A) The physics of positive charges 4 points E field lines radiate out (no need to explicitly say positive charges) OR E goes from high to low V B) 4 points where dl is directed towards sphere AND E field lines Definition of potential radiate out (nb. Students get credit for this statement in A and B). difference OR E goes from high to low V (full credit for B) C) Logic 2 points And since V(r=0)=0 then V( ) is negative. +1 for having correct answer without explicitly stating logic. A, C, D chose answer A, C or D as appropriate

17 Q10 (sketch E cylinder). 10 points. Pretest Q5 Charge distribution 3 points +1.5 for correct sign (positive on top) +1 for charge residing on surface only (Q=0 inside cylinder) +0.5 for nonuniform distribution with charge concentrated at the poles E inside cylinder E outside cylinder (give full credit unless diagram clearly shows a uniform distribution) 3 points Full credit given if E=0 inside (whether explicitly stated, or drawn with no field lines inside). No credit if E=0 everywhere. 4 points See below for grading rubric for common answers. +1 for an E that is approximately perpendicular to surface (or at least not not perpendicular) +1 for going to E 0 far away from cylinder +2 for field lines starting on + and ending on charges -1 for no evidence of E 0 ( see B and E ) 0 if the field looks like E 0 everywhere outside A 1 pt - E is expelled like a magnetic field B 2 pt - looks like dipole (as if external E not present) C 0 pt - E constant everywhere outside D 0 pt - E constant everywhere inside and outside E 0 pts - E radial outwards (E out : 4 pts) (E out 1 pt) (E out 2 pts) (E out 0 pts) (E out 0 pts) (E out 0 pts)

18 Q11. 6 points. (BC s on E and V) V 3 points Full credit for V(r=R)=constant or V(r=R)=0 or V in =V out = 0 (or constant) for V=0 for r<r +1.5 for V in =V out (with no mention of value) Note V below and V above are satisfactory equivalents for V in and V out. Continuity of E parallel is satisfactory equivalent. +1 for E perp is continuous E 3 points Full credit for E out - E in = +σ/ε 0 or equivalent (such as V out / n V in / n = σ /ε 0 or discontinuity in E perpendicular ) for only mentioning E=0 inside +1 for mentioning E=0 inside and mentioning discontinuity at the surface +1.5 for just writing V / r = σ /ε 0 +1 for E parallel is discontinuous -0.5 for a sign error Q12 (graph E and V of disk). 22 points. Pretest Q6 A gave a value for E instead of boundary condition. B said that E is continuous or E in =E out C said that the derivative of E is discontinuous or V is discontinuous D only gave one condition Q12 Part A. 2 points. A Value of E near disk 2 point Full credit for σ/2ε 0. (or 2pσ) +1 for σ/ε 0. (or pσ) for the same E as an infinite plane +0.5 for E = constant or E=0 Q12 Part B. 2 points. B Behavior of E near infinity 2 points Full credit for kq/r 2 or 1/r 2 or 1/z 2 +1 for point charge 0 for goes to zero Q12 Part C (graph E of disk). 9 points. If a student draws field lines for Q12C or Q12D, grade this question based on those field lines to the best of your ability. Antisymme try of E 3 points Full credit if E is not symmetric (ie., E is negative for negative z) field Behavior of E at infinity Behavior of E at origin 3 points Full credit if it drops towards E=0 far away with a behavior that looks like 1/z for any increase in E while moving away from disk (as in C below) 3 points Full credit for a finite value that is not zero or infinity (though the full answer is σ/2ε 0 ). The graph does not need to look flat at the top (mathematica shows that the range of z in which E is constant is very small) -1 for a graph that has incorrect signs (a negative value for Ez>0 and a

19 positive value for Ez <1. This problem is easiest to grade based on the way that the students graph appears. Below are mistake codes which correspond to different graphing mistakes. Following each code is the correct point assignment. For example, (0,3,0) represents 0 for antisymmetry, 3 for infinity, and 0 for origin. A Point charge (0,3,0) B Symmetric disk (0,3,3) C Two humps (3,2,0) but (3,1,0) if shows E=0 at a finite value of z D One hump (0,3,3) E Antisymmetric point charge (3,3,0) F Antisymmetric disk (3,3,3). Same points & code for a sketch that looks like D but antisymmetric about the origin. G Only graphed + z range. A. Point charge (0,3,0). B. Symmetric disk (0,3,3). C. Two humps (3,2,0). Give (3,1,0) if E=0 before z reaches infinity, as drawn. D. One hump (0,3,3). If antisymmetric about origin, grade and code as F. E. Antisymmetric point charge (3,3,0). F. Antisymmetric disk (3,3,3). Correct answer. Give (3,3,2) for this sketch inverted (negative disk). Q12 Part D (graph V of disk). 9 points. Symmetry 3 points V should be symmetric about the origin of V Behavior of V at infinity 3 points V should drop to zero, looking like 1/r (or explanation includes mention of 1/r). -1 if looks linear everywhere Behavior of V at origin -1 if really looks like 1/r 2 (comparing to their E field) 3 points Full credit for a kink at z=0. +2 for finite, nonzero value at z=0 (but no kink). Zero value OK if

20 Consistent with E going to finite value at infinity. 0 points if V is constant everywhere. -2 for a negative value or negative infinity. Negative value OK if decreasing towards a finite negative value at infinity. 0 points 0 points if consistent; -1 point for an inconsistency in the symmetry This problem is easiest to grade based on the way that the students graph appears. Below are mistake codes corresponding to different graphing mistakes. Following each code is the correct point assignment. For example, (0,3,0) represents 0 for antisymmetry, 3 for infinity, and 0 for origin. Be sure to subtract a point from symmetry and origin if it s inconsistent with E. Sketches on next page. A Inconsistent with E B Same as E C One hump (3,3,2) D Disk (correct answer) (3,3,3) E Point charge (3,3,0) F Linear point charge (3,2,3) G Antisymmetric; (0,3,3) for finite value at z=0; (0,3,0) for infinite value at z=0 H Bowl (3,0,0) C. One hump (3,3,2). Consistent with E-Field C. D. Disk (3,3,3). Correct answer. Same credit for this graph shifted up or down. Consistent with E-field F. Give (3,3,1) for this sketch inverted (negative disk). E. Point charge (3,3,0). Consistent with E-field E F. Linear point charge (3,2,3). Consistent with E-field F. G. Antisymmetric (0,3,0). Inconsistent with all E shown. Consistent with an E-field A that was negative everywhere. Give (0,3,3) if V is not infinite at origin. H. Bowl (3,0,0).

21 Q13 (Cartesian BC s). 5 points. Correct answer Not graded 0 for B or C 1 for A (correct answer) Score for problem 5 points Must have correct answer and reasoning for full credit. Correct reasoning includes saying that sin=0 at two places and this matches the boundary conditions. Must also include the fact that the exponential solutions will not match the boundary conditions. The pieces of a correct answer are: +4 that sin/cos are zero in two places, which matches boundary conditions +1 for exponentials don t match the boundary conditions Some common wrong answers are: +1 if only say that sin(x=0)=0 +1 for right answer with some reasoning, but poor reasoning +1 for D=0 with sin/cos +1 For stating that the boundary conditions in the x direction are homogenous. +2 for easier/convenient to match x to sin/cos, or periodic in x 0 point for restating boundary conditions Q14 (Dielectric). 8 points. Describe what happens to the dielectric. 4 points Full credit requires that they mention that the dielectric polarizes, giving rise to bound charge on the surfaces. Polarization (max +2, min 0): +2 for either saying the dielectric polarizes or correctly describing the microscopic process of polarization or for a drawing clearly illustrating polarization Bound charge (max +2, min 0): +2 for bound charge on the surfaces -0.5 for describing bound charge of the wrong sign, -1 for indicating that mobile charge are free to move (or just -0.5 if fail to say bound ) +1.5 for drawing with charges shown on top and bottom surfaces w/o specifying they are bound Limiting values of charges and net E-field +/- 0 pts for indicating bound volume charge (can t be determined) 4 points +2 for E=0 inside (+0.5 for E decreases without limiting value) +2 for bound charge is equal to sigma (+0.5 for sigma increases ) +1 for writing only same as in conductor A E increases in dielectric B bound surface charge goes to infinity C Draw charges but don t specify they are bound

22 Q15 (Circle BC) Score 5 points +2.5 points for a correct answer (I or III) -2.5 points for an incorrect answer (II or IV) no + or for one of these two without clear choice (Minimum zero points) Q16 (Multipole). 5 points. Correct Not graded multiple choice A chose II (plus one correct answer) B chose IV (plus one correct answer) C chose II and IV 0 non-zero dipole moment 1 dipole moment is zero (correct answer) Score 5 points A complete answer is that there are two or four oppositely directed dipoles (+2.5) and the addition of the two dipole moments gives zero (+2.5). (The fact that it s a quadrupole does not prove the dipole moment is zero). Also full credit for stating that it is a pure quadrupole and so that is the only term in the multipole expansion. Partial points for the following wrong answers: +2.5 for only saying it s two opposite dipoles or the superposition of two dipoles +0 for Q=0 +1 for saying it s a quadrupole (with no other explanation) +3 for saying that the quadrupole is the first term in the multipole expansion and thus the dipole moment is zero +1 for saying something about the sum of charge and distance (without explicitly saying that the dipole moment is zero by this calculation) s A Quadrupole B No net charge Q17 (Ampere). 7 points. Pretest Q7 Correct 2 points Full credit for saying that B is maximum at the edge of the cylinder Reason 5 points A full explanation is that I through is maximum at the radius, and B dl gets larger as you move away from the cylinder. Full credit for showing graph of B for all r (linear inside cylinder and drops as 1/r outside, continuous at surface). +3 for I through is maximum +2 for B dl increases as you move away +1 only if they just say that it goes as 1/r outside or simply invoke the right hand rule +1 for only mentioning or writing Ampere s Law +1-2 for saying surface with a poor explanation A B decreases with distance B right hand rule C- B max in center of wire