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AP PHYSICS C ELECTRICITY & MAGNETISM 4 SCORING GUIDELINES General Notes about 4 AP Physics Scoring Guidelines 1. The solutions contain the most common method(s) of solving the free-resonse questions, and the allocation of oints for these solutions. Other methods of solution also receive aroriate credit for correct work.. Generally, double enalty for errors is avoided. For examle, if an incorrect answer to art (a) is correctly substituted into an otherwise correct solution to art (b), full credit will usually be awarded. One excetion to this may be cases when the numerical answer to a later art should be easily recognized as wrong, e.g. a seed faster than the seed of light in vacuum. 3. Imlicit statements of concets normally receive credit. For examle, if use of the equation exressing a articular concet is worth one oint, and a student s solution contains the alication of that equation to the roblem but the student does not write the basic equation, the oint is still awarded. 4. The scoring guidelines tyically show numerical results using the value 1 m s is of course also accetable. g = 9.8 m s, but use of 5. Numerical answers that differ from the ublished answer due to differences in rounding throughout the question tyically receive full credit. The excetion is usually when rounding makes a difference in obtaining a reasonable answer. For examle, suose a solution requires subtracting two numbers that should have five significant figures and that differ starting with the fourth digit (e.g..95 and.78). Rounding to three digits will lose the accuracy required to determine the difference in the numbers, and some credit may be lost. Coyright 4 by College Entrance Examination Board. All rights reserved.
AP PHYSICS C ELECTRICITY & MAGNETISM 4 SCORING GUIDELINES Question 1 15 oints total Distribution of oints (a) i. 3 oints For field lines inside the shell that oint outward and are reasonably close to being 9 to both the line charge and the shell surface, with obviously more lines on the right side than the left For field lines outside the shell that are radial and noncontinuous with those inside the shell, reasonably close to 9 to the shell surface, and aroximately evenly saced For no field lines inside the shell, given that there are field lines drawn elsewhere ii. oints For only negative charges on the inside surface of the shell, with obviously more charges on the right side For only ositive charges on the outside of the shell, aroximately evenly saced Coyright 4 by College Entrance Examination Board. All rights reserved. 3
AP PHYSICS C ELECTRICITY & MAGNETISM 4 SCORING GUIDELINES (b) 4 oints Question 1 (continued) Distribution of oints 4 V a 3 V b V c 1 V d 3 V e For giving the lowest number to V d (i.e. lacing it at the highest otential) For giving V c a higher number than V d (i.e. lacing it at a lower otential) For giving V b and V e the same number For giving the highest number to V a (i.e. lacing it at the lowest otential) (c) i. oints Using Gauss s law: Q Ú Ei d A = ε Q EÚ da = ε For the correct exression for the value of the integral of the area A = r (where is the length of the Gaussian surface) For the correct exression for the total charge enclosed, which comes only from the line charge Q = l Substituting and solving for the field: l E( r ) = ε l E = ε r Coyright 4 by College Entrance Examination Board. All rights reserved. 4
AP PHYSICS C ELECTRICITY & MAGNETISM 4 SCORING GUIDELINES (c) ii. (continued) oints Question 1 (continued) Distribution of oints For any indication that the total field or the total charge enclosed is the sum of contributions from the line charge and the shell Qtot = Qline + Qshell or Etot = Eline + Eshell The field from the line charge is the same as calculated in art i. l Eline = εr For the shell, the charge enclosed is that between r and r 1 Q ( r ) = r - r shell 1 For substitution of the correct enclosed charge for the shell into Gauss s law r Eshell ( r ) = ( r - r1 ε ) ( ) r E = r - r ε shell 1 r ( ) l r E = + r - r ε ε tot 1 r r iii. oints For any indication that the total field or the total charge enclosed is the sum of contributions from the line charge and the shell Qtot = Qline + Qshell or Etot = Eline + Eshell The field from the line charge is the same as calculated in art i. l Eline = εr For the shell, the charge enclosed is that between and r Q ( r ) = r - r shell 1 r 1 For substitution of the correct enclosed charged for the shell into Gauss s law r Eshell ( r ) = ( r - r1 ε ) ( ) r E = r - r ε shell 1 r ( ) l r E = + r - r ε ε tot 1 r r Coyright 4 by College Entrance Examination Board. All rights reserved. 5
AP PHYSICS C ELECTRICITY & MAGNETISM 4 SCORING GUIDELINES Question 15 oints total Distribution of oints (a) Once the switch is closed, V = V. R1 C From the grah, one can see that at t =, V. Therefore, all the voltage dro occurs across resistor R. For the correct answer, with units V = V R C = (b) From the grah, the maximum voltage across the caacitor (and thus also R 1 ), is 1 V. The remaining voltage dro occurs across R. V = V -1 V R For the correct answer, with units V = 8 V R (c) 3 oints A long time after the switch is closed, I =. For any indication that I R1 R Using Ohm s law to calculate the current in R 1 : 3 IR = VR R1 = ( 1 V) ( 15 kw) =.8 1 A 1 1 Using Ohm s law to calculate R : R R C = I -3 ( ) ( ) R = V I = 8 V.8 1 A For the correct numerical answer For the correct units R = 1 kw NOTE: There was a discreancy in this question between information in the grah and the circuit diagram. The value of the time constant as shown in the grah was about a factor of ten too large when comared to the value determined from the information in the diagram. Determination of R by equating the exression for the time constant in terms of resistance and caacitance to the time constant from the grah yields a negative resistance. Since it was not the fault of the students that the correct analysis yielded a negative value of resistance, full credit could be earned for this method. Coyright 4 by College Entrance Examination Board. All rights reserved. 6
AP PHYSICS C ELECTRICITY & MAGNETISM 4 SCORING GUIDELINES (d) 3 oints Question (continued) Distribution of oints Using the relationshi for the energy stored in a caacitor: 1 U = CV For substituting the correct value of voltage For substituting the correct value of caacitance 1 U = ( m F )( 1 V ) For the correct answer, with correct units -3 U = 1.44 1 J (e) 4 oints For a correct value for the initial current From art (a), V V immediately after the switch is closed R = -3 Imax R = ( V) ( 1 kw) = 1 A For a correct value for the final current From art (c), I R -3 =.8 1 A a long time after the switch is closed For a correctly curving grah For labeling the vertical axis (including units) Only the general shae of the grah and the endoints were graded. The time constant of the grah was not evaluated. Coyright 4 by College Entrance Examination Board. All rights reserved. 7
AP PHYSICS C ELECTRICITY & MAGNETISM 4 SCORING GUIDELINES (f) 3 oints Question (continued) Distribution of oints For indicating that the energy would be greater For a correct and comlete exlanation For examle: Consider the circuit a long time after the switch is closed, when there is no current in the caacitor. If R is relaced with a smaller resistance, then the total resistance decreases. This results in a larger current through the resistors. Therefore, the voltage across R 1, and thus across the caacitor, increases. Since U = 1 CV, energy increases. One of the two exlanation oints could be earned for an incomlete but correct exlanation. oints Coyright 4 by College Entrance Examination Board. All rights reserved. 8
AP PHYSICS C ELECTRICITY & MAGNETISM 4 SCORING GUIDELINES Question 3 15 oints total Distribution of oints (a) 4 oints The magnetic field from the wire with current I is given by: mi B = (this equation can be remembered or derived from Amere s law) r Using the equation for the flux: f = Ú Bi d A m For correctly including the dimension 4 of the loo when re-exressing the area integral as an integral over r da = 4 dr For the correct limits on the integral f m 4 4 m m I I dr = Ú 4 dr = r Ú r For correctly erforming the integration f m 4 I m m I = ln r = ln 4 ln For the correct answer mi fm = ln 4 ( - ) (b) 3 oints For correctly indicating that the induced current is counterclockwise For a comlete correct exlanation For examle: The decreasing current means that the magnetic field, which oints out of the lane of the age, is decreasing in magnitude. Using Lenz s law, the induced current thus needs to create a field to counteract this change. The current must therefore be counterclockwise to create a field directed out of the age oints Only one oint was awarded for incomlete exlanations, such as merely saying that Lenz s law or the right-hand rule justifies the answer. Coyright 4 by College Entrance Examination Board. All rights reserved. 9
AP PHYSICS C ELECTRICITY & MAGNETISM 4 SCORING GUIDELINES (c) 4 oints Question 3 (continued) Distribution of oints For a correct relationshi between the current, resistance, and derivative of the flux I = R = ( R)( - df dt) e loo 1 For using the exression for flux from art (a) mi fm = ln 4 For substituting I() t in the flux equation mi -kt fm = ( ln 4 ) e For correctly differentiating this exression to determine the emf df mi -kt e =- =-- ( k) ( ln 4 ) e dt kmi -kt e = ( ln 4 ) e Substituting into the exression for current: kmi -kt Iloo = ( ln 4 ) e R (d) 4 oints For using a correct exression for ower P = I R For substituting the exression for current from art (c) Ê kmi -kt ˆ Ê kmi ˆ 1 -kt P = Á ( ln 4 ) e R = ( ln 4 ) Ë R Á Ë e R For indicating that the dissiated energy is the integral of the ower U = Ú P dt Ê km I ˆ 1 U = Á R ( ln 4 ) Ë Ú -kt For correctly integrating the exression e ˆ ( ln 4) ( ) Ê kmi 1 1 U = Á - e Ë R k Ê kmi ˆ 1 1 U = Á - Ë R k Ê mi ˆ U = Á ( ln 4) Ë dt -kt ( ln 4) ( )( - 1) k R Coyright 4 by College Entrance Examination Board. All rights reserved. 1