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AP® Physics C: Electricity and Magnetism 2015 Scoring Guidelines © 2015 The College Board. College Board, Advanced Placement Program, AP, AP Central, and the acorn logo are registered trademarks of the College Board. Visit the College Board on the Web: www.collegeboard.org. AP Central is the official online home for the AP Program: apcentral.collegeboard.org. AP® PHYSICS C: ELECTRICITY AND MAGNETISM 2015 SCORING GUIDELINES Question 1 15 points total Distribution of points (a) i. 1 point For at least one arrow between the plates pointing downward from the positive plates toward the negative plate and no extraneous arrows pointing in any other direction ii. 1 point For drawing an appropriate Gaussian surface (enclosing at least the inner edge of one of the plates) that can be used to determine the electric field between the plates iii. 1 point 1 point 3 points For using a correct statement of Gauss’s law Q Ú E dA = 1 point For applying Gauss’s law, using an enclosed charge and surface area consistent with the surface drawn in part (a)-ii q E ( AGS ) = enc ( AGS is the area of the end of the Gaussian surface between the 1 point e0 e0 plates) qenc qenc ˆ Ê E = e0 AGS Ë using s = AGS ¯ Q s E = e0 using s = A For a correct answer with work shown Q E = e0 A ( ) © 2015 The College Board. Visit the College Board on the Web: www.collegeboard.org. 1 point AP® PHYSICS C: ELECTRICITY AND MAGNETISM 2015 SCORING GUIDELINES Question 1 (continued) Distribution of points (b) 1 point Comparing the equation for electric field between parallel plates to the given equation: Q Q = E = ke0 A e0k0e- x D A For an answer consistent with part (a)-iii k = k0 e - x D 1 point (c) i. 1 point Using the equation relating the electric field to potential difference dV E =dx For a correct differential equation Ê ˆ Q dV = -Áx D dx Ë eke A ¯˜ 1 point 0 0 Q e0k0e - x D A Alternate Solution: Using the equation relating the electric field to potential difference: DV = - Ú E dr dV = dx For a correct differential equation Q DV = - dx e0k0e- x D A Q DV = dx e0k0e - x D A Ú Ú © 2015 The College Board. Visit the College Board on the Web: www.collegeboard.org. Alternate Point 1 point AP® PHYSICS C: ELECTRICITY AND MAGNETISM 2015 SCORING GUIDELINES Question 1 (continued) Distribution of points (c) (continued) 4 points ii. Separating the variables in the differential equation from part (c)(i): Q dV = dx e0k0e - x D A Q ˆ xD dV = Ê e dx Ë e0k0 A ¯ For using the correct limits of integration in attempting to integrate the equation above VD Ú V0 1 point D Q ˆ xD dV = Ê e dx Ë e0k0 A ¯ Ú 0 For correctly integrating the equation D Q [V ]VV0D = ÊË e k A ˆ¯ ÈÎ De x D ˘˚0 0 0 1 point QD (VD - V0 ) = ÊË e k A ˆ¯ (e D D - e0 ) 0 0 For an expression that gives the correct absolute value of the potential difference between the plates For having the potential difference be positive QD ˆ DV = Ê (e - 1) Ë e0k0 A ¯ 1.72QD DV = e0k0 A (d) 1 point 1 point 1 point Using the equation for capacitance: Q Q C = = DV QD Ê ˆ e -1 ( ) Ë e0k0 A ¯ For an answer consistent with part (c)-ii e0k0 A C = D ( e - 1) ek A C = 0 0 1.72 D © 2015 The College Board. Visit the College Board on the Web: www.collegeboard.org. 1 point AP® PHYSICS C: ELECTRICITY AND MAGNETISM 2015 SCORING GUIDELINES Question 1 (continued) Distribution of points (e) 3 points For selecting UV > UC For correctly comparing the capacitance or the potential difference with the varying dielectric constant to the capacitance or the potential difference with the uniform dielectric constant For correctly comparing the two stored energies consistent with the comparison of the capacitances or potential differences Q2 Example: According to the equation from part (d), CC > CV . Since U = , if the 2C charge stored on the two capacitors is the same, then UV > UC . © 2015 The College Board. Visit the College Board on the Web: www.collegeboard.org. 1 point 1 point 1 point AP® PHYSICS C: ELECTRICITY AND MAGNETISM 2015 SCORING GUIDELINES Question 2 15 points total Distribution of points (a) i. 2 points Using Ohm’s law: V = IR For a correct application of Kirchhoff’s loop rule e = Ir + IR I = e (r + R) For a correct expression for the measured voltage across the variable resistor V ii. = 1 point e (r + R ) 1 point R 1 point For an expression of 1 V as a function of 1 R consistent with answer from part (a)(i) 1 = er R1 + e1 V 1 point ( ) (b) 4 points For correctly labeling both axes with variables and units For correctly scaling both axes with an acceptable and appropriate scale For correctly plotting the data points For correctly drawing a straight line that best represents the data © 2015 The College Board. Visit the College Board on the Web: www.collegeboard.org. 1 point 1 point 1 point 1 point AP® PHYSICS C: ELECTRICITY AND MAGNETISM 2015 SCORING GUIDELINES Question 2 (continued) Distribution of points (c) i. 2 points For using a value for the y-intercept consistent with the straight line drawn in part (b) y = mx + b b = 0.080 1 V For a correct substitution into the equation from part (a)(ii) (r + R) 1 1 = V e R 1 r 1 = e R + e1 V b=1e e = 1 (0.080 1 V) e = 12.5 V ii. 1 point 1 point 2 points For calculating the slope using the straight line from part (b) and not data points y - y1 (0.151 - 0.100) m= 2 = = 0.0463 W V x2 - x1 (1.50 - 0.40) For a correct substitution into equation from part (a)(ii) r m = 1 point 1 point e r = me r = (0.0463 W V )(12.5 V ) r = 0.58 W (d) 2 points For using Ohm’s law e = Ir For a correct substitution of values from part (c): (12.5 V ) = I (0.58 W) I = 21.6 A (e) 1 point 1 point 2 points For selecting “The voltmeter with high resistance” For a correct justification Example: The voltmeter acts like a resistor in a circuit with the battery. It will measure the potential difference across its own internal resistance. The higher its internal resistance, the closer its potential difference will be to the emf of the battery. © 2015 The College Board. Visit the College Board on the Web: www.collegeboard.org. 1 point 1 point AP® PHYSICS C: ELECTRICITY AND MAGNETISM 2015 SCORING GUIDELINES Question 3 15 points total (a) Distribution of points 3 points For properly using a correct equation to calculate the magnetic flux F B = B dA 1 point For correct substitution of the area and trigonometric function 1 point Ú FB = BA(cos q ) = Bpr (cos q ) 2 For correct substitution into the above equation 1 point F B = 4 (1 - 0.2t ) p (0.10 m ) (cos60∞) F B = 0.063 - 0.013t (or in terms of a and b, F B = a (1 - bt )(0.016) ) 2 (b) 2 points For using a correct equation to solve for the emf FB e = - d dt Substituting the expression from part (a): e = dtd (0.063 - 0.013t ) For an answer consistent with part (a) e = 0.013 V Note: Any sign on the answer is ignored. 1 point 1 point (c) i. 1 point For a substitution into Ohm’s law consistent with the answer from part (b) V = IR (0.013 V ) I = (50 W) 1 point I = 2.6 ¥ 10-4 A ii. 2 points The correct choice is “Counterclockwise”. For a justification that incorporates that the original magnetic field is changing For a justification that correctly relates the induced current to the direction of a new magnetic field created by that current Example: Looking down at the loop from P, the vertical component of the magnetic field of the loop is upward and decreasing. To oppose this change, the current in the loop must create a magnetic field that is directed upward at point P. This requires a counterclockwise current in the loop. © 2015 The College Board. Visit the College Board on the Web: www.collegeboard.org. 1 point 1 point AP® PHYSICS C: ELECTRICITY AND MAGNETISM 2015 SCORING GUIDELINES Question 3 (continued) Distribution of points (d) 2 points For using a correct equation to calculate the energy dissipated E = Pt = I 2 Rt or E = Pt = e 2 R t 1 point For a substitution into either of the above equations with the answer from part (c)(i) or part (b) 1 point ( ( ) 2 E = 2.6 ¥ 10-4 A (50 W )( 4.0 s) ) or E = ( 0.013 V)2 (4.0 s) (50 W) E = 1.4 ¥ 10-5 J (e) 4 points For selecting both “Zero” for the net magnetic force and “Nonzero” for the Net magnetic torque For indicating the forces on directly opposite sides of the loop are in opposite directions For concluding that the forces cancel & net force is zero For indicating that the torques add & net torque is nonzero Example: Since the current changes direction relative to the magnetic field as you go around the loop, the magnetic field will exert force of equal magnitude but opposite direction on opposite sides of the loop. These forces all cancel out resulting in zero net force. However, since this force is up on one side of the loop and down on the other side of the loop, this will create torques that rotate the loop in the same direction. Therefore, the net torque is not zero. Units 1 point 1 point 1 point 1 point 1 point For correct units on at least two parts with a calculated numerical answer and no incorrect units © 2015 The College Board. Visit the College Board on the Web: www.collegeboard.org. 1 point