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Unformatted text preview: Physics 8.02 Final Exam Solutions Spring 2005 Some (possibly useful) Relations: enclosed o closed surface Q d ε ⋅ = ∫∫ E A G G w total closed open path surface external self open surface d d d dt d d dt ⋅ = − ⋅ ⎡ ⎤ = − + ⋅ ⎣ ⎦ ∫ ∫∫ ∫∫ E s B A B B G G G G G G G v A closed surface d ⋅ = ∫∫ B A G G w closed path thru o E E open surface d d I dt where d µ µ ε ⋅ = + Φ Φ = ⋅ ∫ ∫∫ o o B s E A G G G G v 2 1 2 elec u E = ε 2 1 2 mag u B = µ 2 ˆ 4 µ π × = = ∫ ∫ s r B B source source Id d r G G G q q q = + × F E v G G G B G B d I d = × F s G G G 1 ˆ 4 2 dq r πε = dE r G 2 1 ˆ 4 V dq r πε = ∫ E r G 2 1 2 1 P 2 1 P The electric potential at point P minus that at point P is given by V( P ) V( P ) d − = − ⋅ ∫ E s G G i N i i 1 i, q 1 V 4 r πε = = ∆ = P ∑ V IR ∆ = 2 2 Joule Heating P I R V = = ∆ / R Q C V = ∆ 2 2 capacitor 1 2 2 Q U C V C = ∆ = self field self one turn N d ⋅ = ∫∫ B A L I G G 2 inductor 1 2 U L = I 1 LC = ω 0 0 1 c = µ ε 1 f T = 2 2 f T ω = π = π 2 k = π λ c T f k = λ = λ = ω 1 = × µ S E G B G G DOUBLE SLIT: constructive: sin , 0, 1, 2, 3, ... d m m θ = λ = ± ± ± destructive: 1 sin , 0, 1, 2, 3, .. 2 d m m ⎛ ⎞ θ = + λ = ± ± ± ⎜ ⎟ ⎝ ⎠ SINGLE SLIT: destructive: sin , 1, 2, 3, ... a n n θ = λ = ± ± ± Areas, Volumes, etc.: 1) The area of a circle of radius r is π r 2 Its circumference is 2 π r 2) The surface area of a sphere of radius r is 4 π r 2 . Its volume is 3 3 r 4 π 3) The area of the sides of a cylinder of radius r and height h is 2 π r h. Its volume is π r 2 h USEFUL INTEGRALS: d c dx d c = − ∫ ( ) 2 2 1 2 d c r dr d c = − ∫ 1 ln d c d dr r c ⎡ ⎤ = ⎢ ⎥ ⎣ ⎦ ∫ 2 1 1 d c dr r c 1 d ⎡ ⎤ ⎛ = − ⎜ ⎟ ⎢ ⎥ ⎞ ⎣ ⎦ ⎝ ∫ ⎠ MIT PHYSICS DEPARTMENT ` page 2 8.02 Final Exam Spring 2005 FAMILY (last) NAME GIVEN (first) NAME Student ID Number Your Section (check one): ___MW 10 am ____MW 12 noon ____MW 2 pm ___TTh 10 am ____TTh 12 noon ____TTh 2 pm Your Group (e.g. 10A): _________ Problem Score Grader 1 (40 points) 2 (40 points) 3 (40 points) 4 (40 points) 5 (40 points) 6 (40 points) TOTAL MIT PHYSICS DEPARTMENT ` page 3 Problem 1: Eight Conceptual Questions (40 points out of 240 total). Circle your choice for the correct answer to the question. Question A (5 points): Consider a cubic volume with sides of length a oriented as shown in the figure. An electric field fills the space both inside and outside the cube and this electric field everywhere points in the +x direction. The magnitude o the field is independent of y and z but varies with x . The value of the electric field at x = 0 is f 2 3 ˆ / x o Volts meter a ε = = E i and the value of the electric field at x = a is 2 7 ˆ / x o Volts meter a ε = = E i . The total charge contained inside the cube is (a) 10 Coulombs (b) 4 Coulombs (c) 0 Coulombs (d) +4 Coulombs CORRECT (e) +7 Coulombs (f) Not enough information given to answer...
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This note was uploaded on 04/07/2008 for the course 8 8.02 taught by Professor Hudson during the Spring '07 term at MIT.
 Spring '07
 Hudson

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