Final_2005Spring_Solutions

Final_2005Spring_Solutions - Physics 8.02 Final Exam...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

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.

Page1 / 19

Final_2005Spring_Solutions - Physics 8.02 Final Exam...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online