Exam1_2006Spr_Solutions

Exam1_2006Spr_Solutions - Physics 8.02 Exam One Solutions...

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

View Full Document Right Arrow Icon
Physics 8.02 Exam One Solutions Spring 2006 Please Remove this Tear Sheet from Your Exam 23 ˆ 44 oo qq rr πε == Er G G r ˆ points source q observer r r r= from to G inside-free o closed surface Q κ ε ⋅= ∫∫ EdA G G w points from inside to outside dA G q = FE GG = = b a a b b to a from moving V V V s d E G G 0 = path closed s d E G G WU q =∆ = ∆ V point charge 4 o q V r = = pairs all j i o j i q q U r r G G 4 2 1 2 ov UE ⎡⎤ = ⎢⎥ ⎣⎦ ∫∫∫ x o l d V E x = - V x , E y = - V y , E z = - V z r V E r =− for spherical symmetry Q C V = U = 1 2 C V 2 = Q 2 2C C 12 parallel CC =+ series C = + Circumferences, Areas, Volumes: Circle of radius r : Area = π r 2 Circumference = 2 π r Sphere of radius r Surface Area = 4 r 2 Volume = 3 4 3 r Cylinder of radius r and height h Side surface area = 2 r h End cap surface area = 2 r 2 Volume = r 2 h Definition of trig functions sin is opposite/hypotenuse; cos is adjacent/hypotenuse; tangent is opposite over adjacent; Properties of 30, 45, and 60 degrees ( π /6, π /4, and π /3 radians) : sin( π /6) = cos( π /3) = 1/2, sin( π /3) = cos( π /6) = 2 / 3 ; sin( π /4) = cos( π /4) = 2 / 1 ; Integrals that may be useful a b dr b a = ) / ln( 1 a b dr r b a = = b a dr r b a 1 1 1 2 Some potentially useful numbers 2 9 2 0 Breakdown of air E ~ 3 x 10 6 1N m 91 0 4C e k × V/m Speed of light c = 3 x 10 8 m s / Electron charge e = 1.6 x 10 -19 C Avagadro’s number N A = 6.02 x 10 23 mol -1
Background image of page 1

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

View Full DocumentRight Arrow Icon
MIT PHYSICS DEPARTMENT page 2 of 14 8.02 Exam One Spring 2006 FAMILY (last) NAME GIVEN (first) NAME Student ID Number Your Section: ___L01 MW 9 am ____L02 MW 11 am ____L03 MW 1 pm ____L04 MW 3 pm ___L05 TR 9 am ____L06 TR 11 am ____L07 TR 1 pm ____L08 TR 3 pm Your Group (e.g. 10A): ______________ Problem Score Grader 1 (20 points) 2 (5 points) 3 (25 points) 4 (25 points) 5 (25 points) TOTAL 8.02 Exam #1 Solutions Spring 2006
Background image of page 2
MIT PHYSICS DEPARTMENT page 3 of 14 Problem 1: Five Short Questions. Circle your choice for the correct answer. Question A (4 points out of 20 points): A negative charge –q sits at the origin. What is the electric flux generated by this charge through a square area (side length 2 a ) centered on the x-axis at x = - a . That is, for the picture at right, calculate: GG ˆ SS dd A = ∫∫ En E Φ= EA G 1. 0 6 E q ε 2. 0 4 E q 3. 0 E q 4. 0 4 E q 5. 0 6 E q 6. 0 E 7. 0 6 E q Φ=− 8. 0 4 E q 9. 0 E q 10. 0 4 E q 11. 0 6 E q 12. I don’t know (this answer is worth 1 point) Question B (4 points out of 20 points): Consider two electric dipoles, each consisting of two equal and opposite point charges at the ends of an insulating rod of length d . The dipoles sit along the x- axis a distance r apart, oriented as shown below. Their finite separation r >> d > 0 At the instant pictured, the dipole ON THE LEFT will feel: 1. a force to the left and a torque trying to make it rotate clockwise 2. a force to the left and a torque trying to make it rotate counterclockwise 3. a force to the left and no torque 4. a force to the right and a torque trying to make it rotate clockwise.
Background image of page 3

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

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 14

Exam1_2006Spr_Solutions - Physics 8.02 Exam One Solutions...

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