207_test3key_spr11

207_test3key_spr11 - Dr. Huerta Phy 207 Test 3 ANSWER KEY...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Dr. Huerta Phy 207 Test 3 ANSWER KEY April 20 2011 Section HJ, MWF 3:35 – 4:25 p. m. Signature: Name: 1 2 3 4 5 ID number: DO THE TWO MULTIPLE CHOICE PROBLEMS [1.] AND [2.] AND DO TWO OUT OF PROBLEMS [3.], [4.], [5.] CROSS OUT THE BOX OF THE PROBLEM YOU WILL NOT DO Your signature signifies that you will obey the HONOR CODE You may be asked to show your photo ID during the exam. PUT AN X NEXT TO YOUR DISCUSSION SECTION: [ ] Dr. Mezincescu 1O, T 9:30 - 10:20 p.m. [ ] Dr. Alvarez 1Q, T 12:30 - 1:20 p.m. [ ] Dr. Huerta 1R, T 2:00 - 2 :50 p.m. You may be asked to show your photo ID during the exam. FORMULAE SHEET Z dx x 2 + a 2 = ln( x + p x 2 + a 2 ) , Z xdx ( x 2 + a 2 ) 3 / 2 = - 1 ( x 2 + a 2 ) 1 / 2 , Z dx ( x 2 + a 2 ) 3 / 2 = x a 2 ( x 2 + a 2 ) 1 / 2 e = 1 . 6 × 10 - 19 C, m e = 9 . 1 × 10 - 31 kg, m p = 1 . 67 × 10 - 27 kg, k = 1 4 π± 0 = 9 × 10 9 N · m 2 C 2 . Coulomb’s Law, Electric Fields (N/C), Electric Field Flux Φ , and Gauss’ Law ± ± ~ F ± ± = ± ± q 1 q 2 ± ± 4 π± 0 r 2 ~ F q = q ~ E, | ~ E Q | = kQ r 2 , Φ closed surface = H ~ E · d ~ A = q inside ± 0 Electric Field and sheets of surface charge density σ : On each side of an infinite sheet E n = σ 2 ± 0 . However, outside the surface of a conductor, E n = σ ± 0 . Force, Potential Energy and Torque for an Electric Dipole The Force, torque, and potential energy of an electric dipole ~ p immersed in uniform electric field ~ E are ~ F total = 0 , U = - ~ p · ~ E, ~ τ = ~ p × ~ E. Harmonic Oscillator: F x = - kx , ω = p k/m Electric Potential and Energy : W agent = ( K + U ) f - ( K + U ) i , W field = U i - U f U q ( ~ r ) = qV ( ~ r ) , V i - V f = R f i ~ E · d~s, ~ E = - ~ V = - ∂V ∂x ˆ ı - ∂V ∂y ˆ - ∂V ∂z ˆ k, E r = - ∂V ∂r Capacitance Q = CV, C = κ± o A d , series 1 C eq = 1 C 1 + 1 C 2 , parallel C eq = C 1 + C 2 Physics 207 TEST 1 April 20 2011
Background image of page 1

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

View Full DocumentRight Arrow Icon
Dr. Huerta Phy 207 Test 3 ANSWER KEY April 20 2011 Circuits i = R ~ J · ˆ ndA, ~ J = q + n + v D + + q - n - v D - = σ ~ E, σ = 1 /ρ, V = iR, R = ρL/A Resistors : in series R eq = R 1 + R 2 , in parallel 1 R eq = 1 R 1 + 1 R 2 Kirchhoff’s rules : (1) at a junction, current in equals current out, (2) sum of voltage rises around a loop equals zero. Magnetic forces : ~ F = q~v × ~ B, R = mv qB , ~ F = Z i ~ ds × B, ~ τ = μ × ~ B. Gauss’ Law for ~ B , Biot-Savart and Ampere’s laws : I S ~ B · ndA = 0 , d ~ B = μ 0 i 4 π d~s × ~ r r 3 , I C ~ B · ~ ds = μ 0 i in Faraday’s Law : Φ B = Z ~ B · ˆ ndA, E = - d Φ B dt E = I ( ~ E + ~u × ~ B ) · ~ ds, E = - L d i dt , E 2 = - M di 1 dt Self inductance : A coil of length , and cross sectional area A , with n turns per unit length has a self inductance L = μ 0 n 2 ‘A . R-L circuit
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 8

207_test3key_spr11 - Dr. Huerta Phy 207 Test 3 ANSWER KEY...

This preview shows document pages 1 - 3. Sign up to view the full document.

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