Sp06-Final-Smoot-Exa - Department of Physics University of California Berkeley Final Examination Physics 7B Lecture 1 Prof Smoot 8 AM 11 AM 12 May

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Unformatted text preview: Department of Physics, University of California, Berkeley Final Examination Physics 7B, Lecture 1, Prof. Smoot 8 AM - 11 AM, 12 May 2006 Name: SID No: Discussion Section: Name of TA: Problem 1 Problem 2 Problem 3 Problem 4 Problem 5 Problem 6 Problem 7 Problem 8 Score: 1 ∇ V = ∂V ∂x ˆx + ∂V ∂y ˆy + ∂V ∂z ˆ z ∇ V = ∂V ∂r ˆ r + 1 r ∂V ∂φ ˆ φ + ∂V ∂z ˆ z ∇ V = ∂V ∂r ˆ r + 1 r ∂V ∂θ ˆ θ + 1 rsinθ ∂V ∂φ ˆ φ Some useful Integrals: dx √ x 2 + a 2 = log x + √ x 2 + a 2 or sinh- 1 x a dx x 2 + a 2 = 1 a arctan ( x a ) dx ( x 2 + a 2 ) 3 / 2 = x a 2 √ x 2 + a 2 2 Answer all eight problems. Write clearly and explain your work. Partial credit will be given for incomplete solutions provided your logic is reasonable and clear. Cross out any parts that you don’t want to be graded. Enclose your answers with boxes. Express all numerical answers in SI units . Answers with no explanation or disconnected comments will not be credited. If you obtain an answer that is questionable, explain why you think it is wrong. Constants and Conversion factors Avogadro number, N A 6 . 022 × 10 23 Permittivity of vacuum, 8 . 85 × 10- 12 F · m- 1 Permeability of vacuum, μ 4 π × 10- 7 T · m · A- 1 Speed of light in vacuum, c 1 / √ μ = 3 × 10 8 m/s Electron: Charge, q e- 1 . 602 × 10- 19 C Mass, m e 9 . 91 × 10- 31 kg Universal gas constant, R 8.315 J · mol- 1 · K- 1 = 1.99 cal · mol- 1 · K- 1 Boltzmann constant, k 1 . 381 × 10- 23 J · K- 1 Stefan-Boltzmann constant, σ 5 . 67 × 10- 8 W · m- 2 · K- 4 Acceleration due to gravity, g 9.8 m · s- 2 Water: Specific Heat 1 kcal · kg- 1 · ◦ C- 1 Water: Heat of fusion 80 kcal · kg- 1 Heat of vaporization 539 kcal · kg- 1 1 atm 1 . 013 × 10 5 N · m- 2 1 kcal 4 . 18 × 10 3 J 1 hp 746 W 1 liter 10 3 cm 3 Equations and formulae: Coulomb s law : F = 1 4 π q 1 q 2 r 2 ˆ r Lorentz F orce Law : G F = q ( G E + Gv × G B ) Electric field : d E = 1 4 π dq r 2 ˆ r P otential : dV = 1 4 π dq r P otential difference : V ab =- b a E · d l P otential energy : U ab = qV ab Electric field and potential : E =-∇ V Electric dipole : p = q d Magnetic dipole : Gμ = NI G A T orque on a dipole : Gτ = p × E Gτ = Gμ × G B P otential energy of a dipole : U =- p · E U =- Gμ · G B Gauss s Law ( s ) E · d A = q enclosed B · d A = 0 Biot- Savart Law : d B = μ 4 π Id l × ˆ r r 2 d F = I G × B F araday Induction Law : B · d =- d dt Φ B where Φ B = B · d A Ampere s Law : E · d = μ o d dt Φ E + μ I enclosed where Φ E = E · d A Energy density : u E = 1 2 E 2 u B = 1 2 μ B 2 P oynting vector : S = 1 μ E × B Displacement Current : I d = d Φ E dt 3 Capacitance : C = q V ab Energy stored in a capacitor : U = 1 2 CV 2 Capacitors in series : 1 C eq = 1 C i Capacitors in parallel ; C eq = C i Inductance : L = Nφ B I Energy stored in an Inductor : U = 1 2 LI 2 Inductors in series : L eq = L i Capacitors in parallel ; 1 L eq = 1 L i Resistors in series...
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This note was uploaded on 02/09/2010 for the course PHYSICS 7B taught by Professor Packard during the Spring '08 term at University of California, Berkeley.

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Sp06-Final-Smoot-Exa - Department of Physics University of California Berkeley Final Examination Physics 7B Lecture 1 Prof Smoot 8 AM 11 AM 12 May

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