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Unformatted text preview: Professor George F. Smoot Department of Physics University of California, Berkeley Midterm Examination 2 Physics 7B, Section 3 6:00 pm  8:00 pm, November 4, 2008 Name: SID No: Discussion Section: Name of TA: Problem 1 Problem 2 Problem 3 Problem 4 Problem 5 Problem 6 Score: 1 Answer all six 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 Some useful equations 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 Charge of electron, q e = e 1 . 602 × 10 19 C Mass of electron, m e 9 . 11 × 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 StefanBoltzmann constant, σ 5 . 67 × 10 8 W · m 2 · K 4 Acceleration due to gravity, g 9.8 m · s 2 Specific heat of water 1 kcal · kg 1 · ◦ C 1 Heat of fusion of water 80 kcal · kg 1 1 atm 1 . 013 × 10 5 N · m 2 1 kcal 4 . 18 × 10 3 J 1 hp 746 W Coulomb s law : F = 1 4 π q 1 q 2 r 2 ˆ r Electric field : d E = 1 4 π dq r 2 ˆ r Electric dipole : p = q d Torque on a dipole : ~ τ = p × E Potential energy of a dipole : U = p · E Gauss s law : I E · d A = q encl Potential difference : V ab = Z b a E · d l Potential : dV = 1 4 π dq r Potential energy : U ab = qV ab Electric field and potential : E =∇ V Capacitance : C = q V ab 2 Capacitors in series : 1 C eq = X 1 C i Capacitors in parallel ; C eq = X C i Energy stored in a capacitor : U = 1 2 CV 2 Energy density : u = 1 2 E 2 Current : I = dq dt Current and current density : dI = j · d A Ohm s law : V = IR Ohm s law : j = σ E ; E = ρ j Resistivity and resistance : R = ρ l A Electric power ( dc ) delivered : P = V I Electric power ( dc ) dissipated : P = V 2 /R = I 2 R Average electric power ( ac ) delivered : P = I rms V rms = I V / 2 Average electric power ( ac ) dissipated : P = I 2 rms R = I 2 R/ 2 rms current : I rms = I / √ 2 rms voltage V rms = V / √ 2 Resistors in series : R eq = X R i Resistors in parallel : 1 R eq = X 1 R i Kirchhoff current rule : X I = 0 at a node Kirchhoff potential rule : X Δ V = X IR around a loop ∇ 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: Z dx √ x 2 + a 2 = log x + √ x 2 + a 2 or sinh 1 x a Z dx x 2 + a 2 = 1 a arctan ( x a ) Z dx ( x 2 + a 2 ) 3 / 2 = x a 2 √ x 2 + a 2 3 Table 1: Dielectric Constants for Some Materials Material...
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This note was uploaded on 02/04/2010 for the course PHYSICS 7B taught by Professor Packard during the Spring '08 term at Berkeley.
 Spring '08
 Packard
 Physics

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