Fa08-MT2-Smoot-Exam

# Fa08-MT2-Smoot-Exam - Professor George F Smoot Department...

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Professor George F. Smoot Department of Physics University of California, Berkeley Mid-term 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

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Capacitors in series : 1 C eq = 1 C i Capacitors in parallel ; C eq = C i Energy stored in a capacitor : U = 1 2 CV 2 Energy density : u = 1 2 0 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 0 V 0 / 2 Average electric power ( ac ) dissipated : P = I 2 rms R = I 2 0 R/ 2 rms current : I rms = I 0 / 2 rms voltage V rms = V 0 / 2 Resistors in series : R eq = R i Resistors in parallel : 1 R eq = 1 R i Kirchhoff current rule : I = 0 at a node Kirchhoff potential rule : Δ V = 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: 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 3

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Table 1: Dielectric Constants for Some Materials Material Dielectric Constant Dielectric Break Down K (V/m) Vacuum 1.0000 > 10 15 Air 1.0006 3 × 10 6 Teflon 2.1 60 × 10 6 Paraffin 2.2 10 × 10 6 Polystyrene 2.6 24 × 10 6 Vinyl (plastic) 2 - 4 50 × 10 6 Paper 3.7 15 × 10 6 Quartz 4.3 8 × 10 6 Oil 4 12 × 10 6 Glass Pyrex 5 14 × 10 6 Porcelain 6 - 8 5 × 10 6 Mica 7 150 × 10 6 Silicon Dioxide 3.9 5 - 15 × 10 6 Silicon 11.68 - × 10 6 Water (liquid) 80
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