HW 4 - segovia (cs39966) – Homework 4 – Spurlock –...

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Unformatted text preview: segovia (cs39966) – Homework 4 – Spurlock – (14441) 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points 46 Ω 13 . 7 μ F 18 V S How long after the switch is closed does the voltage across the resistor drop to V f = 12 V? Correct answer: 0 . 000255524 s. Explanation: R C E S Let : R = 46 Ω , V f = 12 V , C = 13 . 7 μ F = 1 . 37 × 10 − 5 F , and E = 18 V . Since the current in an R C circuit is I = E R e − t/ ( R C ) and the voltage across the resistor is V = I R to find the time at which the voltage is V f we need to solve the equation V f R = E R e − t/ ( R C ) for t . This gives t =- R C log parenleftbigg V f E parenrightbigg =- (46 Ω)(1 . 37 × 10 − 5 F) log parenleftbigg 12 V 18 V parenrightbigg = . 000255524 s . 002 (part 2 of 2) 10.0 points What is the charge on the capacitor at this time? Correct answer: 8 . 22 × 10 − 5 C. Explanation: At this time the charge on the capacitor is given by Q = C V . Since we know the capacitance, and we know that the voltage drop across the capacitor must be the total voltage, E , minus the voltage V f across the resistor, we have Q = C ( E - V f ) = (1 . 37 × 10 − 5 F)(18 V- 12 V) = 8 . 22 × 10 − 5 C . 003 10.0 points At t=0 the switch S is closed with the ca- pacitor is uncharged. i 66 μ F 4 kΩ S 37 V What is the charge on the capacitor when I = 2 mA? Correct answer: 0 . 001914 C. Explanation: i C R S E segovia (cs39966) – Homework 4 – Spurlock – (14441) 2 Let : C = 66 μ F = 6 . 6 × 10 − 5 F , E = 37 V , R = 4 kΩ = 4000 Ω , and I = 2 mA = 0 . 002 A . Applying Kirchhoff’s law E - Q C- I R = 0 , so Q = C ( E - I R ) = (6 . 6 × 10 − 5 F) × [37 V- (0 . 002 A) (4000 Ω)] = . 001914 C . 004 10.0 points A proton moving at 1 . 1 × 10 6 m / s through a magnetic field of 4 . 5 T experiences a magnetic force of magnitude 5 . 6 × 10 − 13 N. The charge of proton is 1 . 60218 × 10 − 19 C and the mass of proton is 1 . 67262 × 10 − 27 kg. What is the angle between the proton’s velocity and the field? Correct answer: 44 . 9193 ◦ . Explanation: Let : E = 5 . 6 × 10 − 13 N , B = 4 . 5 T , v = 1 . 1 × 10 6 m / s , q p = 1 . 60218 × 10 − 19 C , and m p = 1 . 67262 × 10 − 27 kg . The Lorentz force acting on a moving charged particle in a magnetic field is F = q v B sin θ , where q is the charge, v is the speed, B is the magnetic field. So thus the angle between the velocity and the field is θ = arcsin bracketleftbigg F q v B bracketrightbigg = arcsin bracketleftbigg (5 . 6 × 10 − 13 N) (1 . 60218 × 10 − 19 C) × 1 (1 . 1 × 10 6 m / s) (4 . 5 T) bracketrightbigg = 44 . 9193 ◦ ....
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This note was uploaded on 11/10/2011 for the course PHYSICS 303L taught by Professor Hoffman during the Fall '11 term at University of Texas at Austin.

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HW 4 - segovia (cs39966) – Homework 4 – Spurlock –...

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