Homework_07-solutions - toupal (rgt374) Homework 07 Chiu...

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Unformatted text preview: toupal (rgt374) Homework 07 Chiu (58295) 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 3) 10.0 points 3 . 5 M 1 . 2 F 16 . 7 V S The switch is closed at t = 0. Find the charge on the capacitor at 3 . 03 s. Correct answer: 10 . 2994 C. Explanation: R C E S Let : t = 3 . 03 s , R = 3 . 5 M = 3 . 5 10 6 , C = 1 . 2 F = 1 . 2 10- 6 F , and E = 16 . 7 V . At t = 3 . 03 s, q = C E parenleftBig 1 e- t/ ( R C ) parenrightBig = (1 . 2 10- 6 F) (16 . 7 V) braceleftbigg 1 exp bracketleftbigg (3 . 03 s) (3 . 5 10 6 ) (1 . 2 10- 6 F) bracketrightbiggbracerightbigg = 1 . 02994 10- 5 C = 10 . 2994 C . 002 (part 2 of 3) 10.0 points Find the current in the resistor at 3 . 03 s. Correct answer: 2 . 31919 A. Explanation: At t = 3 . 03 s, I = E R e- t/ ( R C ) = 16 . 7 V 3 . 5 10 6 exp bracketleftbigg (3 . 03 s) (3 . 5 10 6 ) (1 . 2 10- 6 F) bracketrightbigg = 2 . 31919 10- 6 A = 2 . 31919 A . 003 (part 3 of 3) 10.0 points At 3 . 03 s the current in the resistor is I (Part 2) and the charge on the capacitor is q (Part 1). What is the power delivered by the bat- tery? Correct answer: 38 . 7304 W. Explanation: In the time interval t , work done by the battery in pushing charge q across the bat- tery is W battery = q E . Correspondingly, the power is d W battery dt = E dq dt = I E . The power dissipated in a resistor is d W resistor dt = I 2 R . The power to create the electric field in a capacitor is d W capacitor dt = I q C . Thus the total power dissipated in the capac- itor and resistor, that is the power delivered toupal (rgt374) Homework 07 Chiu (58295) 2 by the battery is d W battery dt = I parenleftBig I R + q C parenrightBig = (2 . 31919 10- 6 A) bracketleftbigg (2 . 31919 10- 6 A) (3 . 5 10 6 ) + (1 . 02994 10- 5 C) (1 . 2 10- 6 F) bracketrightbigg = 3 . 87304 10- 5 W = 38 . 7304 W . 004 10.0 points Suppose a new particle is discovered, and it is found that a beam of these particles passes undeflected through crossed electric and magnetic fields, where E = 323 V / m and B = 0 . 00781 T. If the electric field is turned off, the particles move in the magnetic field in circular paths of radius r = 1 . 47 cm. Determine q m for the particles from these data. Correct answer: 3 . 60233 10 8 C / kg. Explanation: Let : E = 323 V / m , q e = q p = 1 . 60218 10- 19 C , m e = 9 . 10939 10- 31 kg , m p = 1 . 67262 10- 27 kg , B = 0 . 00781 T , and r = 0 . 0147 m . Since the particle passes undeflected, the electric force on it is equal to the magnetic force. When the electric field is turned off, only the magnetic force is exerted on it as the centripetal force. So we get two equations q E = B q v B q v = mv 2 r ....
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Homework_07-solutions - toupal (rgt374) Homework 07 Chiu...

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