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Unformatted text preview: toupal (rgt374) – Homework 07 – Chiu – (58295) 1 This printout should have 24 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 3) 10.0 points 3 . 5 MΩ 1 . 2 μ 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 μ C. Explanation: R C E S Let : t = 3 . 03 s , R = 3 . 5 MΩ = 3 . 5 × 10 6 Ω , C = 1 . 2 μ 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 Ω) (1 . 2 × 10 6 F) bracketrightbiggbracerightbigg = 1 . 02994 × 10 5 C = 10 . 2994 μ C . 002 (part 2 of 3) 10.0 points Find the current in the resistor at 3 . 03 s. Correct answer: 2 . 31919 μ A. Explanation: At t = 3 . 03 s, I = E R e t/ ( R C ) = 16 . 7 V 3 . 5 × 10 6 Ω × exp bracketleftbigg − (3 . 03 s) (3 . 5 × 10 6 Ω) (1 . 2 × 10 6 F) bracketrightbigg = 2 . 31919 × 10 6 A = 2 . 31919 μ 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 μ W. Explanation: In the time interval Δ t , work done by the battery in pushing charge Δ q across the bat tery is Δ W battery = Δ 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 Ω) + (1 . 02994 × 10 5 C) (1 . 2 × 10 6 F) bracketrightbigg = 3 . 87304 × 10 5 W = 38 . 7304 μ 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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This note was uploaded on 04/26/2009 for the course PHY 58310 taught by Professor Chiu during the Spring '09 term at University of Texas.
 Spring '09
 Chiu
 Physics, Current, Work

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