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gilvin (jg47854) - 8: The Electric Potential - meyers - (21235) 1 This print- out should have 22 questions. Multiple-choice questions may continue on the next column or page - find all choices before answering. so the magnitude of the potential difference is V = E d = d 0 This homework is due Tuesday, March 1, at ( midnight Tucson time. = 1.1 10 10 C/m 2 ) (0.011 m) 8.85419 10 12 C 2 /N m 2 001 10.0 points To recharge a 12 V battery, a battery charger must move 2.4 10 5 C of charge from the negative terminal to the positive terminal. How much work is done by the battery charger? Correct answer: 2.88 10 6 J. Explanation: Given : q = 2.4 10 5 C and V = 12 V . The potential difference is V = W q , W = q V = (2.4 10 5 C) (12 V) = 2.88 10 6 J . 002 10.0 points Two flat conductors are placed with their in- ner faces separated by 11 mm. If the surface charge density on one of the inner faces is 110 pC/m 2 and the other inner face 110 pC/m 2 , what is the magnitude of the electric potential difference between the two conductors? Correct answer: 0.136658 V. Explanation: Let: = 8.85419 10 12 C 2 /N m 2 , = 110 pC/m 2 = 1.1 10 10 C/m 2 , and d = 11 mm = 0.011 m . The electric field between two flat conduc- tors is = 0.136658 V . 003 10.0 points A voltmeter indicates that the difference in potential between two plates is 53 V. The plates are 0.36 m apart. What electric field intensity exists between them? Correct answer: 147.222 N/C. Explanation: Let : V = 53 V and d = 0.36 m . The potential difference is V = E d E = V d = 53 V 0.36 m = 147.222 N/C . 004 10.0 points Two parallel conducting plates are connected to a constant voltage source. The magni- tude of the electric field between the plates is 2022 N/C. If the voltage is quadrupled and the dis- tance between the plates is reduced to 1 5 the original distance, what is the magnitude of the new electric field? Correct answer: 40440 N/C. E = Explanation: gilvin (jg47854) - 8: The Electric Potential - meyers - (21235) 2 Q is the total Let : E = 2022 N/C , charge inside R the sphere. V = 4 V , and d = 1 5 d . The electric field between two parallel con- ducting plates is E = V d , where V is the p Using Gausss Law, you can show that for r voltage between the plates, and d is the dis- tance between the plates, so the new electric field has a magnitude of ( ) r < R the electric field is given by E = , 3 where is the charge density. The charge density is E = V d = 20 E = 4 V = 20 V d d 1. = Q 5 R 3 . Q = 20 (2022 N/C) = 40440 N/C . 005 10.0 points When you touch a friend after walking across a rug on a dry day, you typically draw a spark of about 2 mm.... View Full Document
Physics 2
2- Coulomb 2-solutions
3. Electric Field 1-solutions
Current and current density-solutions (2)
Coulomb's Law-solutions
Electric Field 1-solutions
Gauss's law 2 -solutions
Capacitance-solutions
test 2 practice-solutions
Homework 6
Jackson_1_8_Homework_Solution
Homework3
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