8- The Electric Potential-solutions

# 8- The Electric Potential-solutions - gilvin(jg47854 8 The...

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Unformatted text preview: 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. This homework is due Tuesday, March 1, at midnight Tucson time. 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 E = so the magnitude of the potential difference is V = E d = d = ( 1 . 1 10 10 C / m 2 ) (0 . 011 m) 8 . 85419 10 12 C 2 / N m 2 = . 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 . 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. Explanation: gilvin (jg47854) 8: The Electric Potential meyers (21235) 2 Let : E = 2022 N / C , 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 voltage between the plates, and d is the dis- tance between the plates, so the new electric field has a magnitude of E = V d = 4 V d 5 = 20 parenleftbigg V d parenrightbigg = 20 E = 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. The magnitude of the electric field for which dielectric breakdown occurs in air is about 3 MV / m....
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## This note was uploaded on 04/29/2011 for the course PHYS 202 taught by Professor Doc during the Spring '11 term at North Texas.

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8- The Electric Potential-solutions - gilvin(jg47854 8 The...

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