The Electric Potential-solutions

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

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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 Δ σ ϵ 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: ϵ 0 = 8.85419 × 10 −12 C 2 /N · m 2 , σ = 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 = σ ϵ 0 Explanation:

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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 Gauss’s Law, you can show that for ρ 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 ϵ 0 where ρ is the charge density. The charge density ρ is E = −V d = 20 E = − 4 V = 20 −V d d 1. ρ = Q 5 π 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. 2. ρ =
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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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The Electric Potential-solutions - gilvin(jg47854 8 The...

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