# 4 - karna (pk4534) HW03 li (59050) 1 This print-out should...

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Unformatted text preview: karna (pk4534) HW03 li (59050) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points Air becomes a conductor when the electric field strength exceeds 3 . 00 10 6 N/C. Determine the maximum amount of charge that can be carried by a metal sphere 1.5 m in radius. The value of the Coulomb constant is 8 . 99 10 9 N m 2 / C 2 . Correct answer: 0 . 000750834 C. Explanation: Let : E = 3 . 00 10 6 N / C , k e = 8 . 99 10 9 N m 2 / C 2 , and r = 1 . 5 m . q = E r 2 k C = (3 10 6 N / C) (1 . 5 m) 2 8 . 99 10 9 N m 2 / C 2 = . 000750834 C . 002 (part 1 of 4) 10.0 points A point charge q 1 is concentric with two spher- ical conducting thick shells, as shown in the figure below. The smaller spherical conduct- ing shell has a net charge of q 2 and the larger spherical conducting shell has a net charge of q 3 . q 3 q 2 q 1 R 1 R 2 R 3 R 4 R 5 r 1 r 2 r 3 r 4 Hint: Under static conditions, the charge on a conductor resides on the surface of the conductor. What is the charge Q r 1 on the inner surface of the smaller spherical conducting shell? 1. Q r 1 =- q 1 + q 2 + q 3 2. Q r 1 = + q 1 + q 2 3. Q r 1 = + q 1 + q 2 + q 3 4. Q r 1 =- q 1- q 2 5. Q r 1 =- q 1 + q 2 6. Q r 1 = + q 1- q 2 7. Q r 1 =- q 1 correct 8. Q r 1 =- q 1- q 2- q 3 9. Q r 1 = + q 1 10. Q r 1 = 0 Explanation: Basic Concept: Under static conditions, the electric field inside a conductor is zero. Charge is conserved, neither created or dis- troyed. Solution: The net charge inside a Gaus- sian surface located at r = R 2 must be zero, since the field in the conductor must be zero. karna (pk4534) HW03 li (59050) 2 Therefore, the charge on the inner surface of the smaller spherical conducting shell must be- q 1 . 003 (part 2 of 4) 10.0 points What is the charge Q r 2 on the outer surface of the smaller spherical conducting shell? 1. Q r 2 = + q 1 2. Q r 2 = 0 3. Q r 2 =- q 1 + q 2 4. Q r 2 =- q 1- q 2 5. Q r 2 = + q 1 + q 2 + q 3 6. Q r 2 =- q 1- q 2- q 3 7. Q r 2 = + q 1- q 2 8. Q r 2 =- q 1 9. Q r 2 = + q 1 + q 2 correct 10. Q r 2 = + q 1 + q 2- q 3 Explanation: The net charge inside a Gaussian surface located at r = R 3 must be q 1 + q 2 , since charge is conserved. Therefore, the charge on the outer surface of the smaller spherical conducting shell must be q 1 + q 2 . 004 (part 3 of 4) 10.0 points What is the charge Q r 3 on the inner surface of the larger spherical conducting shell? 1. Q r 3 =- q 1- q 2 correct 2. Q r 3 = + q 1- q 2 3. Q r 3 = + q 1 + q 2 + q 3 4. Q r 3 =- q 1- q 2 + q 3 5. Q r 3 =- q 1- q 2- q 3 6. Q r 3 = 0 7. Q r 3 =- q 1 8. Q r 3 = + q 1 9. Q r 3 =- q 1 + q 2 10. Q r 3 = + q 1 + q 2 Explanation: The net charge inside a Gaussian surface located at r = R 4 must be zero, since the field in the conductor must be zero. Therefore, the charge on the inner surface of the large spherical conducting shell must be- ( q 1 + q 2 ).)....
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## This note was uploaded on 11/02/2009 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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4 - karna (pk4534) HW03 li (59050) 1 This print-out should...

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