test 2 practice-solutions

test 2 practice-solutions - gilvin (jg47854) test 2...

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Unformatted text preview: gilvin (jg47854) test 2 practice meyers (21235) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. This practice set is due at midnight Sun- day, April 3. It wont count toward your total grade, its for exam practice. Worked solu- tions to questions 7 - 11, which we looked at in class, will be posted as a pdf file on the web- site sometime Saturday. Also note that this is not a comprehensive practice exam, just a few problems that might recall previous ma- terial and might provide extra practice. The test covers chapters 28-31. 001 10.0 points Consider the uniformly charged sphere with radius R , shown in the figure below. R r p Q is the total charge inside the sphere Find the value of the electrostatic potential at the same radius r . 1. V = 1 3 k Q r bracketleftBigg 2- parenleftbigg R r parenrightbigg 2 bracketrightBigg 2. V = 1 3 k Q R bracketleftbigg 2- parenleftBig r R parenrightBig 2 bracketrightbigg 3. V = 1 3 k Q R bracketleftBigg 2- parenleftbigg R r parenrightbigg 2 bracketrightBigg 4. V = 1 2 k Q R bracketleftbigg 3- parenleftBig r R parenrightBig 2 bracketrightbigg correct 5. V = 1 2 k Q R parenleftBig r R parenrightBig 2 6. V = 1 2 k Q R bracketleftBigg 3- parenleftbigg R r parenrightbigg 2 bracketrightBigg 7. V = 1 3 k Q r bracketleftbigg 2- parenleftBig r R parenrightBig 2 bracketrightbigg 8. V = 1 2 k Q r bracketleftBigg 3- parenleftbigg R r parenrightbigg 2 bracketrightBigg 9. V = 1 3 k Q R parenleftbigg R r parenrightbigg 2 10. V = 1 2 k Q r bracketleftbigg 3- parenleftBig r R parenrightBig 2 bracketrightbigg Explanation: The electric potential at distance r from the center is V =- integraldisplay r vector E dvectors. Using Gauss Law we find the electric field inside and outside of the sphere... vector E ( r ) = k Qr R 3 r for r < R. If r R , we can treat the whole sphere as a point vector E ( r ) = k Q r 2 r . Hence the potential at distance r is V =- integraldisplay R k Q r 2 dr- integraldisplay r R k Qr R 3 dr = k Q R- k Q 2 R 3 ( r 2- R 2 ) = 1 2 k Q R bracketleftbigg 3- parenleftBig r R parenrightBig 2 bracketrightbigg . 002 (part 1 of 2) 10.0 points Given : k e = 8 . 98755 10 9 N m 2 / C 2 . A uniformly charged insulating rod of length 8 . 1 cm is bent into the shape of a semicircle as in the figure. 8 . 1 c m- 6 C O If the rod has a total charge of- 6 C, find the horizontal component of the electric field at O , the center of the semicircle. gilvin (jg47854) test 2 practice meyers (21235) 2 Hint: Define right as positive. Correct answer:- 5 . 16419 10 7 N / C. Explanation: Let : L = 8 . 1 cm = 0 . 081 m , and q =- 6 C =- 6 10 6 C ....
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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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test 2 practice-solutions - gilvin (jg47854) test 2...

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