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Unformatted text preview: gilvin (jg47854) test 2 practice meyers (21235) 1 This printout should have 21 questions. Multiplechoice 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 2831. 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.
 Spring '11
 doc
 Magnetism

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