test 2 practice-solutions

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

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R ? k Q gilvin (jg47854) - test 2 practice - meyers - (21235) 1 ( ) 2 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 won’t count toward your total k Q 9. V = 1 3 R k Q 10. V = 1 2 r Explanation: R r [ ( ) 2 ] r 3 − R grade, it ’s 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. The electric potential at distance r from the center is r V = − E · ds . Using Gauss’ Law we find the electric field inside and outside of the sphere. .. E (r) = k Q r r for r < R . 001 10.0 points Consider the uniformly charged sphere with radius R, shown in the figure below. Q is the total R 3 If r ≥ R, we can treat the whole sphere as a point E charge inside the sphere p R (r) = k Q r . r 2 Hence the potential at distance r is R r V = − r = k Q Find the value of the electrostatic potential k Q k Q r dr − dr r 2 R R 3 ( r 2 − R 2 ) [ 2 R 3 at the same radius r . [ = 1 2 ( ) 2 ] k Q r 3 − R R 1. V = 1 3 k Q r ( 2 − [ ( ) 2 ] R r 002 (part 1 of 2) 10.0 points ) 2 ] 2. V = 1 3 k Q R 2 − [ r Given : k e = 8.98755 × 10 9 N m 2 /C 2 . R ( ) 2 ] 3. V = 1 3 k Q R A uniformly charged insulating rod of 2 − R r length 8.1 cm is bent into the shape of a 4. V = 1 2 k Q R [ ( ) 2 ] r 3 − correct R ( ) 2 semicircle as in the figure. 5. V = 1 2 k Q R k Q [ r −6 µC R ( ) 2 ] O R 6. V = 1 2 7. V = 1 3 8. V = 1 2 R k Q r k Q r 3 − [ 2 − [ 3 −

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r ( ) 2 ] r R ( ) 2 ] R If the rod has a total charge of −6 µC, find r the horizontal component of the electric field at O, the center of the semicircle.
gilvin (jg47854) - test 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 µC = −6 × 10 −6 C . Due to symmetry E y = dE y = 0 , practice - meyers - (21235) 2 Since the rod has negative charge, the field is pointing to the left (towards the charge dis- tribution). A positive test charge at O would feel an attractive force from the semicircle, pointing to the left. 003 (part 2 of 2) 10.0 points Determine the value of the electric potential V at the center of the semicircle. Correct answer: −2.0915 × 10 6 J/C. Explanation: The potential at a point due to a continuous charge distribution can be found using and V = k dq sin θ dq r . E

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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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