oldhomework 02-solutions

oldhomework 02-solutions - WANG (xw788) oldhomework 02...

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Unformatted text preview: WANG (xw788) oldhomework 02 Turner (59070) 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 3) 10.0 points In an early model for the atom the nucleus of charge q was surrounded by a circular ring of radius r = a of negative charge q . Find the magnitude of the force on the nuclear charge if it is displaced the distance x = 2 a along the axis of the ring. 1. F = 0 2. F = 2 3 3 k e q 2 a 2 3. None of these. 4. F = 1 2 k e q 2 a 2 5. F = 1 3 k e q 2 a 2 6. F = 1 2 k e q 2 a 2 7. F = 3 2 2 k e q 2 a 2 8. F = k e q 2 a 2 9. F = 2 3 3 k e q 2 a 2 correct 10. F = 1 3 k e q 2 a 2 Explanation: By a simple integration, we know that the electric field from the ring in the yz plane with radius a is expressed as dE = k e dq r 2 dE y = 0 dE x = dE cos = k e dq r 2 x r = k e x ( x 2 + a 2 ) 3 / 2 dq . Therefore E = E x = integraldisplay k e x ( x 2 + a 2 ) 3 / 2 dq = k e x ( x 2 + a 2 ) 3 / 2 integraldisplay dq = k e x ( x 2 + a 2 ) 3 / 2 q . Therefore, the force on the nucleus is F = k e x q 2 ( x 2 + a 2 ) 3 / 2 = k e 2 a q 2 [( 2 a ) 2 + a 2 ] 3 / 2 = 2 3 3 k e q 2 a 2 , where x = 2 a , as given in the problem. 002 (part 2 of 3) 10.0 points Will this force 1. be undetermined in direction. 2. restore the nucleus to the center of the ring. correct 3. repel the nucleus from the center of the ring. Explanation: The force between the negatively charged ring and the positively charged nucleus is at- tractive. 003 (part 3 of 3) 10.0 points Find the minimum energy required to move the nucleus from the equilibrium position to infinity for this model for the hydrogen with q = 1 . 602 10 19 C and the radius a = 0 . 27 nm? The value of the Coulomb constant is 8 . 98755 10 9 N m 2 / C 2 . 1 elec- tron Volt is equal to 1 . 602 10 19 C . Correct answer: 5 . 33261 eV. Explanation: WANG (xw788) oldhomework 02 Turner (59070) 2 Let : q = 1 . 602 10 19 C , a = 0 . 27 nm = 2 . 7 10 10 m , 1 eV = 1 . 602 10 19 C and k e = 8 . 98755 10 9 N m 2 / C 2 . The required minimum energy is integraldisplay F dx = integraldisplay bracketleftbigg k e x q 2 ( x 2 + a 2 ) 3 / 2 bracketrightbigg dx = k e q 2 a = ( 8 . 98755 10 9 N m 2 / C 2 ) ( 1 . 602 10 19 C ) 2 2 . 7 10 10 m = 8 . 54285 10 19 J ....
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This note was uploaded on 10/11/2009 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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oldhomework 02-solutions - WANG (xw788) oldhomework 02...

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