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Unformatted text preview: nguyen (jmn727) oldhomework 02 Turner (59070) 1 This printout should have 13 questions. Multiplechoice 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 = k e q 2 a 2 2. None of these. 3. F = 2 3 3 k e q 2 a 2 correct 4. F = 3 2 2 k e q 2 a 2 5. F = 2 3 3 k e q 2 a 2 6. F = 1 2 k e q 2 a 2 7. F = 1 2 k e q 2 a 2 8. F = 0 9. F = 1 3 k e q 2 a 2 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. repel the nucleus from the center of the ring. 2. be undetermined in direction. 3. restore the nucleus to the center of the ring. correct 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 . 4 nm? The value of the Coulomb con stant is 8 . 98755 10 9 N m 2 / C 2 . 1 electron Volt is equal to 1 . 602 10 19 C . Correct answer: 3 . 59951 eV. Explanation: nguyen (jmn727) oldhomework 02 Turner (59070) 2 Let : q = 1 . 602 10 19 C , a = 0 . 4 nm = 4 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 4 10 10 m = 5 . 76642 10 19 J ....
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This note was uploaded on 01/21/2010 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
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