2 - karna (pk4534) HW02 li (59050) 1 This print-out should...

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Unformatted text preview: karna (pk4534) HW02 li (59050) 1 This print-out should have 31 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 3 k e q 2 a 2 5. F = k e q 2 a 2 6. F = 2 3 3 k e q 2 a 2 correct 7. F = 1 3 k e q 2 a 2 8. F = 1 2 k e q 2 a 2 9. F = 1 2 k e q 2 a 2 10. F = 3 2 2 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. repel the nucleus from the center of the ring. 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 . 17 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: 8 . 46945 eV. Explanation: karna (pk4534) HW02 li (59050) 2 Let : q = 1 . 602 10 19 C , a = 0 . 17 nm = 1 . 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 1 . 7 10 10 m = 1 . 35681 10 18 J . 004 (part 1 of 3) 10.0 points A rod of length with uniform charge per unit length , where > 0, is placed a distance d from the origin along the x axis. A similar rod with the same charge is placed along the y axis as in the figure. Consider only a small segment of length x (it is of infinitesimal length) on the horizontal rod, which is placed along the x axis. The segment is at a distance x from the origin....
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This note was uploaded on 11/02/2009 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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2 - karna (pk4534) HW02 li (59050) 1 This print-out should...

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