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# oldhw 02 - nguyen(jmn727 oldhomework 02 Turner(59070 This...

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nguyen (jmn727) – 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 = 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 θ = 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:

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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 0 F dx = integraldisplay 0 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 . 004 (part 1 of 2) 10.0 points Three charges are arranged as shown in the figure.
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oldhw 02 - nguyen(jmn727 oldhomework 02 Turner(59070 This...

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