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Unformatted text preview: WANG (xw788) – 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 = 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 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. 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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 Spring '08
 Turner
 Physics, Charge, Work, Electric charge, KE

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