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Unformatted text preview: karna (pk4534) – HW02 – li – (59050) 1 This printout should have 31 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 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 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. 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 ℓ with uniform charge per unit length λ , where λ > 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 Δ 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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 Spring '08
 Turner
 Physics, Charge, Electrostatics, Correct Answer, Electric charge, Karna

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