CH 16 - Electric Potential-solutions

CH 16 - Electric Potential-solutions - This print-out...

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Unformatted text preview: This print-out should have 7 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Through what potential difference would an electron need to be accelerated for it to achieve a speed of 6 % of the speed of light (2 . 998 × 10 8 m / s), starting from rest? The mass of an electron is 9 . 109 × 10- 31 kg and the charge on an electron is 1 . 602 × 10- 19 C. Ignore special relativity for this problem. Correct answer: 919 . 907 V. Explanation: Let : s = 6% = 0 . 06 , c = 2 . 998 × 10 8 m / s , m e = 9 . 109 × 10- 31 kg , and q e = 1 . 602 × 10- 19 C . The speed of the electron is v = 0 . 06 c = 0 . 06 ( 2 . 998 × 10 8 m / s ) = 1 . 7988 × 10 7 m / s , By conservation of energy 1 2 m e v 2 =- (- q e ) Δ V Δ V = m e v 2 2 q e = ( 9 . 109 × 10- 31 kg ) × ( 1 . 7988 × 10 7 m / s ) 2 2 (1 . 602 × 10- 19 C) = 919 . 907 V . 002 10.0 points Particles A (of mass m and charge Q ) and B (of m and charge 5 Q ) are released from rest with the distance between them equal to 1 . 8969 m. If Q = 29 μ C, what is the kinetic energy of particle B at the instant when the particles are 3 . 8969 m apart?...
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This note was uploaded on 10/03/2010 for the course PH 102 taught by Professor Staff during the Fall '08 term at Alabama.

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CH 16 - Electric Potential-solutions - This print-out...

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