Homework 4-solutions - soto(rrs946 Homework 4 Homann(56605...

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soto (rrs946) – Homework 4 – Hofmann – (56605) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – Fnd all choices be±ore answering. 001 (part 1 of 3) 10.0 points A proton is released ±rom rest in a uni±orm electric Feld o± magnitude 40000 V / m di- rected along the positive x axis. The proton undergoes a displacement o± 0 . 2 m in the di- rection o± the electric Feld as shown in the Fgure. The mass o± a proton is 1 . 672623 × 10 27 kg. + + + + + + + + + + + + + - - - - - - - - - - - - - v A = 0 + 0 . 2 m v 0 40000 V / m A B ²ind the change in the electric potential i± the proton moves ±rom the point A to B . Correct answer: - 8000 V. Explanation: Let : q p = 1 . 60218 × 10 19 C , m p = 1 . 67262 × 10 27 kg , E = 40000 V / m , and x = 0 . 2 m . ²rom the equation Δ V = - ± E · i B A d±s or in a non-calculus way ±or the straight path AB , Δ V = - ± E · ± AB we obtain Δ V = - E x = - (40000 V / m) (0 . 2 m) = - 8000 V . The negative result tells us that the electric potential o± the proton decreases as it moves ±rom A to B . 002 (part 2 of 3) 10.0 points ²ind the change in potential energy o± the proton ±or this displacement. Correct answer: - 1 . 28174 × 10 15 J. Explanation: The change in the potential energy o± the proton is Δ U = q p Δ V = (1 . 60218 × 10 19 C) ( - 8000 V) = - 1 . 28174 × 10 15 J . The negative sign here means that the poten- tial energy o± the proton decreases as it moves in the direction o± the electric Feld. This makes sense since, as the proton accelerates in the direction o± the Feld, it gains kinetic energy and at the same time (the Feld) loses potential energy (the energy is conserved). 003 (part 3 of 3) 10.0 points Apply the principle o± energy conservation to Fnd the speed o± the proton a±ter it has moved 0 . 2 m, starting ±rom rest. Correct answer: 1 . 23799 × 10 6 m / s. Explanation: Conservation o± energy in this case is Δ K + Δ U = 0 . K f - K i + Δ U = 0 1 2 m p v 2 f + Δ U = 0 v f = r - 2 Δ U m p = r - 2 ( - 1 . 28174 × 10 15 J) 1 . 67262 × 10 27 kg = 1 . 23799 × 10 6 m / s .
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soto (rrs946) – Homework 4 – Hofmann – (56605) 2 004 (part 1 of 2) 10.0 points An electron moving parallel to the x axis has an initial speed oF 5 × 10 6 m / s at the origin. Its speed is reduced to 1 × 10 5 m / s at the point x P , 5 cm away From the origin. The mass oF the electron is 9 . 10939 × 10 31 kg and the charge oF the electron is - 1 . 60218 × 10 19 C. Calculate the absolute value oF the poten- tial diference between this point and the ori- gin. Correct answer: 71 . 042 V. Explanation: Let : m e = 9 . 10939 × 10 31 kg , q e = - 1 . 60218 × 10 19 C , v i = 5 × 10 6 m / s , v f = 1 × 10 5 m / s , and d = 5 cm . By conservation oF energy ( K + U ) f = ( K + U ) i U f - U i Δ U = K i - K f , Using the de±nition oF kinetic energy and solv- ing For Δ U yields Δ U = 1 2 m e ( v 2 i - v 2 f ) = 1 2 (9 . 10939 × 10 31 kg) × b ( 5 × 10 6 m / s ) 2 - (1 × 10 5 m / s) 2 B = 1 . 13822 × 10 17 J , ²inally, the potential diference is Δ V = Δ U q e = 1 . 13822 × 10 17 J ( - 1 . 60218 × 10 19 C) = - 71 . 042 V .
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Homework 4-solutions - soto(rrs946 Homework 4 Homann(56605...

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