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ID: 2025546 • Letter: #
Question
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A---------------------->B<br>
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A= 14V, B=32.6V, S = 0.130m
Calculate the magnitude of the acceleration of a proton (charge 1.60×10-19 C and mass 1.67×10-27 kg) if placed in this field. Then, answer the following questions:
a) Call A the beginning point of the displacement deltaS in the figure A, and the end of the displacement, B. What is the potential energy of an electron at point A and point B?
b)What is the magnitude of the electric field in the figure?
c)What is the force on the electron? In which direction is it? Is it the same at point a and at point B? Why?
d)The electron starts from rest at the beginning of the displacement. Argue why the electron should move in the direction of the displacement.
e)Use the Work-Kinetic Energy theorem to find the speed of the electron at the end of the displacement. DO NOT use the equation v = vo + at.
Explanation / Answer
The potential at point A is VA = 14 V The potential at point B is VB = 32.6 V The distance between two points s = 0.130 m The charge of the proton is q =1.60×10-19 C The mass of the proton is m = 1.67×10-27 kg The force on the proton is F = qE = q(V/s) = (1.60×10-19 C)[(32.6 V- 14V)/0.13 m] = 2.3*10-17 N thus the accelertion of the proton is a = F/m = (2.3*10-17 N)/(1.67×10-27 kg) = 1.4*1010 m/s2 (a) The potential energy of the electron at point A is UA = qVA = (-1.60×10-19 C)(14 V) = -2.24*10-18 J The potential energy of the electron at point B is UB = qVB = (-1.60×10-19 C)(32.6 V) = -5.22*10-18 J (b) The magnitude of the eletcric field E = V/s = (VB-VA) / 0.13m = (32.6 V - 14V) / 0.13 m = 143 V/m (c) The force on the electron is F = qE = (1.60×10-19 C)[(32.6 V- 14V)/0.13 m] = 2.28*10-17 N The direction is from B to A because the potential at point B is low and at point A is high. (d) Always the electron move from low potential to high potential. So the electron has to move from B to A. (e) From work energy theorem Fs = (1/2) mv2 Therefore the speed of the electron is v = [2(Fs) / m] = [2(2.28*10-17 N)(0.13 m) /(1.67×10-27 kg)] = 5.95*104 m/s (a) The potential energy of the electron at point A is UA = qVA = (-1.60×10-19 C)(14 V) = -2.24*10-18 J The potential energy of the electron at point B is UB = qVB = (-1.60×10-19 C)(32.6 V) = -5.22*10-18 J (b) The magnitude of the eletcric field = (-1.60×10-19 C)(32.6 V) = -5.22*10-18 J E = V/s = (VB-VA) / 0.13m = (32.6 V - 14V) / 0.13 m = 143 V/m (c) The force on the electron is F = qE = (1.60×10-19 C)[(32.6 V- 14V)/0.13 m] = 2.28*10-17 N The direction is from B to A because the potential at point B is low and at point A is high. (d) Always the electron move from low potential to high potential. So the electron has to move from B to A. (e) From work energy theorem Fs = (1/2) mv2 Therefore the speed of the electron is v = [2(Fs) / m] = [2(2.28*10-17 N)(0.13 m) /(1.67×10-27 kg)] = 5.95*104 m/sRelated Questions
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