This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part. Tutorial Exercise (a) Calculate the speed of a proton that is accelerated from rest through an electric potential difference of 145 V. (b) Calculate the speed of an electron that is accelerated through the same potential difference. Part 1 of 5 - Conceptualize For this moderate potential difference, we expect that the final speed of the particles will be substantially less than the speed of light. Because of the much greater mass of the proton, we expect the speed of the electron to be significantly greater than the speed of the proton. An equal force on both particles will result in a much greater acceleration for the electron. Part 2 of 5 - Categorize We apply conservation of energy to each particle-field system to find the final speed from the kinetic energy of the particle. Part 3 of 5 - Analyze (a) For the proton-field system, energy is conserved as the proton moves from high to low potential. We have Kit Uj = Kf+ Uf. This becomes the following conservation of energy equation. 0 + qvi = =mpup2 + 0 The initial potential energy of the proton is qVi = 1.60 x 10-19 c) v ) ( 4319 ) = [ x 10-17 J. Submit Skip (you cannot come back) Need Help? Read It -/2 Points] DETAILS MY NOTES ASK YOUR TEACHER PRACTICE ANC A uniform electric field of magnitude 270 V/m is directed in the positive x direction. A + 12.0 UC charge moves from the origin to the point (x, y) = (20.0 cm, 50.0 cm). (a) What is the change in the potential energy of the charge field system? (b) Through what potential difference does the charge move