I need help with the practice and exercise I need help with the practice and exe

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I need help with the practice and exercise I need help with the practice and exercise This is the example in the book
No Service 10:16 AM * 35% PRACTICE IT Use the worked example above to help you solve this problem. A proton is released from rest at x =-1.60 cm in a constant electric field with magnitude 1.52 × 103 N/C, pointing in the positive x-direction (a) Calculate the change in potential energy when the proton reaches x = 4.74 cm. -1.54e-15 Your response is off by a multiple of ten. (b) An electron is now fired in the same direction from the same position. What is its change in electric potential energy if it reaches x = 11.50 cm? 4.18e-16 X Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. (c) If the direction of the electric field is reversed and an electron is released from rest at x = 4.00 cm, by how much has its electric potential energy changed when it reaches x = 7.40 cm? EXERCISE HINTS: GETTING STARTED I'M STUCKI Find the change in electric potential energy associated with the electron in part (b) as it goes on from x 0.115 to x =-0.018 m. (Note that the electron must turn around and go back at some point. The location of the turning point is unimportant, because changes in potential energy depend only on the end points of the path.) Need Help?Read It Submit Answer Save ProgressPractice Another Version

Explanation / Answer

x = -1.6 cm
E = 1.52*10^3 N/C


a)
x = 4.74 cm = 0.0474 m
PE = - (1.6*10^-19 C)*(1.52*10^3 N/C )*(0.00474 - (-0.0016))
PE = -1.541888 × 10^-18 J

b)
x = 11.5 cm = 0.115 m
PE = - (- 1.6*10^-19 C)*(1.52*10^3 N/C )*(0.00474 - (-0.115))
PE = 2.912×10^-17 J

c)
PE = - (- 1.6*10^-19 C)*(1.52*10^3 N/C )*(0.074 - 0.04 )
PE = 8.269×10^-18 J

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