Use the worked example above to help you solve this problem. A proton is release

Question

Use the worked example above to help you solve this problem. A proton is released from rest at x = -2. 90 cm in a constant electric field with magnitude 1. 50 times 103 N/C, pointing in the positive x-direction. Calculate the change in potential energy when the proton reaches x = 4. 86 cm. 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. 40 cm? If the direction of the electric field is reversed and an electron is released from rest at x = 3. 70 cm, by how mu Ch has its electric potential energy changed when it reaches x = 6. 90 cm? Find the change in electric potential energy associated with the electron in part (b) as it goes on from x = 0. 114 to x = -0. 019 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. ) Delta PE = J

Explanation / Answer

a)1.8624*10^-17 J b)3.432*10^-17 J c)7.68*10^-17 J PE change=3.192*10^-17 J

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