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ome File Edit View History Bookmarks People Window Help O www.webassign.net/web/Student/Assignment-ResponsesMast?dep 15687454 PRACTICE IT Use the worked example above to help you solve this problem. A proton is released from rest at x -3.00 cm in a constant electric field with magnitude 1.48 x 103 Nyc, pointing in the positive x-direction. (a) Calculate the change in potential energy when the proton reaches 5.45 cm. (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 12.80 cm? (c) If the direction of the electric field is reversed and an released from rest at x 3.40 cm, by how much has its electric potential energy changed when it reaches x 6.90 cm? EXERCISE GETTING STARTED I l'M STUCK! Find the change in electric potential energy associated with the electron in part (b) as it goes on from 0.128 to -0,018 m. (Note that the electron must turn around and go back at some point. The of the turning polnt is unimportant, because changes in potential energy depend only on the end points of the path.) Need Help?Explanation / Answer
part a:
change in potential energy=charge*potential difference
=charge*(-electric field*distance)
=1.6*10^(-19)*(-1.48*10^3*(5.45-(-3))*0.01=-2*10^(-17) J
part b:
for electron , change in potential energy
=charge*(-electric field*distance)
=-1.6*10^(-19)*(-1.48*10^3*(12.8-(-3))*0.01=3.74*10^17 J
part c:
if direction of electric field is reversed, the -ve sign associated with electric field
for calcluation of potential change will be removed.
hence change in potential energy
=-1.6*10^(-19)*(1.48*10^3*(6.9-(3.4))*0.01=-8.288*10^(-18) J
exercise:
using the same formula,
change in potential energy
=-1.6*10^(-19)*(-1.48*10^3*(-0.018-(0.128))=-3.45728*10^(-17) J
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