GOAL Illustrate the concept of electric potential energy. PROBLEM A proton is re
ID: 1869950 • Letter: G
Question
GOAL Illustrate the concept of electric potential energy. PROBLEM A proton is released from rest at x =-2.00 cm in a constant electric field with magnitude 1.50 × 103 N/C, pointing in the positive x-direction. (a) Calculate the change in the electric potential energy associated with the proton when it reaches x = 5.00 cm. (b) An electron is now fired in the same direction from the same position. What is its change in electric potential energy associated with the electron if it reaches x 12.0 cm? (c) If the direction of the electric field is reversed and an electron is released from rest at x 3.00 cm, by how much has the electric potential energy changed when the electron reaches r = 7.00 cm? STRATEGY This problem requires a straightforward substitution of given values into the definition of electric potential energy SOLUTION (A) Calculate the change in the electric potential energy associated with the proton. Apply the definition of electric potential energy (B) Find the change in electric potential energy associated with an electron fired from x =-0.0200 m and reaching x = 0.120 m of electric potential Apply the energ, but in this case note that the electric charge g is negative --(-1.60 x 10'39C)1.50 x 103 NK)10.120 m-(40200m)1 -+3.36 x 10-a,, (C) Find the change in potential energy associated with an electron traveling from x = 3.00 cm to x = 7.00 cm if the direction of the electric field is reversed. the electric fiele points in the negative s-direction (-1.60x10 9.50 x 10-a. CK-.50 x 10/C) (0070 m-0.030 m - REMARKS Notice that the proton (actually the proton-field system) lost potential energy when it moved in the positive x- direction, whereas the electron gained potential energy when it moved in the same direction. Finding changes in potential energy with the field reversed was only a matter of supplying a minus sign, bringing the total number in this case to three! It's important not to drop any of the signsExplanation / Answer
Practice it:
a) change in potential energy = -Workdone
= -F.d
= -F*d*cos(0)
= -q*E*d
= -1.6*10^-19*1.49*10^3*(0.0496 - (-0.025))
= -1.78*10^-17 J
b) change in potential energy = -Workdone
= -F.d
= -F*d*cos(180)
= q*E*d
= 1.6*10^-19*1.49*10^3*(0.118 - (-0.025))
= 3.41*10^-17 J
c)
change in potential energy = -Workdone
= -F.d
= -F*d*cos(0)
= -q*E*d
= -1.6*10^-19*1.49*10^3*(0.074 - 0.031)
= -1.03*10^-17 J
Excercise :
change in potential energy = -Workdone
= -F.d
= -F*d*cos(0)
= -q*E*d
= -1.6*10^-19*1.49*10^3*(0.118 - (-0.018))
= -3.24*10^-17 J
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