HW2 Spring 16 CA Model Of A Red B C b www.webassign net t/web/Student/Assig dep
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HW2 Spring 16 CA Model Of A Red B C b www.webassign net t/web/Student/Assig dep 9255 115 points I Previous Answers SerCP10 16 AE.001 EXAMPLE 16.1 Potential Energy Differences in an Electric Field 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 x 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 12.0 cm? (c) If the direction of the electric field is reversed and an electron is released reaches x 7.00 cm? STRATEGY This problem requires a straightforward substitution of given values into the definition of electric potential energy SOLUTION Calculate the change in the electric potential energy associated with the proton Apply the definition of electric potential energy (1.60 x 10 19 C)(1.50 x 103 N/C) x [0.0500 m (-0.0200 m)] 1.68 x 101 J reaching x 0.120 m Apply the definition of electric potential energy, but in this case note that the (-1.60 x 10-19 c)(1.50 x 102 N/c) x Co.120 m (-o.0200 m)] electric charge a is negative +3.36 x 10 i (C) Find the change in potential energy associated with an electron traveling from x 3.00 cm to 7.00 cm if the direction of the electric field is reversed Substitute, but now the electric field points in the negative x-direction, (-1.60 x 10-15 c)(-1.50 x 103 Nyc) x (0.07o m 0.030 m)] hence carries a minus sign -9.60 x 101 LEARN MORE 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 9:10 PM 2/10/2016Explanation / Answer
PRACTICE IT:
a)
delta PE = - q*Ex*(xf - xi)
= - (1.60*10^-19)*(1.50*10^3)*(0.0464 - (-0.0220))
= -1.64*10^-17 J
b)
delta PE = - q*Ex*(xf - xi)
= - (-1.60*10^-19)*(1.50*10^3)*(0.122 - (-0.0220))
= 3.5*10^-17 J
c)
delta PE = - q*Ex*(xf - xi)
= - (-1.60*10^-19)*(-1.50*10^3)*(0.066 -0.034)
= -7.7*10^-18 J
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