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 10^3 N/C, pointing in the positive x-direction. (a) Calculate the change in potential energy when the proton reaches x = 4.96 cm. The response you submitted has the wrong sign. J (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 = 13.00 cm? Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. J (c) If the direction of the electric field is reversed and an electron is released from rest at x = 3.80 cm, by how much has its electric potential energy changed when it reaches x = 7.20 cm? The response you submitted has the wrong sign. J

Answer 1

(1)

del PE = -q Ex del x

= - (1.6 * 10^-19) ( 1.50 * 10^3 )( 0.0496 m - ( -0.0290)

=-1.88 * 10^-17 J

(2)

del PE = -q Ex del x

= - (-1.6 * 10^-19) ( 1.50 * 10^3 )( 0.13 m - ( -0.0290)

=3.81 * 10^-17 J

(3)

del PE = -q Ex del x

= - (-1.6 * 10^-19) (- 1.50 * 10^3 )( 0.072m - ( 0.0380)

=-8.16 * 10^-18 J