Question

A proton is released from rest at x = -2.9 cm in a constant electric field with magnitude 1.70
x104 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 = 5.9 cm. (b) An electron is now fired in the
same direction from the same position. What is the change in electric potential energy associated
with the electron if it reaches x = 18.0 cm? (c) If the direction of the electric field is reversed and
an electron is released from rest at x = 4.00 cm, by how much has the electric potential energy
changed when the electron reaches x= 9.00 cm?

Answer 1

As $\phi_b-\phi_a=-E_x(x_b-x_a)$ then $\Delta U=-qE_x(x_b-x_a)$ hence

(a) $\Delta U=-1.6\sdot10^{-19}\sdot1.7\sdot10^4(5.9+2.9)\sdot10^{-2}\approx-2.4\sdot10^{-16}J$

(b) $\Delta U=-(-1.6\sdot10^{-19})\sdot1.7\sdot10^4(18+2.9)\sdot10^{-2}\approx5.7\sdot10^{-16}J$

(c) $\Delta U=-(-1.6\sdot10{-19})\sdot(-1.7\sdot10^4)(9-4)\sdot10^{-2}\approx-1.4\sdot10^{-16}J$