Answer to Question #123458 in Physics for mohammed akram

Question

A proton (q=1.6x10 -19 C) moves 2.0 x 10 -6 m in the direction of electric field that has a magnitude of 2.0 N/C. What is the change in the electrical potential energy associated with the proton?

Answer 1

The change in the electrical potential energy of the proton

$$\Delta W_p=qEd\\ =1.6\times 10^{-19}\times 2.0\times 2.0\times 10^{-6}=6.4\times 10^{-25}\;\rm J$$