AP Physics C E and M

Question

An electron moves 0.10 m along the direction of an electric field of a magnitude of 3.0 N/C. What is the change in electric potential energy between the electron’s initial and final points?

Answer 1

Let

s = 0.1 m be the distance travelled by the electron,

E = 3.0 N/C be the magnitude of the electric field,

q = -1.6×10^-19 C be the charge of the electron,

F (unknown) be the force with which the electric field acts on the electron,

ΔU (to be calculated) be the difference in potential energy.

First let's figure out whether ΔU is positive or negative.

The direction of an electrical field is defined as away from a positive charge.

The statement of the problem says that the (negatively charged) electron

moves ALONG the direction of the electrical field.

That mean it moves against the force acting on it, hence will gain

extra potential energy;

that is, the finally calculated change in potential energy ΔU is positive.

Having determinized the sign, we can calculate the magnitude of the

potential energy difference, |ΔU|.

|ΔU| = |F|s = |Eq|s

Substituting actual numbers

ΔU = |ΔU| = 3.0×1.6×10^-19×0.1 = 4.8×10^-19 J