Question

An electron moves 5 m in the direction of an electric field of strength 300 N/C. The change in electrical potential energy is:

Answer 1

First we need to find the voltage difference between the initial and final location of the electron.
Since the electron travelled for a distance d=5 m in an uniform electric field of intensity E=300 N/C, the voltage difference between the final and initial location is

`\Delta V= Ed=(300 N/C)(5 m)=1500 V`

And then, we can calculate the change in potential energy of the electron, which is the product between the charge of the electron and the voltage difference:

`\Delta U = q\Delta V=(-1.6 \cdot 10^(-19)C)(1500 V)=-2.4 \cdot 10^(-16)J`
And the negative sign is due to the fact that we assumed the electron traveled in the natural direction of the electric field, so traveling from a point at lower voltage to a point at higher voltage (the sign of
`\Delta V` is positive), so since it's a negative charge the electron is losing potential energy.