Question

Electrons are continuously being knocked out of air molecules by cosmic ray particles from space. Once the electrons are free, they experience an electric force, F, caused by an electric field, E, that’s in the atmosphere due to charged particles that are already on Earth. If the electric field near Earth is 150 N/C and directed downward, what is change in potential energy, ΔU, when the force causes the electron to move upward through a distance of d=520 m? Through what potential difference does the electron move?

Answer 1

(a) $1.25\cdot 10^{-14} J$

The change in potential energy of the electron is given by:

$\Delta U=q E d$

where

$q=1.6\cdot 10^{-19}C$ is the magnitude of the electron's charge

$E=150 N/C$ is the magnitude of the electric field

d = 520 m is the distance through which the electron has moved

Substituting into the equation, we find

$\Delta U=(1.6\cdot 10^{-19}C)(150 N/C)(520 m)=1.25\cdot 10^{-14} J$

(b) 78 kV

The potential difference the electron has moved through is given by

$\Delta V=Ed$

where

$E=150 N/C$ is the magnitude of the electric field

d = 520 m is the distance through which the electron has moved

Substituting into the equation, we find

$\Delta V=Ed=(150 N/C)(520 m)=78,000 V=78 kV$