Question

Usually the force of gravity on electrons is neglected. To see why, we can compare the force of the Earth’s gravity on an electron with the force exerted on the electron by an electric field of magnitude of 40000 V/m (a relatively small field). What is the force exerted on the electron by an electric field of magnitude of 40000 V/m? The acceleration of gravity is 9.8 m/s 2 , the mass of an electron is 9.10939 × 10−31 kg, and the elementary charge 1.602 × 10−19 C. Answer in units of N.

Answer 1

Answer:

$6.4\cdot 10^{-15} N$

Explanation:

The electric force exerted on the electron is given by:

$F=qE$

where

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

E = 40000 V/m is the electric field

Substituting,

$F=(1.6\cdot 10^{-19} C)(40000 V/m)=6.4\cdot 10^{-15} N$

By comparison, the gravitational force exerted on the electron is:

$F=mg$

where

$m=9.10939\cdot 10^{-31} kg$ is the mass of the electron

g = 9.8 m/s^2 is the acceleration due to gravity

Substituting,

$F=(9.10939\cdot 10^{-31} kg)(9.8 m/s^2)=8.93\cdot 10^{-30}N$