Question

An electron travels 0.01 m across an electric potential difference of 6 x 10^6 V. What is the force exerted on the electron? Type your answer below including the correct units

Answer 1

Answer:

$9.6\cdot 10^{-11}N$

Explanation:

The force experienced by a charged particle immersed in an electric field is

$F=qE$

where

q is the charge of the particle

E is the strength of the electric field

For a uniform field, we can write

$E=\frac{\Delta V}{d}$

where

$\Delta V$ is the potential difference

d is the distance travelled

So we can rewrite the force as

$F=\frac{q\Delta V}{d}$

In this problem:

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

$\Delta V=6\cdot 10^6 V$ is the potential difference

d = 0.01 m is the distance covered

Substituting, we find the force:

$F=\frac{(1.6\cdot 10^{-19})(6\cdot 10^6)}{0.01}=9.6\cdot 10^{-11}N$