Question

What is the electric force (with direction) on an electron in a uniform electric field of strength 3400 N/C that points due east? Take the positive direction to be east.

Answer 1

Answer:

• So, the force its $\ 5.4468 \ 10 ^{-16}$ N to the west.

Explanation:

The force $\vec{F}$ on a charge q made by an electric field $\vec{E}$ its

$\vec{F} = q \vec{E}$

The electric charge of the electron its

$q \ = \ - \ 1.602 \ 10 ^{-19} \ C$.

Taking the unit vector $\hat{i}$ pointing towards the east, the electric field will be:

$\vec{E}= 3400 \ \frac{N}{C} \ \hat{i}$.

So, the force will be:

$\vec{F} = \ - \ 1.602 \ 10 ^{-19} \ C \ * \ 3400 \ \frac{N}{C} \ \hat{i}$

$\vec{F} = \ - \ 5446.8 \ 10 ^{-19} \ N \ \hat{i}$

$\vec{F} = \ - \ 5.4468 \ 10 ^{-16} \ N \ \hat{i}$

$\vec{F} = \ - \ 5.4468 \ 10 ^{-16} \ N \ \hat{i}$

$\vec{F} = \ - \ 5.4468 \ 10 ^{-16} \ N \ \hat{i}$

So, the force its $\ 5.4468 \ 10 ^{-16}$ N to the west.