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

• So, the force its
  `\ 5.4468 \ 10 ^(-16)` N to the west.

Step-by-step 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 \ (N)/(C) \ \hat{i}`.

So, the force will be:

`\vec{F} =  \ - \ 1.602 \ 10 ^(-19) \ C \ * \ 3400 \ (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.