Question

The electric field near the surface of Earth points downward and has a magnitude of 152 N/C. What is the ratio of the magnitude of the upward electric force on an electron to the magnitude of gravitational force on the electron?

Answer 1

`2.7* 10^(12)`

Step-by-step explanation:

We are given that

Electric field =
`E=152N/C`

We have to find the ratio of magnitude of the upward electric force on an electron to the magnitude of gravitational force on the electron.

We know that

F=qE

Charge on an electron,q=
`1.6* 10^(-19)C`

Using the formula

Upward electric force=
`152* 1.6* 10^(-19)` N

Upward electric force=
`F_e=2.43* 10^(-17)` N

Mass of electron=
`m_e=9.1* 10^(-31) kg`

`g=9.8m/s^2`

Gravitational force=
`F=mg`

Using the formula

Gravitational force=
`F_g=9.1* 10^(-31)* 9.8=8.92* 10^(-30) N`

Ratio of Fe to the Fg=
`(2.43* 10^(-17))/(8.92* 10^(-30))`

`(F_e)/(F_g)=2.7* 10^(12)`