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

Answer:

$2.7\times 10^{12}$

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=$1.6\times 10^{-19}C$

Using the formula

Upward electric force=$152\times 1.6\times 10^{-19}$ N

Upward electric force=$F_e=2.43\times 10^{-17}$ N

Mass of electron=$m_e=9.1\times 10^{-31} kg$

$g=9.8m/s^2$

Gravitational force=$F=mg$

Using the formula

Gravitational force=$F_g=9.1\times 10^{-31}\times 9.8=8.92\times 10^{-30} N$

Ratio of Fe to the Fg=$\frac{2.43\times 10^{-17}}{8.92\times 10^{-30}}$

$\frac{F_e}{F_g}=2.7\times 10^{12}$