Answer:
[tex]\frac{F_e}{F_{g}} = 9.87 \times 10^{21}[/tex]
Explanation:
As we know that the charge on electron and proton is same as the charge of an electron
so the electrostatic force between the electron and proton is given as
[tex]F = \frac{kq_1q_2}{r^2}[/tex]
so we will have
[tex]F = \frac{(9\times 10^9)(1.6 \times 10^{-19})^2}{(5.11 \times 10^{-11})^2}[/tex]
[tex]F = 8.82 \times 10^{-8} N[/tex]
Now the weight of an electron is given as
[tex]W = mg[/tex]
[tex]W = (9.11 \times 10^{-31})(9.81)[/tex]
[tex]W = 8.94 \times 10^{-30} N[/tex]
now the ratio of electrical force and weight of the electron is given as
[tex]\frac{F_e}{F_g} = \frac{8.82 \times 10^{-8}}{8.94 \times 10^{-30}}[/tex]
[tex]\frac{F_e}{F_{g}} = 9.87 \times 10^{21}[/tex]
