Question

It is known that the gravitational force of attraction between two protons is much weaker than the electrical repulsion. For two protons at a distance d apart, calculate the ratio of the size of the gravitational attraction to that of the electrical repulsion. Specifically, find the magnitude of Fgravitational/Felectrical. Use the following data: k = 8.99×109 Nm2/C2 charge on one of the protons = 1.60×10-19 C G = 6.67×10-11 Nm2/kg2 mass of one of the protons = 1.67×10-27 kg.

Answer 1

`(F_g)/(F_e)=8.08* 10^(-37)`

Step-by-step explanation:

We are given that

`k=8.99* 10^9 Nm^2/C^2`

Charge on proton=
`q=1.6* 10^(-19) C`

`G=6.67* 10^(-11) Nm^2/Kg^2`

Mass of one proton=
`m=1.67* 10^(-27) Kg`

The distance between two protons=d

Gravitational attraction force=
`F_g=(Gm_1m_2)/(d^2)`

Using the formula

Gravitational attraction force between two protons=
`F_g=(6.67* 10^(-11)* (1.67* 10^(-27))^2)/(d^2)`..(1)

Electric force =
`F_e=(kq_1q_2)/(d^2)`

Electric repulsive force between two protons=
`F_e=(8.99* 10^9* (1.6* 10^(-19))^2)/(d^2)`

`(F_g)/(F_e)=((6.67* 10^(-11)* (1.67* 10^(-27))^2)/(d^2))/((8.99* 10^9* (1.6* 10^(-19))^2)/(d^2))`

`(F_g)/(F_e)=(6.67* (1.67* 10^(-27))^2)/(8.99* 10^9* (1.6* 10^(-19))^2)`

`(F_g)/(F_e)=8.08* 10^(-37)`