Question

Two protons are 9.261 fm apart. (1 fm= 1 femtometer = 1 x 10^-15 m.) What is the ratio of the electric force to the gravitational force on one proton due to the other proton?

Answer 1

`F_e:F_g = 1:1.237 * 10^(36)`

Step-by-step explanation:

Parameters given:

Distance between the protons, r =
`1 * 10^(-15) m`

Electric charge of proton, Q =
`1.602 * 10^(-19) C`

Mass of proton, m =
`1.673 *10^(-27) kg`

The electric force on one proton due to the other given as:

`F_e = (k*Q*Q)/(r^2)`

where k = Coulomb's constant

`F_e = (9 * 10^9 * 1.602 * 10^(-19) * 1.602 * 10^(-19) )/((1 * 10^(-15))^2) \\\\\\F_e = 230.98 N`

The gravitational force on one proton due to the other is given as:

`F_g = (G*m*m)/(r^2)`

where G = gravitational constant

`F_g = (6.67408 * 10^(-11)*1.673 *10^(-27) * 1.673 *10^(-27))/((1 * 10^(-15))^2) \\\\\\F_g = 1.868 * 10^(-34) N`

Therefore, the ratio of electric force,
`F_e`, and gravitational force,
`F_g`, is:

`(F_e)/(F_g) = (230.98)/(1.868 * 10^(-34)) = (1)/(1.237 * 10^(36)) \\\\\\F_e:F_g = 1:1.237 * 10^(36)`

Answer 2

`1.24* 10^(36)`

Step-by-step explanation:

`q` = magnitude of charge on each proton = 1.6 x 10⁻¹⁹ C

`m` = mass of each proton = 1.67 x 10⁻²⁷ kg

r = distance between the two protons = 1 x 10⁻¹⁵ m

Electric force between the two protons is given as

`F_(e) = (kq^(2))/(r^(2))`

`F_(e) = ((9* 10^(9))(1.6* 10^(-19))^(2))/((1* 10^(-15))^(2))`

`F_(e) = 230.4` N

Gravitational force between the two protons is given as

`F_(g) = (Gm^(2))/(r^(2))`

`F_(g) = ((6.67* 10^(-11))(1.67* 10^(-27))^(2))/((1* 10^(-15))^(2))`

`F_(g) = 1.86* 10^(-34)` N

Ratio is given as

`Ratio =(F_(e))/(F_(g))`

`Ratio =(230.4)/(1.86* 10^(-34))`

`Ratio = 1.24* 10^(36)`