Question

Exercise 15.1 (a) Find the magnitude of the electric force between two protons separated by 1 femtometer (10^-15), approximately the distance between two protons in the nucleus of a helium atom. (b) If the protons were not held together by the strong nuclear force, what would be their initial acceleration due to the electric force between them

Answer 1

(a) 230.4 N

(b)
`a = 1.38*10^(29) m/s^(2)`

Step-by-step explanation:

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

Distance between the two protons, d = 1 femto meter =
`1*10^(-15)m`

(a) By use of coulomb's law, the force between the two protons is given by

`F =(k* q* q)/(d^(2))`

Where, k is the Coulombic constant = 9x 10^9 Nm^2/C^2

So, the force between them is given by

`F =(9*10^(9)* 1.6*10^(-19)* 1.6*10^(-19))/(\left (1*10^(-15)  \right )^(2))`

F = 230.4 N

(b) Acceleration is the ratio of force to the mass of proton.

the mass of proton, m = 1.67 x 10^-27 kg

So, acceleration

`a=(230.4)/(1.67*10^(-27))`

`a = 1.38*10^(29) m/s^(2)`