Question

Bare free charges do not remain stationary when close together. To illustrate this, calculate the acceleration of two isolated protons separated by 3.73 nm (a typical distance between gas atoms). (Enter the magnitude in m/s2.)

Answer 1

Acceleration,
`a=9.91* 10^(15)\ m/s^2`

Step-by-step explanation:

It is given that,

Separation between the protons,
`r=3.73\ nm=3.73* 10^(-9)\ m`

Charge on protons,
`q=1.6* 10^(-19)\ C`

Mass of protons,
`m=1.67* 10^(-27)\ kg`

We need to find the acceleration of two isolated protons. It can be calculated by equating electric force between protons and force due to motion as :

`ma=(kq^2)/(r^2)`

`a=(kq^2)/(mr^2)`

`a=(9* 10^9* (1.6* 10^(-19))^2)/(1.67* 10^(-27)* (3.73* 10^(-9))^2)`

`a=9.91* 10^(15)\ m/s^2`

So, the acceleration of two isolated protons is
`9.91* 10^(15)\ m/s^2`. Hence, this is the required solution.