Question

Two electrons are initially at rest separated by a distance of 2nm. At time t=0, they start to move apart due to Coulombic repulsion. How long does it take for them to separate to a distance of 20nm?

Answer 1

`t=2.5* 10^(-14)\ s`

Step-by-step explanation:

We know that charge on electron

`q=1.6* 10^(-19)\ C`

r= 2 nm

We know that force between two charge given

`F=K(Q_1Q_2)/(r^2)`

Now by putting the value

`F=9*10^9(1.6* 10^(-19)* 1.6* 10^(-19))/((2* 10^(-9))^2)`

`F=5.67* 10^(-11)\ N`

We know that mass of electron

The mass of electron

`m=9.1* 10^(-31)\ kg`

F= m a

a= Acceleration of electron

a= F/m

`a=(5.67* 10^(-11))/(9.1* 10^(-31))\ m/s^2`

`a=6.2* 10^(19) m/s^2`

`S=ut+(1)/(2)at^2`

initial velocity given that zero ,u=0

`20* 10^(-9)=(1)/(2)* 6.2* 10^(19) t^2`

`t=\sqrt {(40* 10^(-9))/(6.2* 10^(19))}`

`t=2.5* 10^(-14)\ s`