Answer:
[tex]t=2.5\times 10^{-14}\ s[/tex]
Explanation:
We know that charge on electron
[tex]q=1.6\times 10^{-19}\ C[/tex]
r= 2 nm
We know that force between two charge given
[tex]F=K\dfrac{Q_1Q_2}{r^2}[/tex]
Now by putting the value
[tex]F=9\times10^9\dfrac{1.6\times 10^{-19}\times 1.6\times 10^{-19}}{(2\times 10^{-9})^2}[/tex]
[tex]F=5.67\times 10^{-11}\ N[/tex]
We know that mass of electron
The mass of electron
[tex]m=9.1\times 10^{-31}\ kg[/tex]
F= m a
a= Acceleration of electron
a= F/m
[tex]a=\dfrac{5.67\times 10^{-11}}{9.1\times 10^{-31}}\ m/s^2[/tex]
[tex]a=6.2\times 10^{19} m/s^2[/tex]
[tex]S=ut+\dfrac{1}{2}at^2[/tex]
initial velocity given that zero ,u=0
[tex]20\times 10^{-9}=\dfrac{1}{2}\times 6.2\times 10^{19} t^2[/tex]
[tex]t=\sqrt {\dfrac{40\times 10^{-9}}{6.2\times 10^{19}}}[/tex]
[tex]t=2.5\times 10^{-14}\ s[/tex]
