Explanation:
An electron is released from rest, u = 0
We know that charge per unit area is called the surface charge density i.e. [tex]\sigma=\dfrac{q}{A}=2.1\times 10^{-7}\ C/m^2[/tex]
Distance between the plates, [tex]d=1.2\times 10^{-2}\ m[/tex]
Let E is the electric field,
[tex]E=\dfrac{\sigma}{\epsilon_o}[/tex]
[tex]E=\dfrac{2.1\times 10^{-7}}{8.85\times 10^{-12}}[/tex]
E = 23728.81 N/C
Now, [tex]ma=qE[/tex]
[tex]a=\dfrac{qE}{m}[/tex]
[tex]a=\dfrac{1.6\times 10^{-19}\times 23728.81}{9.1\times 10^{-31}}[/tex]
[tex]a=4.17\times 10^{15}\ m/s^2[/tex]
Let v is the speed of the electron just before it reaches the positive plate. So, third equation of motion becomes :
[tex]v^2=2ad[/tex]
[tex]v^2=2\times 4.17\times 10^{15}\times 1.2\times 10^{-2}[/tex]
[tex]v=10.003\times 10^6\ m/s[/tex]
Hence, this is the required solution.
