Respuesta :
Answer:
The speed of electron is [tex]8.7\times10^{6}\ m/s[/tex]
Explanation:
Given that,
Separation of the plate = 1.20 cm
Suppose the field is [tex]E=1.80\times10^{4}\ N/C[/tex].
If the electron is accelerated from rest near the negative plate and passes through a tiny hole in the positive plate.
What the speed does it leave the hole?
We need to calculate the acceleration
Using formula of electric force
[tex]F = qE[/tex]
[tex]ma=qE[/tex]
[tex]a=\dfrac{qE}{m}[/tex]
We need to calculate the speed of electron
Using equation of motion
[tex]v^2=u^2+2as[/tex]
[tex]v^2=2as[/tex]
Put the value of acceleration in the formula
[tex]v^2=2\times\dfrac{qE}{m}\times s[/tex]
Put the value into the formula
[tex]v^2=2\times\dfrac{1.6\times10^{-19}\times1.80\times10^{4}\times1.20\times10^{-2}}{9.11\times10^{-31}}[/tex]
[tex]v=\sqrt{2\times\dfrac{1.6\times10^{-19}\times1.80\times10^{4}\times1.20\times10^{-2}}{9.11\times10^{-31}}}[/tex]
[tex]v=8.7\times10^{6}\ m/s[/tex]
Hence, The speed of electron is [tex]8.7\times10^{6}\ m/s[/tex]
