Answer:
Acceleration, [tex]a=1.12\times 10^{14}\ m/s^2[/tex]
Step-by-step explanation:
We have,
Initial speed of electron, u = 0
Final speed of electron, [tex]v=3\times 10^6\ m/s[/tex]
Distance, d = 0.04 m
It is required to find the average acceleration of the electron. Let it is a. Using equation of motion as :
[tex]v^2-u^2=2ad[/tex]
a = average acceleration
[tex]v^2=2ad\\\\a=\dfrac{v^2}{2d}\\\\a=\dfrac{(3\times 10^6)^2}{2\times 0.04}\\\\a=1.12\times 10^{14}\ m/s^2[/tex]
So, the average acceleration of the electron is [tex]1.12\times 10^{14}\ m/s^2[/tex].
