Explanation:
It is given that,
An electron is released from rest in a weak electric field of, [tex]E=2.3\times 10^{-10}\ N/C[/tex]
Vertical distance covered, [tex]s=1\ \mu m=10^{-6}\ m[/tex]
We need to find the speed of the electron. Let its speed is v. Using third equation of motion as :
[tex]v^2-u^2=2as[/tex]
[tex]v^2=2as[/tex].............(1)
Electric force is [tex]F_e[/tex] and force of gravity is [tex]F_g[/tex]. As both forces are acting in downward direction. So, total force is:
[tex]F=mg+qE[/tex]
[tex]F=9.1\times 10^{-31}\times 9.8+1.6\times 10^{-19}\times 2.3\times 10^{-10}[/tex]
[tex]F=4.57\times 10^{-29}\ N[/tex]
Acceleration of the electron, [tex]a=\dfrac{F}{m}[/tex]
[tex]a=\dfrac{4.57\times 10^{-29}\ N}{9.1\times 10^{-31}\ kg}[/tex]
[tex]a=50.21\ m/s^2[/tex]
Put the value of a in equation (1) as :
[tex]v=\sqrt{2as}[/tex]
[tex]v=\sqrt{2\times 50.21\times 10^{-6}}[/tex]
v = 0.010 m/s
So, the speed of the electron is 0.010 m/s. Hence, this is the required solution.