Respuesta :
Answer:
Electric field acting on the electron is 127500 N/C.
Explanation:
It is given that,
Mass of an electron, [tex]m=9.11\times 10^{-31}\ kg[/tex]
Charge on electron, [tex]q=1.6\times 10^{-19}\ C[/tex]
Initial speed of electron, u = 0
Final speed of electron, [tex]v=3\times 10^7\ m/s[/tex]
Distance covered, s = 2 cm = 0.02 m
We need to find the electric field required. Firstly, we will find the acceleration of the electron from third equation of motion as :
[tex]a=\dfrac{v^2-u^2}{2s}[/tex]
[tex]a=\dfrac{(3\times 10^7)^2-0}{2\times 0.02}[/tex]
[tex]a=2.25\times 10^{16}\ m/s^2[/tex]
According to Newton's law, force acting on the electron is given by :
F = ma
[tex]F=9.1\times 10^{-31}\times 2.25\times 10^{16}[/tex]
[tex]F=2.04\times 10^{-14}\ N[/tex]
Electric force is given by :
F = q E, E = electric field
[tex]E=\dfrac{F}{q}[/tex]
[tex]E=\dfrac{2.04\times 10^{-14}}{1.6\times 10^{-19}}[/tex]
E = 127500 N/C
So, the electric field is 127500 N/C. Hence, this is the required solution.
