Answer:
0.0256 m
Explanation:
initial speed of electron, u = 3 x 10^6 m/s
Final speed of electron, v = 0 m/s
Electric field strength, E = 1 x 10^3 N/C
Charge on electron, q = 1.6 x 10^-19 C
mass of electron, m = 9.1 x 10^-31 kg
The electron experiences an electric force due to electric field which is given by
F = q E
Where, f be the electric force, and E be the strength of electric field.
Let a be the acceleration of electron
So, [tex]a= \frac{F}{m}=\frac{qE}{m}[/tex]
where, m is the mass of electron
By substituting the values, we get
[tex]a=\frac{-1.6\times10^{-19}\times10^{3}}{9.1\times10^{-31}}=-1.758 \times 10^{14}m/s^2[/tex]
Let the distance traveled by the electron before coming to rest is s.
Use third equation of motion
[tex]v^{2}=u^{2}+2as[/tex]
[tex]0^{2}=\left (3\times 10^6 \right )^{2}-2\times 1.758 \times10^{14}\times s[/tex]
s = 0.0256 m
