Answer:
The electric field strength is [tex] E=159250N/C[/tex]
Explanation:
Because the electric field is uniform and the electron is moving parallel to it, then the net force is constant so the acceleration (a) is constant too, that allow us to use the next kinematic equation to find the acceleration of the electron.
[tex]v^{2}=v_{0}^{2}+2a\varDelta x [/tex]
with v the final velocity, v0 the initial velocity and Δx the distance
[tex]a=\frac{v^{2}-v_{0}^{2}}{2\Delta x}=\frac{(4.0\times10^{7})^{2}-(2.0\times10^{7})^{2}}{2(0.021m)} [/tex]
[tex]a=2.8\times10^{16} \frac{m}{s^{2}} [/tex]
With the acceleration we can apply Newton's second Law that relates acceleration and net force (F):
[tex]F=ma [/tex]
with m the mass of the electron ([tex]9.1\times10^{-31}kg [/tex]).
So, the force on the moving electron is:
[tex]F=(9.1\times10^{-31})(2.8\times10^{16}) [/tex]
[tex] F=2.5\times10^{-14} N[/tex]
Finally knowing that electric force and electric field are related with the equation [tex] E=\frac{F}{e}[/tex]
with e the charge of the electron
Using the calculated value of the force and the charge of the electron ([tex]1.6\times10^{-19}C [/tex]):
[tex]E=\frac{2.5\times10^{-14}}{1.6\times10^{-19} } [/tex]
[tex] E=159250N/C[/tex]
