Answer:
[tex]6.4\times 10^5 m/s[/tex]
Explanation:
We are given that
Electric field,E=1.6 N/C
Initial velocity,u=[tex]4.5\times 10^5 m/s[/tex]
We have to find the speed of electron when it reaches point B 0.375 m east of point A.
Charge on electron,q=[tex]1.6\times 10^{-19} C[/tex]
[tex]a=\frac{eE}{m}[/tex]
Where [tex]m=9.1\times 10^{-31} kg[/tex]
Using the formula
[tex]a=\frac{1.6\times 10^{-19}\times 1.6}{9.1\times 10^{-31}}=2.8\times 10^{11}m/s^2[/tex]
s=0.375 m
[tex]v^2=u^2+2as[/tex]
[tex]v=\sqrt{u^2+2as}[/tex]
[tex]v=\sqrt{(4.5\times 10^5)^2+2(2.8\times 10^{11})(0.375)}[/tex]
[tex]v=6.4\times 10^5 m/s[/tex]
