Answer:
[tex]1.8\times 105 N/C[/tex]
Explanation:
We are given that
[tex]u=2\times 10^7 m/s[/tex]
[tex]v=4\times 10^7 m/s[/tex]
d=1.9 cm=[tex]\frac{1.9}{100}=0.019 m[/tex]
Using 1m=100 cm
We have to find the electric field strength.
[tex]v^2-u^2=2as[/tex]
Using the formula
[tex](4\times 10^7)^2-(2\times 10^7)^2=2a(0.019)[/tex]
[tex]16\times 10^{14}-4\times 10^{14}=0.038a[/tex]
[tex]0.038a=12\times 10^{14}[/tex]
[tex]a=\frac{12}{0.038}\times 10^{14}=3.16\times 10^{16}m/s^2[/tex]
[tex]q=1.6\times 10^{-19} C[/tex]
Mass of electron,m[tex]=9.1\times 10^{-31} kg[/tex]
[tex]E=\frac{ma}{q}[/tex]
Substitute the values
[tex]E=\frac{9.1\times 10^{-31}\times 3.16\times 10^{16}}{1.6\times 10^{-19}}[/tex]
[tex]E=1.8\times 105 N/C[/tex]
