Answer:
Acceleration,[tex]a=5.27\times 10^{17}\ m/s^2[/tex]
Explanation:
Given that,
Electric field strength, [tex]E=3\times 10^6\ N/C[/tex]
Mass of the electron, [tex]m=9.109\times 10^{-31}\ kg[/tex]
Charge on electron, [tex]q=1.602\times 10^{-19}\ C[/tex]
Let a is the acceleration experienced by an electron. It can be calculated as :
[tex]ma=qE[/tex]
[tex]a=\dfrac{qE}{m}[/tex]
[tex]a=\dfrac{1.602\times 10^{-19}\times 3\times 10^6}{9.109\times 10^{-31}}[/tex]
[tex]a=5.27\times 10^{17}\ m/s^2[/tex]
Hence, this is the required solution.