Respuesta :
Answer : [tex]4.483\times 10^{15}\ m/s^2[/tex].
Explanation:
It is given that,
Electric field strength, [tex]E=2.55\times 10^{4}\ N/C[/tex]
We know that,
Charge of electron, [tex]q=1.6\times 10^{-19}\ C[/tex]
Mass of electron, [tex]m=9.1\times 10^{-31}\ kg[/tex]
From the definition of electric field, [tex]F=qE[/tex]...............(1)
According to Newton's second law, F = ma..........(2)
From equation (1) and (2)
[tex]ma=qE[/tex]
[tex]a=\dfrac{qE}{m}[/tex]
[tex]a=\dfrac{1.6\times 10^{-19}\ C\times 2.55\times 10^{4}\ N/C }{9.1\times 10^{-31}\ kg}[/tex]
[tex]a=0.4483\times 10^{16}\ m/s^2[/tex]
or
[tex]a=4.483\times 10^{15}\ m/s^2[/tex]
So, the horizontal component of acceleration of an electron is [tex]4.483\times 10^{15}\ m/s^2[/tex].
Hence, it is the required solution.
