(a) The acceleration of the electron due to the field a= 3.5*10^(15) m/s^2
(b) The speed of the electron will be v=17.5*10^(5) m/s
Acceleration is defined as the change of the velocity with the time. Acceleration is a vector quantity and is defined by both the magnitude and the direction.
We know that the force act on the electron in the electric field is given as
F=qE=ma
Here,
q= charge on electron = 1.6*10^-19 C
E = Electric field = 2*10^4 N/m
m= Mass of electron= 9.1*10^-31 kg
a= acceleration of electron
(a) The acceleration can be given as:
[tex]a=\dfrac{qE}{m}[/tex]
[tex]a=\dfrac{1.6\times 10^{-19}\times 2\times 10^{4}}{9.1\times 10^{-31}}[/tex]
[tex]a=3.5\times 10^{15}\ \frac{m}{s}[/tex]
(b) The velocity of the electron will be given as
v= final velocity
u=initial velocity=0
t= time suppose=5 sec
Now from the formula
v=u+at
[tex]v=0+3.5\times 10^{15} \times 5[/tex]
[tex]v=17.5\times 10^{15}\ \dfrac{m}{s}[/tex]
Hence
(a) The acceleration of the electron due to the field a= 3.5*10^(15) m/s^2
(b) The speed of the electron will be v=17.5*10^(5) m/s
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