Answer:
[tex]v= \frac{qE}{m}t+u[/tex]
Explanation:
Given that, the electron is moving parallel to the electric field.
Then the electron experience a force which is given by,
[tex]F=qE[/tex]
Therefore,
[tex]ma=qE\\a=\frac{qE}{m}[/tex]
And q=e, and the acceleration is defines as,
[tex]a=\frac{d^{2}r }{dt^{2} }[/tex]
Here, r is the the path traced by an electron.
Now equate the acceleration.
[tex]\frac{d^{2}r }{dt^{2} }=\frac{eE}{m}[/tex]
Now integrate both side with respect to t.
[tex]\frac{dr}{dt}= \frac{eE}{m}t+C[/tex]
Now, consider at t=0 the initial velocity of electron is u.
[tex]u=\frac{eE}{m}(0)+C\\C=u[/tex]
Therefore the velocity of an electron is,
[tex]v= \frac{eE}{m}t+u[/tex]
