Answer
The electron travels a distance of $~~5.0\times 10^{-5}~m~~$ during this time interval.
Work Step by Step
Since the electric field is in the same direction as the electron's direction of motion, the electrostatic force on the electron will cause it to decelerate.
We can find the magnitude of the force on the electron:
$F = \vert q \vert E$
$F = (1.6\times 10^{-19}~C)(50~N/C)$
$F = 8.0\times 10^{-18}~N$
We can find the magnitude of deceleration:
$F = ma$
$a = \frac{F}{m}$
$a = \frac{8.0\times 10^{-18}~N}{9.109\times 10^{-31}~kg}$
$a = 8.78\times 10^{12}~m/s^2$
We can find distance traveled during the interval of $1.5~ns$:
$x = v_0~t+\frac{1}{2}at^2$
$x = (4.0\times 10^4~m/s)(1.5\times 10^{-9}~s)+\frac{1}{2}(-8.78\times 10^{12}~m/s^2)(1.5\times 10^{-9}~s)^2$
$x = (6.0\times 10^{-5}~m)-(9.8775\times 10^{-6}~m)$
$x = 5.0\times 10^{-5}~m$
The electron travels a distance of $~~5.0\times 10^{-5}~m~~$ during this time interval.
