Answer
The electron will travel a distance of $~~7.1~cm~~$ before stopping momentarily.
Work Step by Step
We can find the magnitude of the force on the electron:
$F = q~E$
$F = (1.6\times 10^{-19}~C)(1.00\times 10^3~N/C)$
$F = 1.6\times 10^{-16}~N$
We can find the magnitude of deceleration:
$F = ma$
$a = \frac{F}{m}$
$a = \frac{1.6\times 10^{-16}~N}{9.109\times 10^{-31}~kg}$
$a = 1.7565\times 10^{14}~m/s^2$
We can find the distance the electron travels until the velocity is $0$:
$v^2 = v_0^2+2ax$
$x = \frac{v^2 - v_0^2}{2a}$
$x = \frac{0 - (5.00\times 10^6~m/s)^2}{(2)(-1.7565\times 10^{14}~m/s^2)}$
$x = 0.071~m$
$x = 7.1~cm$
The electron will travel a distance of $~~7.1~cm~~$ before stopping momentarily.
