Answer
$t = 2.85\times 10^{-8}~s$
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 time $t$ when the velocity is $0$:
$v = v_0+at$
$t = \frac{v - v_0}{a}$
$t = \frac{0 - (5.00\times 10^6~m/s)}{-1.7565\times 10^{14}~m/s^2}$
$t = 2.85\times 10^{-8}~s$
