Answer
At $t = 2.30~ns$:
$x = -1.08\times 10^{-4}~m$
$y=0$
Work Step by Step
Since the direction of the electric field is in the positive x-direction, the force exerted on the electron is in the negative x-direction. We can find the force exerted on the electron:
$F = E~q = (232~N/C)(-1.6\times 10^{-19}~C) = -3.712\times 10^{-17}~N$
We can find the acceleration:
$a = \frac{F}{m} = \frac{-3.712\times 10^{-17}~N}{9.109\times 10^{-31}~kg} = -4.075\times 10^{13}~m/s^2$
We can find the x-coordinate at $t = 2.30~ns$:
$x = \frac{1}{2}at^2$
$x = \frac{1}{2}(-4.075\times 10^{13}~m/s^2)(2.30\times 10^{-9}~s)^2$
$x = -1.08\times 10^{-4}~m$
Since there is no electric force in the y-direction, the y-component of the position is zero.
At $t = 2.30~ns$:
$x = -1.08\times 10^{-4}~m$
$y=0$
