How far should an electron be fired so it stops before reaching the plate?

[Fontseeker]

Homework Statement

In the figure below an electron is shot directly toward the center of a large metal plate that has surface charge density -1.20 × 10^-6 C/m2. If the initial kinetic energy of the electron is 1.60 × 10^-17 J and if the electron is to stop (due to electrostatic repulsion from the plate) just as it reaches the plate, how far from the plate must the launch point be?

Relevant Equations

F=ma
E=(1/2)mV^2

I am trying to solve the following problem:

My attempt at the problem:

$$m = 9.11 * 10^{-31}kg$$
$$q = 1.6 * 10^{-19} C$$

$$V_0 = \sqrt{\frac{2KE}{m}} = \sqrt{\frac{2 * 1.6 * 10^{-17}}{1.6 * 10^{-19}}} = 5,926,739 m/s$$
$$F_{net} = -F_{e} = ma$$
$$a = \frac{-F_{e}}{m} = ma$$
$$E = \frac{\phi}{2\epsilon_0}$$
$$F_e = \frac{E}{q} = \frac{\phi}{2\epsilon_0q}$$
$$a = -\frac{q\phi}{2m\epsilon_0} = \frac{1.6 * 10^{-19} * -1.20 * 10^{-6}}{2 * 9.11 * 10^{-31} * 8.85 * 10^{-12}} = -1.191 * 10^{16} m/s^2$$
$$V_f^2 = V_0^2 + 2ad$$
$$V_f = 0$$
$$-2ad = V_0^2$$
$$d = \frac{V_0^2}{2a} = -\frac{5,926,739}{2 * -1.191 * 10^{16}} = 0.001475m$$

I got that the answer is 0.001475m, but this is not correct. Does anyone know what I am doing wrong?

 

[haruspex]

What is $\frac{10^{-17}}{10^{-19}}$?
But you are making work for yourself in finding the velocity. Just deal with energy.

 

[Fontseeker]

What is $\frac{10^{-17}}{10^{-19}}$?
But you are making work for yourself in finding the velocity. Just deal with energy.

that's a typo, should be:
$$\sqrt{\frac{2 * 1.6 * 10^{-17}}{9.11 * 10^{-31}}} = 5,936,739$$

I've dealt with energy and I get the same thing. Either way, it should work with velocity.

 

[haruspex]

that's a typo, should be:
$$\sqrt{\frac{2 * 1.6 * 10^{-17}}{9.11 * 10^{-31}}} = 5,936,739$$

I've dealt with energy and I get the same thing. Either way, it should work with velocity.

Yes, I'm getting the same answer.
Do you know what the supposed answer is?