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 4
 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?
 Homework 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?
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?