- #1

Fontseeker

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- 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?

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?