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**1. Homework Statement**

A uniform, upward electric field

**E**of magnitude 2.00x10^3 N/C has been set up between two horizontal plates by charging the lower plate positively and the upper plate negatively. The plates have length L = 10.0 cm and separation d = 2.00 cm. An electron is then shot between the plates from the left edge of the lower plate. The initial velocity

**Vi**of the electron makes an angle theta = 45.0 degrees with the lower plate and has a magnitude of 6.00x10^6 m/s. (A) Will the electron strike one of the plates? (B) If so, which plate and how far horizontally from the left edge will the electron strike?

**2. Homework Equations**

[tex]

F = ma_{y}

[/tex]

[tex]

a_{y} = \frac{F}{m} = \frac{eE}{m}

[/tex]

**3. The Attempt at a Solution**

Okay, I think I have this. The components (X and Y) of the velocity are

[tex]

V_{x_{0}} = (6.00 * 10^6 \frac{m}{s})(\cos {\frac{\pi}{4}}) = 4.24 * 10^6 \frac{m}{s} \hat{i}

[/tex]

and

[tex]

V_{y_{0}} = (6.00 * 10^6 \frac{m}{s})(\sin {\frac{\pi}{4}}) = 4.24 * 10^6 \frac{m}{s} \hat{j}

[/tex]

I will start with finding the acceleration,

[tex]

a_{y} = \frac{F}{m} = \frac{eE}{m}

[/tex]

[tex]

a_{y} = \frac{(-1.6*10^-19 C)*(2 * 10^3 N/C}{9.109 * 10^-31 kg}

a_{y} = -3.5 * 10^14 \frac{m}{s^2}

[/tex]

Using substitution and solving for t:

[tex]

t = \frac{-2 * V_{x}}{a} = \frac{-2(4.24 * 10^6 m/s}{-3.5 * 10^14 m/s^2} = 2.4 * 10^{-8} s

[/tex]

So in that length of time, the electron will travel distance:

[tex]

\Delta X = V_{x_{o}} * t = (4.24 * 10^{6} m/s) * (2.4 * 10^{-8} s) = .10176 m

[/tex]

Since 10.176 cm is greater than the 10cm length of the plates, am I done at this point?