This problem involves analyzing the motion of an electron in an electric field, focusing on its trajectory and the angle it makes after passing through the plates. Here's the solution:

Given:

1. Initial acceleration voltage (horizontal motion): $V_{\text{acc}} = 5510 \, \text{V}$
2. Distance between plates: $d = 1.30 \, \text{cm} = 0.0130 \, \text{m}$
3. Length of plates: $L = 6.50 \, \text{cm} = 0.0650 \, \text{m}$
4. Potential difference across the plates (vertical motion): $V_{\text{plates}} = 250.0 \, \text{V}$
5. Charge of an electron: $e = 1.60 \times 10^{-19} \, \text{C}$
6. Mass of an electron: $m = 9.11 \times 10^{-31} \, \text{kg}$

Solution:

1. Horizontal Velocity ($v_x$)

The electron gains horizontal velocity due to the initial acceleration from rest through the $5510 \, \text{V}$ potential difference. Using the energy conservation principle:

$\frac{1}{2} m v_x^2 = e V_{\text{acc}}$

$v_x = \sqrt{\frac{2 e V_{\text{acc}}}{m}}$

Substitute the known values:

$v_x = \sqrt{\frac{2 (1.60 \times 10^{-19}) (5510)}{9.11 \times 10^{-31}}}$

$v_x \approx 4.39 \times 10^7 \, \text{m/s}$

2. Electric Field ($E_y$)

The electric field between the plates is given by:

$E_y = \frac{V_{\text{plates}}}{d}$

Substitute the values:

$E_y = \frac{250.0}{0.0130} \approx 1.92 \times 10^4 \, \text{N/C}$

3. Vertical Acceleration ($a_y$)

The vertical acceleration is caused by the electric field. Using $F = ma$ and $F = eE_y$:

$a_y = \frac{e E_y}{m}$

Substitute the values:

$a_y = \frac{(1.60 \times 10^{-19}) (1.92 \times 10^4)}{9.11 \times 10^{-31}}$

$a_y \approx 3.37 \times 10^{15} \, \text{m/s}^2$

4. Time in the Plates ($t$)

The electron spends time $t$ traveling through the plates, determined by its horizontal velocity:

$t = \frac{L}{v_x}$

Substitute the values:

$t = \frac{0.0650}{4.39 \times 10^7} \approx 1.48 \times 10^{-9} \, \text{s}$

5. Vertical Velocity ($v_y$)

The vertical velocity is given by:

$v_y = a_y t$

Substitute the values:

$v_y = (3.37 \times 10^{15})(1.48 \times 10^{-9})$

$v_y \approx 4.99 \times 10^6 \, \text{m/s}$

6. Final Angle ($\theta$)

The angle of the trajectory is:

$\theta = \tan^{-1}\left(\frac{v_y}{v_x}\right)$

Substitute the values:

$\theta = \tan^{-1}\left(\frac{4.99 \times 10^6}{4.39 \times 10^7}\right)$

$\theta \approx \tan^{-1}(0.1137) \approx 6.50^\circ$

Final Answer:

The electron will be traveling at an angle of approximately $\theta = 6.50^\circ$ after it passes between the plates.

Would you like a breakdown of any specific step or additional clarification?

Expanding Questions:

1. What is the significance of neglecting gravity in this problem?
2. How would the result change if the horizontal velocity $v_x$ were higher or lower?
3. Can we calculate the trajectory of the electron graphically?
4. What is the role of the electron's charge in determining its motion?
5. How does the distance between the plates affect the angle of the electron's motion?

Tip:

In problems involving electric fields and charged particles, always ensure units are consistent (meters, volts, etc.) to avoid errors in calculation.