To solve the problem, we need to consider the forces acting on the electron as it passes between the deflection plates and calculate how much it is deflected by the time it reaches the screen $S$.

Given Data:

• Initial speed of the electron: $v_0 = 6.50 \times 10^6 \, \text{m/s}$
• Potential difference between the plates: $V = 18.0 \, \text{V}$
• Distance between the plates: $d = 2.0 \, \text{cm} = 0.02 \, \text{m}$
• Length of the plates: $L = 6.0 \, \text{cm} = 0.06 \, \text{m}$
• Distance from the plates to the screen: $D = 12.0 \, \text{cm} = 0.12 \, \text{m}$

Steps to Solution:

1. Electric Field between Plates: The electric field $E$ between the plates is given by: $E = \frac{V}{d} = \frac{18.0 \, \text{V}}{0.02 \, \text{m}} = 900 \, \text{V/m}$

2. Force on the Electron: The force on the electron due to the electric field is: $F = eE = (1.6 \times 10^{-19} \, \text{C}) \times 900 \, \text{V/m} = 1.44 \times 10^{-16} \, \text{N}$

3. Acceleration of the Electron: The acceleration $a$ of the electron in the vertical direction (y-direction) is: $a = \frac{F}{m} = \frac{1.44 \times 10^{-16} \, \text{N}}{9.11 \times 10^{-31} \, \text{kg}} \approx 1.58 \times 10^{14} \, \text{m/s}^2$

4. Time to Pass Through Plates: The time $t$ it takes for the electron to pass through the plates is: $t = \frac{L}{v_0} = \frac{0.06 \, \text{m}}{6.50 \times 10^6 \, \text{m/s}} \approx 9.23 \times 10^{-9} \, \text{s}$

5. Vertical Displacement While Between Plates: The vertical displacement $y_1$ while the electron is between the plates can be calculated using: $y_1 = \frac{1}{2} a t^2 = \frac{1}{2} \times 1.58 \times 10^{14} \, \text{m/s}^2 \times (9.23 \times 10^{-9} \, \text{s})^2 \approx 6.73 \times 10^{-3} \, \text{m} = 6.73 \, \text{mm}$

6. Vertical Velocity After Leaving the Plates: The vertical velocity $v_y$ after the electron leaves the plates is: $v_y = at = 1.58 \times 10^{14} \, \text{m/s}^2 \times 9.23 \times 10^{-9} \, \text{s} \approx 1.46 \times 10^6 \, \text{m/s}$

7. Time to Reach the Screen: The time $t_2$ it takes for the electron to travel from the end of the plates to the screen is: $t_2 = \frac{D}{v_0} = \frac{0.12 \, \text{m}}{6.50 \times 10^6 \, \text{m/s}} \approx 1.85 \times 10^{-8} \, \text{s}$

8. Vertical Displacement After Leaving the Plates: The additional vertical displacement $y_2$ during this time is: $y_2 = v_y \times t_2 = 1.46 \times 10^6 \, \text{m/s} \times 1.85 \times 10^{-8} \, \text{s} \approx 2.70 \times 10^{-2} \, \text{m} = 27.0 \, \text{mm}$

9. Total Vertical Displacement: The total vertical displacement $y$ is the sum of $y_1$ and $y_2$: $y = y_1 + y_2 = 6.73 \, \text{mm} + 27.0 \, \text{mm} = 33.7 \, \text{mm}$

Final Answer:

The electron will strike the fluorescent screen at a position approximately 33.7 mm below the axis.

Would you like more details or have any questions? Here are some related questions you might explore:

1. How would the displacement change if the potential difference was doubled?
2. What would happen if the initial velocity of the electron was increased?
3. How does the mass of the electron affect its final position on the screen?
4. What if the distance between the plates was increased? How would that affect the result?
5. How does the orientation of the plates affect the direction of deflection?

Tip: Always remember that in uniform electric fields, charged particles experience constant acceleration, similar to how objects fall under gravity.