Conventional TV tube. Early television sets depended on images being built up on the screen by the deflection of electrons directed toward the screen from the rear of the tube. The figure below shows such a deflection system. The length of the plates is 2.59 cm and the deflecting electric field between the two plates is 9.00 × 105 N/C vertically upward. If an electron enters the space between the plates with a horizontal speed of 3.86 × 10' m/s, what is the vertical displacement of Ay at the end of the plates?

To find the vertical displacement \( \Delta y \) of the electron as it passes through the electric field, we need to consider the following steps and use the known values:

### Given:
- Length of plates, \( L = 2.59 \, \text{cm} = 2.59 \times 10^{-2} \, \text{m} \)
- Electric field strength, \( E = 9.00 \times 10^5 \, \text{N/C} \)
- Horizontal velocity of electron, \( v_x = 3.86 \times 10^7 \, \text{m/s} \)
- Charge of electron, \( e = 1.60 \times 10^{-19} \, \text{C} \)
- Mass of electron, \( m_e = 9.11 \times 10^{-31} \, \text{kg} \)

### Step 1: Find the force on the electron
The electric force on the electron due to the field is given by:
\[
F = eE
\]
Substituting the values:
\[
F = (1.60 \times 10^{-19}) \times (9.00 \times 10^5) = 1.44 \times 10^{-13} \, \text{N}
\]

### Step 2: Calculate the vertical acceleration \( a_y \)
The vertical acceleration due to this force is given by Newton's second law:
\[
a_y = \frac{F}{m_e}
\]
Substituting the values:
\[
a_y = \frac{1.44 \times 10^{-13}}{9.11 \times 10^{-31}} = 1.58 \times 10^{17} \, \text{m/s}^2
\]

### Step 3: Find the time \( t \) the electron spends between the plates
The electron moves horizontally with a constant speed, so the time to pass through the plates is:
\[
t = \frac{L}{v_x}
\]
Substituting the values:
\[
t = \frac{2.59 \times 10^{-2}}{3.86 \times 10^7} = 6.71 \times 10^{-10} \, \text{s}
\]

### Step 4: Calculate the vertical displacement \( \Delta y \)
The vertical displacement can be calculated using the kinematic equation for motion under constant acceleration:
\[
\Delta y = \frac{1}{2} a_y t^2
\]
Substituting the values:
\[
\Delta y = \frac{1}{2} (1.58 \times 10^{17}) (6.71 \times 10^{-10})^2
\]
First, calculate \( t^2 \):
\[
t^2 = (6.71 \times 10^{-10})^2 = 4.50 \times 10^{-19}
\]
Now, substitute this into the equation for \( \Delta y \):
\[
\Delta y = \frac{1}{2} \times 1.58 \times 10^{17} \times 4.50 \times 10^{-19} = 3.55 \times 10^{-2} \, \text{m}
\]

Thus, the vertical displacement \( \Delta y \) is:
\[
\Delta y = 0.0355 \, \text{m} \, = 3.55 \, \text{cm}
\]

### Final Answer:
The vertical displacement of the electron at the end of the plates is \( \Delta y = 3.55 \, \text{cm} \).

---

Would you like further details or clarification on any step?

### Related questions:
1. What would happen to the vertical displacement if the horizontal velocity doubled?
2. How does the electron’s charge affect its trajectory in an electric field?
3. What would happen to the displacement if the electric field was reversed?
4. How would the motion differ if the particle was a proton instead of an electron?
5. What is the significance of the length of the plates in affecting the electron’s path?

### Tip:
For problems involving electric fields and particle motion, always break the motion into perpendicular components (horizontal and vertical) and analyze each independently.

Conventional TV tube. Early television sets depended on images being built up on the screen by the deflection of electrons directed toward the screen from the rear of the tube. The figure below shows such a deflection system. The length of the plates is 2.59 cm and the deflecting electric field between the two plates is 9.00 × 105 N/C vertically upward. If an electron enters the space between the plates with a horizontal speed of 3.86 × 10^7 m/s, what is the vertical displacement of Ay at the end of the plates?

Explore how to calculate the vertical displacement of an electron passing through an electric field in a deflection system. This problem involves kinematics, electric fields, and Newton's Second Law. Key data includes a plate length of 2.59 cm, a horizontal velocity of 3.86 × 10^7 m/s, and a field strength of 9.00 × 10^5 N/C. Suitable for undergraduate physics students.

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