Question 1185357
Here are the solutions to your physics problems:

**1. Electric Force on a Point Charge:**

*   **Formula:** Force (F) = Electric Field (E) * Charge (q)
*   **Calculation:** F = (12 N/C) * (2.5 x 10⁻⁷ C) = 3.0 x 10⁻⁶ N

**2. Electron in a Cathode Ray Tube:**

**(a) Diagram:**

```
Negative Plate (-)                  Positive Plate (+)
--------------------------------------------------
|                                                  |
|  e⁻ (electron) -->                             |
|                                                  |
--------------------------------------------------
```

**(b) Kinetic Energy:**

*   **Formula:** Kinetic Energy (KE) = Charge (q) * Potential Difference (V)
*   **Calculation:** KE = (1.6 x 10⁻¹⁹ C) * (2 x 10⁴ V) = 3.2 x 10⁻¹⁵ J

**(c) Velocity:**

*   **Formula:** KE = (1/2) * mass (m) * velocity² (v²)
*   **Rearrange for velocity:** v = √(2 * KE / m)
*   **Calculation:** v = √(2 * 3.2 x 10⁻¹⁵ J / 9.11 x 10⁻³¹ kg) = √(7.025 x 10¹⁵) ≈ 8.38 x 10⁷ m/s

**3. Proton Moved Against an Electric Field:**

*   **Explanation:** When a positive charge (like a proton) is moved *against* an electric field, work is done *on* the charge by the external force. This work increases the electric potential energy of the proton.  The change in electric potential energy is *positive*.

*   **Diagram:**

```
+ (High Potential)
|
|  Force (F) <--- Proton (+)
|
- (Low Potential)
```

The electric field points from high potential to low potential. The force on the proton due to the field is in the same direction. To move the proton *against* the field, an external force must act in the opposite direction. This external force does positive work, increasing the proton's potential energy.

**4. Electron Traveling Through Plates:**

1.  **Time to traverse plates:**
    *   **Formula:** time (t) = distance (d) / velocity (v)
    *   **Calculation:** t = (0.10 m) / (1.5 x 10⁶ m/s) = 6.67 x 10⁻⁸ s

2.  **Vertical acceleration:**
    *   **Formula:** Force (F) = Electric Field (E) * Charge (q) = mass (m) * acceleration (a)
    *   **Rearrange for acceleration:** a = (E * q) / m
    *   **Calculation:** a = (100 N/C * 1.6 x 10⁻¹⁹ C) / (9.11 x 10⁻³¹ kg) ≈ 1.76 x 10¹³ m/s²

3.  **Vertical velocity at exit:**
    *   **Formula:** final velocity (vf) = initial velocity (vi) + acceleration (a) * time (t)
    *   **Calculation:** vf = 0 + (1.76 x 10¹³ m/s²) * (6.67 x 10⁻⁸ s) ≈ 1.17 x 10⁶ m/s

4.  **Final velocity (magnitude):**
    *   The horizontal velocity remains constant at 1.5 x 10⁶ m/s.
    *   **Calculation:**  Final Velocity = √((1.5 x 10⁶ m/s)² + (1.17 x 10⁶ m/s)²) ≈ 1.9 x 10⁶ m/s

5. **Final Velocity (direction):**
    * The angle below the horizontal can be found by taking the inverse tangent of the vertical velocity divided by the horizontal velocity.

Therefore, the final velocity of the electron as it leaves the plates is approximately 1.9 x 10⁶ m/s.