Question 1185357: 1.At a certain point P in an electric field, the magnitude of the electric field is 12 N/C. Calculate the magnitude of the electric force that would be exerted on a point charge of 2.5 x 10-7 C, located at P.
2.An electron leaves the negative terminal of a cathode ray tube and travels towards the positive plate starting from rest. The electric potential difference between the plates is 2x104 Volts. [mass of electron = 9.11x10-31 kg and the charge on an electron is 1.6x10-19 C]
(a)Draw a diagram.
(b)Calculate the electron’s kinetic energy when it reaches the positive plate.
(c)Calculate the electron’s velocity when it reaches the positive plate.
3.When a proton is moved against an electric field by an unbalanced force, what happens to the change in electric potential energy? Is it positive or negative? Explain using a diagram and the concept of work.
4.An electron is travelling horizontally at a speed of 1.5x106 m/s through two horizontal plates. It starts at the negative plate at the top. The plates are 10cm long and the electric field strength between the plates is 100N/C. Calculate the final velocity of the electron as it leaves the plates.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! 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.
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