A proton is projected horizontally midway between two parallel plates that are separated by 0.47 cm. The uniform electric field has magnitude 6.0 x 105 N/C between the plates. If the plates are 5.6 cm long, find the minimum speed of the proton if it just misses the lower plate as it emerges from the field. The gravitational force on the proton can be neglected. 6.2 x 106 m/s 12 x 106 m/s 11 x 106 m/s 7.4 x 106 m/s 8.7 x 106 m/s E

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### Projectile Motion of a Proton in an Electric Field

#### Problem Statement:

A proton is projected horizontally midway between two parallel plates that are separated by 0.47 cm. The uniform electric field has a magnitude of \(6.0 \times 10^5 \, \text{N/C}\) between the plates. If the plates are 5.6 cm long, find the minimum speed of the proton if it just misses the lower plate as it emerges from the field. The gravitational force on the proton can be neglected.

The available options for the minimum speed of the proton are:
- \(6.2 \times 10^6 \, \text{m/s}\)
- \(12 \times 10^6 \, \text{m/s}\)
- \(11 \times 10^6 \, \text{m/s}\)
- \(7.4 \times 10^6 \, \text{m/s}\)
- \(8.7 \times 10^6 \, \text{m/s}\)

#### Diagram Explanation:

The diagram shows:

1. **Two Parallel Plates**: Represented as two horizontal black lines, separated by a distance.
2. **Electric Field (\(\vec{E}\))**: The electric field between the plates is shown with downward arrows, indicating the direction of the field from the upper plate to the lower plate.
3. **Proton's Path**: The proton is represented as a black dot on the left side with a horizontal arrow indicating its initial horizontal velocity.

#### Calculation Approach:

To solve this problem, the following steps should be considered:

1. **Determine the Electric Force**:
   Use the formula for the force on a charge in an electric field: 
   \[
   F = qE
   \]
   where \(q\) is the charge of the proton \((q = 1.6 \times 10^{-19} \, \text{C})\) and \(E\) is the electric field strength \((6.0 \times 10^5 \, \text{N/C})\).

2. **Calculate the Acceleration**:
   Use Newton’s Second Law to find the acceleration:
   \[
   a = \frac{F}{m}
   \]
   where \(m\) is the mass of the proton \((m = 1.67 \times 10^{-27}
Transcribed Image Text:### Projectile Motion of a Proton in an Electric Field #### Problem Statement: A proton is projected horizontally midway between two parallel plates that are separated by 0.47 cm. The uniform electric field has a magnitude of \(6.0 \times 10^5 \, \text{N/C}\) between the plates. If the plates are 5.6 cm long, find the minimum speed of the proton if it just misses the lower plate as it emerges from the field. The gravitational force on the proton can be neglected. The available options for the minimum speed of the proton are: - \(6.2 \times 10^6 \, \text{m/s}\) - \(12 \times 10^6 \, \text{m/s}\) - \(11 \times 10^6 \, \text{m/s}\) - \(7.4 \times 10^6 \, \text{m/s}\) - \(8.7 \times 10^6 \, \text{m/s}\) #### Diagram Explanation: The diagram shows: 1. **Two Parallel Plates**: Represented as two horizontal black lines, separated by a distance. 2. **Electric Field (\(\vec{E}\))**: The electric field between the plates is shown with downward arrows, indicating the direction of the field from the upper plate to the lower plate. 3. **Proton's Path**: The proton is represented as a black dot on the left side with a horizontal arrow indicating its initial horizontal velocity. #### Calculation Approach: To solve this problem, the following steps should be considered: 1. **Determine the Electric Force**: Use the formula for the force on a charge in an electric field: \[ F = qE \] where \(q\) is the charge of the proton \((q = 1.6 \times 10^{-19} \, \text{C})\) and \(E\) is the electric field strength \((6.0 \times 10^5 \, \text{N/C})\). 2. **Calculate the Acceleration**: Use Newton’s Second Law to find the acceleration: \[ a = \frac{F}{m} \] where \(m\) is the mass of the proton \((m = 1.67 \times 10^{-27}
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