A parallel-plate capacitor is constructed with two disks spaced 2.00 mm apart. charged to a potential difference of 500 V. A proton is shot through a small hole in the negative plate (point A) with a speed of 2.0 x 105 m/s. a. Use the law of conservation of energy and determine the potential at the point where the speed of the proton is zero (point B, shown in the diagram) How far is the point B from the negative plate of the capacitor? b. OV E 5 Before: After: B =0 mm 500 V Vy=0 m/s X X d = 2.00 mm
A parallel-plate capacitor is constructed with two disks spaced 2.00 mm apart. charged to a potential difference of 500 V. A proton is shot through a small hole in the negative plate (point A) with a speed of 2.0 x 105 m/s. a. Use the law of conservation of energy and determine the potential at the point where the speed of the proton is zero (point B, shown in the diagram) How far is the point B from the negative plate of the capacitor? b. OV E 5 Before: After: B =0 mm 500 V Vy=0 m/s X X d = 2.00 mm
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![### Parallel-Plate Capacitor and Proton Velocity
**Problem Statement:**
A parallel-plate capacitor is constructed with two disks spaced 2.00 mm apart. It is charged to a potential difference of 500 V. A proton is shot through a small hole in the negative plate (point A) with a speed of \(2.0 \times 10^5 \, \text{m/s}\).
**Tasks:**
a. Use the law of conservation of energy and determine the potential at the point where the speed of the proton is zero (point B, shown in the diagram).
b. How far is the point B from the negative plate of the capacitor?
#### Diagram Explanation:
The diagram at the bottom of the text shows a simplified representation of the parallel-plate capacitor and the motion of the proton between two plates:
1. **Plates**:
- The left plate is at 0 V.
- The right plate is at 500 V.
2. **Proton Path**:
- A proton starts at point A on the left plate with an initial speed \( v_i = 2.0 \times 10^5 \, \text{m/s} \).
- It moves towards point B, where its speed is reduced to zero \( v_f = 0 \, \text{m/s} \).
3. **Distance**:
- The plates are separated by a distance of \( d = 2.00 \, \text{mm} \).
An electric field \( E \) acts between the plates, pointing from the positive plate (right) to the negative plate (left).
#### Steps for Solution:
- **Conservation of Energy**:
- At point A, the proton has both kinetic and electric potential energy.
- At point B, where the speed is zero, the proton has only electric potential energy.
Using the law of conservation of energy:
\[
K.E. + P.E._{\text{initial}} = P.E._{\text{final}}
\]
\[
\frac{1}{2} m v_i^2 + q V_A = q V_B
\]
where:
- \( m \) is the mass of the proton.
- \( v_i \) is the initial speed.
- \( q \) is the charge of the proton.
- \( V_A \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4a6e4794-5644-431d-9e96-b37fbaf0b60a%2F48d51b18-d5c4-4487-a047-8172e33dbc45%2Fv6dfvh_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Parallel-Plate Capacitor and Proton Velocity
**Problem Statement:**
A parallel-plate capacitor is constructed with two disks spaced 2.00 mm apart. It is charged to a potential difference of 500 V. A proton is shot through a small hole in the negative plate (point A) with a speed of \(2.0 \times 10^5 \, \text{m/s}\).
**Tasks:**
a. Use the law of conservation of energy and determine the potential at the point where the speed of the proton is zero (point B, shown in the diagram).
b. How far is the point B from the negative plate of the capacitor?
#### Diagram Explanation:
The diagram at the bottom of the text shows a simplified representation of the parallel-plate capacitor and the motion of the proton between two plates:
1. **Plates**:
- The left plate is at 0 V.
- The right plate is at 500 V.
2. **Proton Path**:
- A proton starts at point A on the left plate with an initial speed \( v_i = 2.0 \times 10^5 \, \text{m/s} \).
- It moves towards point B, where its speed is reduced to zero \( v_f = 0 \, \text{m/s} \).
3. **Distance**:
- The plates are separated by a distance of \( d = 2.00 \, \text{mm} \).
An electric field \( E \) acts between the plates, pointing from the positive plate (right) to the negative plate (left).
#### Steps for Solution:
- **Conservation of Energy**:
- At point A, the proton has both kinetic and electric potential energy.
- At point B, where the speed is zero, the proton has only electric potential energy.
Using the law of conservation of energy:
\[
K.E. + P.E._{\text{initial}} = P.E._{\text{final}}
\]
\[
\frac{1}{2} m v_i^2 + q V_A = q V_B
\]
where:
- \( m \) is the mass of the proton.
- \( v_i \) is the initial speed.
- \( q \) is the charge of the proton.
- \( V_A \
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