Consider a parallel-plate capacitor with plate separation d, plate area A, whose plates have charge ±Q. A particle of charge q < 0 and mass m is released from rest at the negative plate of the capacitor and allowed to accelerate towards the positive plate. With what speed does the charge strike the positive plate? Answer in terms of d, A, Q, q, m, and/or €0.
Consider a parallel-plate capacitor with plate separation d, plate area A, whose plates have charge ±Q. A particle of charge q < 0 and mass m is released from rest at the negative plate of the capacitor and allowed to accelerate towards the positive plate. With what speed does the charge strike the positive plate? Answer in terms of d, A, Q, q, m, and/or €0.
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Consider a parallel-plate capacitor with plate separation d, plate area A, whose plates have charge ±Q. A particle of charge q < 0 and mass m is released from rest at the negative plate of the capacitor and allowed to accelerate towards the positive plate. With what speed does the charge strike the positive plate? Answer in terms of d, A, Q, q, m, and/or e0. (see image)
![**Physics Problem: Particle Acceleration in a Parallel-Plate Capacitor**
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**Problem Statement:**
Consider a parallel-plate capacitor with plate separation \( d \), plate area \( A \), whose plates have charge \( \pm Q \). A particle of charge \( q < 0 \) and mass \( m \) is released from rest at the negative plate of the capacitor and allowed to accelerate towards the positive plate. With what speed does the charge strike the positive plate? Answer in terms of \( d \), \( A \), \( Q \), \( q \), \( m \), and/or \( \epsilon_0 \) (the permittivity of free space).
---
**Explanation and Solution:**
To solve for the speed \( v \) of the particle when it strikes the positive plate:
1. **Electric Field Calculation:**
- The electric field \( E \) between the plates of a capacitor is given by:
\[
E = \frac{V}{d}
\]
where \( V \) is the voltage across the plates and \( d \) is the distance between the plates.
- The voltage \( V \) can be related to the charge \( Q \) and the capacitance \( C \) by:
\[
V = \frac{Q}{C}
\]
- For a parallel-plate capacitor, the capacitance \( C \) is given by:
\[
C = \frac{\epsilon_0 A}{d}
\]
- Combining these, we have:
\[
V = \frac{Qd}{\epsilon_0 A}
\]
- Thus, the electric field \( E \) is:
\[
E = \frac{Q}{\epsilon_0 A}
\]
2. **Force on the Particle:**
- The force \( F \) on the particle with charge \( q \) in the electric field \( E \) is:
\[
F = qE = q \left( \frac{Q}{\epsilon_0 A} \right)
\]
3. **Acceleration of the Particle:**
- Using Newton’s second law, the acceleration \( a \) of the particle is:
\[
a = \frac{F}{m} = \frac{qQ}{m \epsilon_0 A}](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F651b911d-2c53-4f54-b4b6-29cb6ac234d2%2F61c079aa-8b89-44b5-9c4e-67c4f5047e91%2Fnqf3c1m_processed.png&w=3840&q=75)
Transcribed Image Text:**Physics Problem: Particle Acceleration in a Parallel-Plate Capacitor**
---
**Problem Statement:**
Consider a parallel-plate capacitor with plate separation \( d \), plate area \( A \), whose plates have charge \( \pm Q \). A particle of charge \( q < 0 \) and mass \( m \) is released from rest at the negative plate of the capacitor and allowed to accelerate towards the positive plate. With what speed does the charge strike the positive plate? Answer in terms of \( d \), \( A \), \( Q \), \( q \), \( m \), and/or \( \epsilon_0 \) (the permittivity of free space).
---
**Explanation and Solution:**
To solve for the speed \( v \) of the particle when it strikes the positive plate:
1. **Electric Field Calculation:**
- The electric field \( E \) between the plates of a capacitor is given by:
\[
E = \frac{V}{d}
\]
where \( V \) is the voltage across the plates and \( d \) is the distance between the plates.
- The voltage \( V \) can be related to the charge \( Q \) and the capacitance \( C \) by:
\[
V = \frac{Q}{C}
\]
- For a parallel-plate capacitor, the capacitance \( C \) is given by:
\[
C = \frac{\epsilon_0 A}{d}
\]
- Combining these, we have:
\[
V = \frac{Qd}{\epsilon_0 A}
\]
- Thus, the electric field \( E \) is:
\[
E = \frac{Q}{\epsilon_0 A}
\]
2. **Force on the Particle:**
- The force \( F \) on the particle with charge \( q \) in the electric field \( E \) is:
\[
F = qE = q \left( \frac{Q}{\epsilon_0 A} \right)
\]
3. **Acceleration of the Particle:**
- Using Newton’s second law, the acceleration \( a \) of the particle is:
\[
a = \frac{F}{m} = \frac{qQ}{m \epsilon_0 A}
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