A parallel plate capacitor with plates of area A and plate separation d is charged so that the potential difference between its plates is V. If the capacitor is then isolated and its plate separation is decreased to d/2, what happens to the potential difference between the plates? 6.
A parallel plate capacitor with plates of area A and plate separation d is charged so that the potential difference between its plates is V. If the capacitor is then isolated and its plate separation is decreased to d/2, what happens to the potential difference between the plates? 6.
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
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Question
![**Physics Problem: Parallel Plate Capacitor**
**Problem Statement:**
A parallel plate capacitor with plates of area \( A \) and plate separation \( d \) is charged so that the potential difference between its plates is \( V \). If the capacitor is then isolated and its plate separation is decreased to \( d/2 \), what happens to the potential difference between the plates?
**Diagram Explanation:**
There are no graphs or diagrams included in this text.
**Concepts:**
- **Capacitance of a Parallel Plate Capacitor:**
\[
C = \frac{\varepsilon_0 A}{d}
\]
where \( \varepsilon_0 \) is the permittivity of free space, \( A \) is the area of one plate, and \( d \) is the distance between the plates.
- **Charge on the Capacitor:**
\[
Q = CV
\]
- **Effect of Changing Plate Separation:**
Once the capacitor is isolated, the charge \( Q \) remains constant. If the plate separation is reduced to \( d/2 \), the new capacitance becomes:
\[
C' = \frac{\varepsilon_0 A}{d/2} = \frac{2\varepsilon_0 A}{d}
\]
- **New Potential Difference:**
Given \( Q = C'V' \) and \( Q \) remains constant,
\[
V' = \frac{Q}{C'} = \frac{CV}{C'} = \frac{V}{2}
\]
So the new potential difference when the plate separation is decreased to \( d/2 \) is half of the original potential difference \( V \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3f128b0e-33c3-4912-93ee-e3e692562233%2F7bd4354a-2415-40d7-aebf-7311d8aa344c%2F2dzhajl_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Physics Problem: Parallel Plate Capacitor**
**Problem Statement:**
A parallel plate capacitor with plates of area \( A \) and plate separation \( d \) is charged so that the potential difference between its plates is \( V \). If the capacitor is then isolated and its plate separation is decreased to \( d/2 \), what happens to the potential difference between the plates?
**Diagram Explanation:**
There are no graphs or diagrams included in this text.
**Concepts:**
- **Capacitance of a Parallel Plate Capacitor:**
\[
C = \frac{\varepsilon_0 A}{d}
\]
where \( \varepsilon_0 \) is the permittivity of free space, \( A \) is the area of one plate, and \( d \) is the distance between the plates.
- **Charge on the Capacitor:**
\[
Q = CV
\]
- **Effect of Changing Plate Separation:**
Once the capacitor is isolated, the charge \( Q \) remains constant. If the plate separation is reduced to \( d/2 \), the new capacitance becomes:
\[
C' = \frac{\varepsilon_0 A}{d/2} = \frac{2\varepsilon_0 A}{d}
\]
- **New Potential Difference:**
Given \( Q = C'V' \) and \( Q \) remains constant,
\[
V' = \frac{Q}{C'} = \frac{CV}{C'} = \frac{V}{2}
\]
So the new potential difference when the plate separation is decreased to \( d/2 \) is half of the original potential difference \( V \).
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