A capacitor is the only device in a series circuit with a battery. While remaining hooked up the plate separation is cut in half. What happens to the energy stored in the capacitor?
A capacitor is the only device in a series circuit with a battery. While remaining hooked up the plate separation is cut in half. What happens to the energy stored in the capacitor?
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
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A capacitor is the only device in a series circuit with a battery. While remaining hooked up
the plate separation is cut in half. What happens to the energy stored in the capacitor?
![### Understanding the Effect of Plate Separation on Energy Stored in a Capacitor
**Question:**
"A capacitor is the only device in a series circuit with a battery. While remaining hooked up the plate separation is cut in half. What happens to the energy stored in the capacitor?"
**Explanation:**
When a capacitor is connected in a series circuit with a battery, the voltage across the capacitor is determined by the battery. The energy stored in the capacitor (E) is given by the formula:
\[ E = \frac{1}{2} C V^2 \]
where:
- \( E \) is the energy stored in the capacitor,
- \( C \) is the capacitance, and
- \( V \) is the voltage across the capacitor.
The capacitance \( C \) of a parallel plate capacitor is given by:
\[ C = \frac{\epsilon_0 A}{d} \]
where:
- \( \epsilon_0 \) is the permittivity of free space,
- \( A \) is the area of the plates, and
- \( d \) is the separation between the plates.
When the plate separation \( d \) is cut in half, the capacitance \( C \) will double, since capacitance is inversely proportional to the separation of the plates. As a result, the new capacitance \( C' \) becomes \( 2C \).
Assuming the voltage \( V \) provided by the battery remains constant, the new energy stored in the capacitor \( E' \) will be:
\[ E' = \frac{1}{2} C' V^2 = \frac{1}{2} (2C) V^2 = 2 \left(\frac{1}{2} C V^2\right) = 2E \]
**Conclusion:**
If the plate separation is cut in half while the capacitor remains connected to the battery, the energy stored in the capacitor will be doubled.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F225ea937-ec91-4f60-983d-e6a97001b794%2Fa9031f6f-abf2-4f8f-9e5f-56b8e3e546ec%2Fz39uguv_processed.png&w=3840&q=75)
Transcribed Image Text:### Understanding the Effect of Plate Separation on Energy Stored in a Capacitor
**Question:**
"A capacitor is the only device in a series circuit with a battery. While remaining hooked up the plate separation is cut in half. What happens to the energy stored in the capacitor?"
**Explanation:**
When a capacitor is connected in a series circuit with a battery, the voltage across the capacitor is determined by the battery. The energy stored in the capacitor (E) is given by the formula:
\[ E = \frac{1}{2} C V^2 \]
where:
- \( E \) is the energy stored in the capacitor,
- \( C \) is the capacitance, and
- \( V \) is the voltage across the capacitor.
The capacitance \( C \) of a parallel plate capacitor is given by:
\[ C = \frac{\epsilon_0 A}{d} \]
where:
- \( \epsilon_0 \) is the permittivity of free space,
- \( A \) is the area of the plates, and
- \( d \) is the separation between the plates.
When the plate separation \( d \) is cut in half, the capacitance \( C \) will double, since capacitance is inversely proportional to the separation of the plates. As a result, the new capacitance \( C' \) becomes \( 2C \).
Assuming the voltage \( V \) provided by the battery remains constant, the new energy stored in the capacitor \( E' \) will be:
\[ E' = \frac{1}{2} C' V^2 = \frac{1}{2} (2C) V^2 = 2 \left(\frac{1}{2} C V^2\right) = 2E \]
**Conclusion:**
If the plate separation is cut in half while the capacitor remains connected to the battery, the energy stored in the capacitor will be doubled.
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