An air filled parallel plate capacitor of C-23.5 nF is connected across a battery of AV-160 V. Find the energy U stored in the capacitor (in µJ).
An air filled parallel plate capacitor of C-23.5 nF is connected across a battery of AV-160 V. Find the energy U stored in the capacitor (in µJ).
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![**Problem Statement:**
An air-filled parallel plate capacitor with a capacitance of \( C = 23.5 \, \text{nF} \) is connected across a battery with a potential difference of \( \Delta V = 160 \, \text{V} \). Find the energy \( U \) stored in the capacitor (in microjoules, \( \mu \text{J} \)).
**Solution:**
To find the energy stored in the capacitor, use the formula:
\[ U = \frac{1}{2} C (\Delta V)^2 \]
Given:
- \( C = 23.5 \, \text{nF} = 23.5 \times 10^{-9} \, \text{F} \)
- \( \Delta V = 160 \, \text{V} \)
Substitute these values into the formula:
\[ U = \frac{1}{2} \times 23.5 \times 10^{-9} \times (160)^2 \]
\[ U = \frac{1}{2} \times 23.5 \times 10^{-9} \times 25600 \]
\[ U = 0.5 \times 23.5 \times 10^{-9} \times 25600 \]
\[ U = 300.8 \times 10^{-6} \]
\[ U = 300.8 \, \mu\text{J} \]
**Conclusion:**
The energy stored in the capacitor is \( 300.8 \, \mu\text{J} \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F440b7269-170b-4024-8d0e-4b8a2f594bf7%2Fff42f459-dff0-43ac-959f-065eb0eb6195%2Fulml6jg_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
An air-filled parallel plate capacitor with a capacitance of \( C = 23.5 \, \text{nF} \) is connected across a battery with a potential difference of \( \Delta V = 160 \, \text{V} \). Find the energy \( U \) stored in the capacitor (in microjoules, \( \mu \text{J} \)).
**Solution:**
To find the energy stored in the capacitor, use the formula:
\[ U = \frac{1}{2} C (\Delta V)^2 \]
Given:
- \( C = 23.5 \, \text{nF} = 23.5 \times 10^{-9} \, \text{F} \)
- \( \Delta V = 160 \, \text{V} \)
Substitute these values into the formula:
\[ U = \frac{1}{2} \times 23.5 \times 10^{-9} \times (160)^2 \]
\[ U = \frac{1}{2} \times 23.5 \times 10^{-9} \times 25600 \]
\[ U = 0.5 \times 23.5 \times 10^{-9} \times 25600 \]
\[ U = 300.8 \times 10^{-6} \]
\[ U = 300.8 \, \mu\text{J} \]
**Conclusion:**
The energy stored in the capacitor is \( 300.8 \, \mu\text{J} \).
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