2 Find the capaci tance ot a parallel plate leapacitor if 2000 uC of change is deposited on its plates when 204 are applied acrass the plates.
2 Find the capaci tance ot a parallel plate leapacitor if 2000 uC of change is deposited on its plates when 204 are applied acrass the plates.
Introductory Circuit Analysis (13th Edition)
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ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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![**Problem 2**: Find the capacitance of a parallel plate capacitor if 2000 µC of charge is deposited on its plates when 20V are applied across the plates.
**Solution Explanation**:
To solve for the capacitance \( C \) of a parallel plate capacitor, we use the formula:
\[ C = \frac{Q}{V} \]
where \( Q \) is the charge stored on the plates, and \( V \) is the voltage applied across the plates.
Given:
- \( Q = 2000 \, \mu C = 2000 \times 10^{-6} \, C \)
- \( V = 20 \, V \)
Substitute the given values into the formula:
\[ C = \frac{2000 \times 10^{-6} \, \text{C}}{20 \, \text{V}} \]
Simplify the expression:
\[ C = \frac{2000 \times 10^{-6}}{20} \, \text{F} \]
\[ C = \frac{2000}{20} \times 10^{-6} \, \text{F} \]
\[ C = 100 \times 10^{-6} \, \text{F} \]
\[ C = 100 \, \mu F \]
Therefore, the capacitance of the parallel plate capacitor is:
\[ \boxed{100 \, \mu F} \]
This formula and calculation are fundamental in understanding how capacitance is influenced by the amount of charge stored and the voltage applied across a capacitor's plates. This knowledge is essential for electrical engineering and physics students studying electrical circuits and components.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2405e5cb-20d1-4277-a72a-e2ca429d5d5b%2F3e02c352-3f3a-4a14-8dd1-68d5d23f2bf6%2Fh5a2spb_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem 2**: Find the capacitance of a parallel plate capacitor if 2000 µC of charge is deposited on its plates when 20V are applied across the plates.
**Solution Explanation**:
To solve for the capacitance \( C \) of a parallel plate capacitor, we use the formula:
\[ C = \frac{Q}{V} \]
where \( Q \) is the charge stored on the plates, and \( V \) is the voltage applied across the plates.
Given:
- \( Q = 2000 \, \mu C = 2000 \times 10^{-6} \, C \)
- \( V = 20 \, V \)
Substitute the given values into the formula:
\[ C = \frac{2000 \times 10^{-6} \, \text{C}}{20 \, \text{V}} \]
Simplify the expression:
\[ C = \frac{2000 \times 10^{-6}}{20} \, \text{F} \]
\[ C = \frac{2000}{20} \times 10^{-6} \, \text{F} \]
\[ C = 100 \times 10^{-6} \, \text{F} \]
\[ C = 100 \, \mu F \]
Therefore, the capacitance of the parallel plate capacitor is:
\[ \boxed{100 \, \mu F} \]
This formula and calculation are fundamental in understanding how capacitance is influenced by the amount of charge stored and the voltage applied across a capacitor's plates. This knowledge is essential for electrical engineering and physics students studying electrical circuits and components.
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