0.32 C is stored on a parallel plate capacitor that has plates of area 4.5 m² separated by 0.018 mm of nylon. What is the applied voltage across the capacitor? The value of the nylon is 3.4 The final answer needs to be in kv
0.32 C is stored on a parallel plate capacitor that has plates of area 4.5 m² separated by 0.018 mm of nylon. What is the applied voltage across the capacitor? The value of the nylon is 3.4 The final answer needs to be in kv
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
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![**Problem Statement:**
0.32 C is stored on a parallel plate capacitor that has plates of area 4.5 m² separated by 0.018 mm of nylon. What is the applied voltage across the capacitor? The dielectric constant of the nylon is 3.4.
The final answer needs to be in kV.
**Solution:**
To find the applied voltage across the capacitor, we can use the relationship between the charge (Q), the voltage (V), and the capacitance (C) of the capacitor.
The formula for the capacitance (C) of a parallel plate capacitor with a dielectric material is given by:
\[ C = \frac{{\kappa \cdot \epsilon_0 \cdot A}}{{d}} \]
where:
- \(\kappa\) is the dielectric constant of the material (in this case, nylon which is 3.4),
- \(\epsilon_0\) is the permittivity of free space (\(8.854 \times 10^{-12} \, \text{F/m}\)),
- \(A\) is the area of the plates (4.5 m²),
- \(d\) is the separation between the plates (0.018 mm or \(0.018 \times 10^{-3} \, \text{m}\)).
1. Calculating the capacitance:
\[ C = \frac{3.4 \cdot 8.854 \times 10^{-12} \cdot 4.5}{0.018 \times 10^{-3}} \]
\[ C = \frac{135.615 \times 10^{-12} \times 4.5}{0.018 \times 10^{-3}} \]
\[ C = \frac{610.2675 \times 10^{-12}}{0.018 \times 10^{-3}} \]
\[ C \approx 33.9037 \times 10^{-9} \text{F} \]
\[ C \approx 33.9037 \text{nF} \]
2. Using the relationship \(Q = C \cdot V\), we can solve for the voltage \(V\):
\[ V = \frac{Q}{C} \]
Given that \(Q = 0.32 \, \text{C}\) and \(C = 33.9037 \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7be57ba7-aca5-4231-94a7-c86a6ecf18d3%2F70c44151-d496-4681-935f-338ef938e4f7%2Fkfkgte_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
0.32 C is stored on a parallel plate capacitor that has plates of area 4.5 m² separated by 0.018 mm of nylon. What is the applied voltage across the capacitor? The dielectric constant of the nylon is 3.4.
The final answer needs to be in kV.
**Solution:**
To find the applied voltage across the capacitor, we can use the relationship between the charge (Q), the voltage (V), and the capacitance (C) of the capacitor.
The formula for the capacitance (C) of a parallel plate capacitor with a dielectric material is given by:
\[ C = \frac{{\kappa \cdot \epsilon_0 \cdot A}}{{d}} \]
where:
- \(\kappa\) is the dielectric constant of the material (in this case, nylon which is 3.4),
- \(\epsilon_0\) is the permittivity of free space (\(8.854 \times 10^{-12} \, \text{F/m}\)),
- \(A\) is the area of the plates (4.5 m²),
- \(d\) is the separation between the plates (0.018 mm or \(0.018 \times 10^{-3} \, \text{m}\)).
1. Calculating the capacitance:
\[ C = \frac{3.4 \cdot 8.854 \times 10^{-12} \cdot 4.5}{0.018 \times 10^{-3}} \]
\[ C = \frac{135.615 \times 10^{-12} \times 4.5}{0.018 \times 10^{-3}} \]
\[ C = \frac{610.2675 \times 10^{-12}}{0.018 \times 10^{-3}} \]
\[ C \approx 33.9037 \times 10^{-9} \text{F} \]
\[ C \approx 33.9037 \text{nF} \]
2. Using the relationship \(Q = C \cdot V\), we can solve for the voltage \(V\):
\[ V = \frac{Q}{C} \]
Given that \(Q = 0.32 \, \text{C}\) and \(C = 33.9037 \
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