A parallel-plate capacitor has circular plates of 10.3 cm radius and 1.70 mm separation. (a) Calculate the capacitance. (b) What charge will appear on the plates if a potential difference of 173 V is applied? (a) Number i Units (b) Number i Units
A parallel-plate capacitor has circular plates of 10.3 cm radius and 1.70 mm separation. (a) Calculate the capacitance. (b) What charge will appear on the plates if a potential difference of 173 V is applied? (a) Number i Units (b) Number i Units
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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![**Capacitance of a Parallel-Plate Capacitor**
**Problem Statement:**
A parallel-plate capacitor has circular plates of 10.3 cm radius and 1.70 mm separation.
**Tasks:**
(a) Calculate the capacitance.
(b) What charge will appear on the plates if a potential difference of 173 V is applied?
**Solution:**
### (a) Calculation of Capacitance
To calculate the capacitance \( C \) of a parallel-plate capacitor, the formula is:
\[ C = \frac{\varepsilon_0 A}{d} \]
where:
- \( \varepsilon_0 \) is the permittivity of free space (\( \varepsilon_0 = 8.854 \times 10^{-12} \, \text{F/m} \)),
- \( A \) is the area of one of the plates,
- \( d \) is the separation between the plates.
Given data:
- Radius of the plates \( r = 10.3 \, \text{cm} = 0.103 \, \text{m} \),
- Separation \( d = 1.70 \, \text{mm} = 0.0017 \, \text{m} \).
The area \( A \) of a circular plate is calculated using the formula for the area of a circle:
\[ A = \pi r^2 \]
Plugging in the values:
\[ A = \pi (0.103 \, \text{m})^2 = 3.337 \times 10^{-2} \, \text{m}^2 \]
Now, use the capacitance formula:
\[ C = \frac{8.854 \times 10^{-12} \, \text{F/m} \times 3.337 \times 10^{-2} \, \text{m}^2}{0.0017 \, \text{m}} \]
\[ C = 1.73 \times 10^{-12} \, \text{F} = 1.73 \, \text{pF} \]
*Input field for capacitance:*
(a) Number \( \underline{\hspace{2cm}} \) Units \( \underline{\hspace{2cm}} \)
### (b) Calculation of Charge
To find the charge \( Q \) on the plates](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb8b65224-21df-49da-a8da-84f6a1af571d%2Fa062149c-a47e-4a9d-9c51-fde2abc5d49e%2Fog03b7o_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Capacitance of a Parallel-Plate Capacitor**
**Problem Statement:**
A parallel-plate capacitor has circular plates of 10.3 cm radius and 1.70 mm separation.
**Tasks:**
(a) Calculate the capacitance.
(b) What charge will appear on the plates if a potential difference of 173 V is applied?
**Solution:**
### (a) Calculation of Capacitance
To calculate the capacitance \( C \) of a parallel-plate capacitor, the formula is:
\[ C = \frac{\varepsilon_0 A}{d} \]
where:
- \( \varepsilon_0 \) is the permittivity of free space (\( \varepsilon_0 = 8.854 \times 10^{-12} \, \text{F/m} \)),
- \( A \) is the area of one of the plates,
- \( d \) is the separation between the plates.
Given data:
- Radius of the plates \( r = 10.3 \, \text{cm} = 0.103 \, \text{m} \),
- Separation \( d = 1.70 \, \text{mm} = 0.0017 \, \text{m} \).
The area \( A \) of a circular plate is calculated using the formula for the area of a circle:
\[ A = \pi r^2 \]
Plugging in the values:
\[ A = \pi (0.103 \, \text{m})^2 = 3.337 \times 10^{-2} \, \text{m}^2 \]
Now, use the capacitance formula:
\[ C = \frac{8.854 \times 10^{-12} \, \text{F/m} \times 3.337 \times 10^{-2} \, \text{m}^2}{0.0017 \, \text{m}} \]
\[ C = 1.73 \times 10^{-12} \, \text{F} = 1.73 \, \text{pF} \]
*Input field for capacitance:*
(a) Number \( \underline{\hspace{2cm}} \) Units \( \underline{\hspace{2cm}} \)
### (b) Calculation of Charge
To find the charge \( Q \) on the plates
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