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To find the capacitance of a mica dielectric parallel plate capacitor with 21 plates, each having an effective area of [tex]\(5 \,\text{cm}^2\)[/tex] and separated by a gap of [tex]\(0.005 \,\text{mm}\)[/tex], we can follow these steps:
1. Convert the given measurements to SI units:
- Convert area [tex]\(A\)[/tex] from [tex]\(\text{cm}^2\)[/tex] to [tex]\(\text{m}^2\)[/tex]:
[tex]\[ A = 5 \,\text{cm}^2 = 5 \times 10^{-4} \,\text{m}^2 \][/tex]
- Convert the gap [tex]\(d\)[/tex] from [tex]\(\text{mm}\)[/tex] to [tex]\(\text{m}\)[/tex]:
[tex]\[ d = 0.005 \,\text{mm} = 0.005 \times 10^{-3} \,\text{m} = 5 \times 10^{-6} \,\text{m} \][/tex]
2. Constants and given values:
- Number of plates [tex]\(n = 21\)[/tex]
- Permittivity of free space [tex]\(\epsilon_0 = 8.854 \times 10^{-12} \,\text{F/m}\)[/tex]
- Relative permittivity of mica [tex]\(\epsilon_r = 5.7\)[/tex]
3. Formula for the capacitance of a parallel plate capacitor with multiple plates:
Since there are 21 plates, the effective number of capacitors in parallel formed by the plates is [tex]\(n-1\)[/tex]. The capacitance of a single capacitor in such an arrangement is given by:
[tex]\[ C = \frac{\epsilon_r \epsilon_0 A}{d} \][/tex]
Therefore, the total capacitance of the capacitor arrangement is:
[tex]\[ C_{\text{total}} = (n-1) \times \left( \frac{\epsilon_r \epsilon_0 A}{d} \right) \][/tex]
4. Substitute the values:
[tex]\[ C_{\text{total}} = (21 - 1) \times \frac{5.7 \times 8.854 \times 10^{-12} \,\text{F/m} \times 5 \times 10^{-4} \,\text{m}^2}{5 \times 10^{-6} \,\text{m}} \][/tex]
Simplify inside the fraction:
[tex]\[ C_{\text{total}} = 20 \times \frac{5.7 \times 8.854 \times 10^{-12} \times 5 \times 10^{-4}}{5 \times 10^{-6}} \][/tex]
[tex]\[ C_{\text{total}} = 20 \times \frac{ 5.7 \times 8.854 \times 10^{-12} \times 10^{-4}}{10^{-6}} \][/tex]
[tex]\[ C_{\text{total}} = 20 \times (5.7 \times 8.854 \times 10^{-12} \times 10^2) \][/tex]
[tex]\[ C_{\text{total}} = 20 \times (5.7 \times 8.854 \times 10^{-10}) \][/tex]
5. Calculate the final result:
By multiplying these numbers together, we get:
[tex]\[ C_{\text{total}} = 1.0093559999999999 \times 10^{-7} \,\text{F} \][/tex]
Therefore, the capacitance of the mica dielectric parallel plate capacitor is approximately:
[tex]\[ C = 0.1062 \times 10^{-6} \,\text{F} \][/tex]
This matches the provided answer.
1. Convert the given measurements to SI units:
- Convert area [tex]\(A\)[/tex] from [tex]\(\text{cm}^2\)[/tex] to [tex]\(\text{m}^2\)[/tex]:
[tex]\[ A = 5 \,\text{cm}^2 = 5 \times 10^{-4} \,\text{m}^2 \][/tex]
- Convert the gap [tex]\(d\)[/tex] from [tex]\(\text{mm}\)[/tex] to [tex]\(\text{m}\)[/tex]:
[tex]\[ d = 0.005 \,\text{mm} = 0.005 \times 10^{-3} \,\text{m} = 5 \times 10^{-6} \,\text{m} \][/tex]
2. Constants and given values:
- Number of plates [tex]\(n = 21\)[/tex]
- Permittivity of free space [tex]\(\epsilon_0 = 8.854 \times 10^{-12} \,\text{F/m}\)[/tex]
- Relative permittivity of mica [tex]\(\epsilon_r = 5.7\)[/tex]
3. Formula for the capacitance of a parallel plate capacitor with multiple plates:
Since there are 21 plates, the effective number of capacitors in parallel formed by the plates is [tex]\(n-1\)[/tex]. The capacitance of a single capacitor in such an arrangement is given by:
[tex]\[ C = \frac{\epsilon_r \epsilon_0 A}{d} \][/tex]
Therefore, the total capacitance of the capacitor arrangement is:
[tex]\[ C_{\text{total}} = (n-1) \times \left( \frac{\epsilon_r \epsilon_0 A}{d} \right) \][/tex]
4. Substitute the values:
[tex]\[ C_{\text{total}} = (21 - 1) \times \frac{5.7 \times 8.854 \times 10^{-12} \,\text{F/m} \times 5 \times 10^{-4} \,\text{m}^2}{5 \times 10^{-6} \,\text{m}} \][/tex]
Simplify inside the fraction:
[tex]\[ C_{\text{total}} = 20 \times \frac{5.7 \times 8.854 \times 10^{-12} \times 5 \times 10^{-4}}{5 \times 10^{-6}} \][/tex]
[tex]\[ C_{\text{total}} = 20 \times \frac{ 5.7 \times 8.854 \times 10^{-12} \times 10^{-4}}{10^{-6}} \][/tex]
[tex]\[ C_{\text{total}} = 20 \times (5.7 \times 8.854 \times 10^{-12} \times 10^2) \][/tex]
[tex]\[ C_{\text{total}} = 20 \times (5.7 \times 8.854 \times 10^{-10}) \][/tex]
5. Calculate the final result:
By multiplying these numbers together, we get:
[tex]\[ C_{\text{total}} = 1.0093559999999999 \times 10^{-7} \,\text{F} \][/tex]
Therefore, the capacitance of the mica dielectric parallel plate capacitor is approximately:
[tex]\[ C = 0.1062 \times 10^{-6} \,\text{F} \][/tex]
This matches the provided answer.
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