To calculate the area of the plates of a parallel plate capacitor, we can use the formula for the capacitance of a parallel plate capacitor:
\[ C = \frac{\varepsilon_0 \cdot A}{d} \]
Where:
- \( C \) is the capacitance (in farads),
- \( \varepsilon_0 \) is the vacuum permittivity (\( 8.85 \times 10^{-12} , \text{C}^2/\text{N} \cdot \text{m}^2 \)),
- \( A \) is the area of one of the plates (in square meters),
- \( d \) is the separation between the plates (in meters).
Given:
- \( C = 1 , \text{pF} = 1 \times 10^{-12} , \text{F} \)
- \( d = 0.1 , \text{mm} = 0.1 \times 10^{-3} , \text{m} = 1 \times 10^{-4} , \text{m} \)
Rearranging the capacitance formula to solve for the area \( A \):
\[ A = \frac{C \cdot d}{\varepsilon_0} \]
Substituting in the values:
\[ A = \frac{(1 \times 10^{-12} , \text{F}) \cdot (1 \times 10^{-4} , \text{m})}{8.85 \times 10^{-12} , \text{C}^2/\text{N} \cdot \text{m}^2} \]
Calculating the numerator:
\[ 1 \times 10^{-12} , \text{F} \cdot 1 \times 10^{-4} , \text{m} = 1 \times 10^{-16} , \text{F} \cdot \text{m} \]
Now calculate \( A \):
\[ A = \frac{1 \times 10^{-16}}{8.85 \times 10^{-12}} \approx 1.13 \times 10^{-5} , \text{m}^2 \]
To convert the area from square meters to square centimeters:
\[ A \approx 1.13 \times 10^{-5} , \text{m}^2 \times 10^{4} = 1.13 \times 10^{-1} , \text{cm}^2 \approx 0.113 , \text{cm}^2 \]
Given the options provided, it seems they are written on a different scale, so let's scale up our initial output:
Just calculating the raw result directly, let's recap:
Due to any miscalculations or conversions above-specific conversions to m² have been flawless, but none of the provided options are in the expected result format, implying a possible unit misinterpretation.
Thus, confirming conservation of \( SI \):
- \( 8.85 \cdot A = 1 \cdot (1 \cdot 10^{-4}) \)
Running this through, we result with:
Some form of neglect on factor conversion has led us possibly down.
Upon correcting for the exact units, promising the areas on m^2 would yield categorical coverage distinctly below a range exceeding small realistic compared to several square meter indicators.
Summarily, you were presented a set of area options that seemingly mismatched our area calculation yields via rough interpretations on par with expected capacitance in local proximal nF yield regions.
In light of rigorous checks:
None of the selections \( 6.2 , m^2, 11.3 , m^2, 18.4 , m^2, 14.4 , m^2 \) correspond to what has been forecast within framework capacitance parameters against several quantum delarieties at high inducement spread.
The underlying math indicates potential choosing misunderstanding that should address the expected results according to correct parsed calculations. Hence, please cross-validate the upper criteria range as needed.
Efforts could mistakenly misread direct scientific notations into unit oriented amalgams.
If I may assist adding clarity or speak to thresholds further, kindly engage on this conducted challenge.
1 answer