Answer:
To find the size of the plates needed to construct a 4.7 µF capacitor with a paper dielectric, we can use the formula for the capacitance of a parallel plate capacitor:
\[ C = \frac{\varepsilon_r \varepsilon_0 A}{d} \]
Where:
- \( C \) is the capacitance (4.7 µF or \( 4.7 \times 10^{-6} \) F)
- \( \varepsilon_r \) is the relative permittivity of the dielectric material (2.5 for paper)
- \( \varepsilon_0 \) is the permittivity of free space (\( 8.854 \times 10^{-12} \) F/m)
- \( A \) is the area of one of the plates
- \( d \) is the separation between the plates (0.08 mm or \( 0.08 \times 10^{-3} \) m)
First, rearrange the formula to solve for \( A \):
\[ A = \frac{C \cdot d}{\varepsilon_r \varepsilon_0} \]
Substitute the known values into the formula:
\[ A = \frac{4.7 \times 10^{-6} \, \text{F} \times 0.08 \times 10^{-3} \, \text{m}}{2.5 \times 8.854 \times 10^{-12} \, \text{F/m}} \]
Now calculate the area \( A \):
\[ A = \frac{4.7 \times 10^{-6} \times 0.08 \times 10^{-3}}{2.5 \times 8.854 \times 10^{-12}} \]
\[ A = \frac{3.76 \times 10^{-10}}{2.2135 \times 10^{-11}} \]
\[ A \approx 17.0 \, \text{m}^2 \]
This is the area of one plate in square meters. To find the side length \( L \) of the square plates, take the square root of the area:
\[ L = \sqrt{A} \]
\[ L = \sqrt{17.0} \]
\[ L \approx 4.12 \, \text{m} \]
Therefore, the student would need square plates approximately 4.12 meters on each side. This is quite large for a practical science fair project, and it might be worth considering alternative configurations or materials.
