To find the capacitance [tex]\( C \)[/tex] of a parallel plate capacitor, we use the formula:
[tex]\[ C = \frac{\varepsilon_0 A}{d} \][/tex]
Where:
- [tex]\( \varepsilon_0 \)[/tex] is the permittivity of free space [tex]\( \varepsilon_0 = 8.85 \cdot 10^{-12} \, \text{C}^2 / \text{N} \cdot \text{m}^2 \)[/tex]
- [tex]\( A \)[/tex] is the area of the plates [tex]\( A = 5.25 \cdot 10^{-5} \, \text{m}^2 \)[/tex]
- [tex]\( d \)[/tex] is the separation between the plates [tex]\( d = 3.14 \cdot 10^{-6} \, \text{m} \)[/tex]
Substituting the known values into the formula:
[tex]\[ C = \frac{(8.85 \cdot 10^{-12} \, \text{C}^2 / \text{N} \cdot \text{m}^2) \cdot (5.25 \cdot 10^{-5} \, \text{m}^2)}{3.14 \cdot 10^{-6} \, \text{m}} \][/tex]
Performing the multiplication in the numerator first:
[tex]\[ \varepsilon_0 \cdot A = (8.85 \cdot 10^{-12}) \cdot (5.25 \cdot 10^{-5}) = 46.4625 \cdot 10^{-17} \, \text{C}^2 / \text{N} \cdot \text{m} \][/tex]
Next, dividing by the separation [tex]\( d \)[/tex]:
[tex]\[ C = \frac{46.4625 \cdot 10^{-17}}{3.14 \cdot 10^{-6}} = 1.4796974522292991 \cdot 10^{-10} \, \text{F} \][/tex]
So, the capacitance [tex]\( C \)[/tex] of the capacitor is approximately:
[tex]\[ \boxed{1.48 \cdot 10^{-10} \, \text{F}} \][/tex]
