Answered

A capacitor has plates with an area of [tex]$5.25 \cdot 10^{-5} \, m^2$[/tex], separated by [tex]$3.14 \cdot 10^{-6} \, m$[/tex]. What is the capacitance?

[tex]\[
\begin{array}{c}
{[?] \times 10^{[?]} \, F } \\
\text { Hint: } C = \frac{\varepsilon_0 A}{d} \\
\varepsilon_0 = 8.85 \cdot 10^{-12} \, \frac{C^2}{N \cdot m^2}
\end{array}
\][/tex]



Answer :

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]