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# A parallel-plate capacitor is made from two aluminum-foil sheets, each $6.3 \mathrm{~cm}$ wide and $5.4 \mathrm{~m}$ long. Between the sheets is a Teflon strip of the same width and length that is $0.035 \mathrm{~mm}$ thick. What is the capacitance of this capacitor? (The dielectric constant of Teflon is $2.1$.)

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$\textbf{Given:}$

$\text{Width} = 6.3\ \text{cm} = 6.3 \times 10^{-2}\ \text{m}$

$\text{Length} = 5.4\ \text{m}$

$d = \text{Thickness} = 0.035 \times 10^{-3}\ \text{m}$

$k =2.1$

$\textbf{Approach:}$

Firstly, we will find the area of parallel plate capacitor followed by the capacitance using the equation $C = \frac{k \epsilon_0 A}{d}$

$\textbf{Calculations:}$

Area of parallel plate capacitor is given by :

\begin{align*} A & = \text{Length} \times \text{Width}\\ & = (5.4) \cdot (6.3 \times 10^{-2})\\ & = 34.02 \times 10^{-2}\ m^2 \end{align*}

The distance between the plates of capacitor will be equal to thickness of Teflon strip as Teflon strip is completely filled between the plates of capacitor.

$d = \text{Thickness of Teflon strip}$

$d = 0.035 \times 10^{-3}\ m$

Now, capacitance of parallel plate capacitor is given by :

\begin{align*} C & = \frac{k \epsilon_0 A}{d}\\ & = \frac{(2.1)(8.85 \times 10^{-12})(34.02 \times 10^{-2})}{(0.035 \times 10^{-3})}\\ & = 180646.2 \times 10^{-12}\ F\\ & = 0.1806462 \times 10^{-6}\ F \end{align*}

$\boxed{C = 0.18 \mu F}$

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