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A parallel-plate capacitor is made from two aluminum-foil sheets, each 6.3 cm6.3 \mathrm{~cm} wide and 5.4 m5.4 \mathrm{~m} long. Between the sheets is a Teflon strip of the same width and length that is 0.035 mm0.035 \mathrm{~mm} thick. What is the capacitance of this capacitor? (The dielectric constant of Teflon is 2.12.1.)

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Given:\textbf{Given:}

Width=6.3 cm=6.3×102 m\text{Width} = 6.3\ \text{cm} = 6.3 \times 10^{-2}\ \text{m}

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

d=Thickness=0.035×103 md = \text{Thickness} = 0.035 \times 10^{-3}\ \text{m}

k=2.1k =2.1

Approach:\textbf{Approach:}

Firstly, we will find the area of parallel plate capacitor followed by the capacitance using the equation C=kϵ0AdC = \frac{k \epsilon_0 A}{d}

Calculations:\textbf{Calculations:}

Area of parallel plate capacitor is given by :

A=Length×Width=(5.4)(6.3×102)=34.02×102 m2\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=Thickness of Teflon stripd = \text{Thickness of Teflon strip}

d=0.035×103 md = 0.035 \times 10^{-3}\ m

Now, capacitance of parallel plate capacitor is given by :

C=kϵ0Ad=(2.1)(8.85×1012)(34.02×102)(0.035×103)=180646.2×1012 F=0.1806462×106 F\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*}

C=0.18μF\boxed{C = 0.18 \mu F}

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