Question

A parallel-plate capacitor is made from two aluminum-foil sheets, each 7.5 cm wide and 5.6 m long. Between the sheets is a Teflon strip of the same width and length that is 4.1×10^−2 mm thick. What is the capacitance of this capacitor? (The dielectric constant of Teflon is 2.1.)

Answer 1

Answer:

$0.19\mu F$

Explanation:

Length of the sheet = 7.5 cm =0.075m

Width of the sheet =5.6 m

So area of the sheet A= 5.6×0.075=0.42$m^2$

Value of $\varepsilon _0=8.85\times 10^{-12}$

Distance d =$4.1\times 10^{-2}mm=4.1\times 10^{-5}m$

Dielectric constant K = 2.1 given in question

Capacitance is given by $C=\frac{K\epsilon _0A}{d}=\frac{2.1\times 8.85\times 10^{-12}\times 0.42}{4.1\times 10^{-5}}=0.19\mu F$