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

`0.19\mu F`

Step-by-step 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* 10^(-12)`

Distance d =
`4.1* 10^(-2)mm=4.1* 10^(-5)m`

Dielectric constant K = 2.1 given in question

Capacitance is given by
`C=(K\epsilon _0A)/(d)=(2.1* 8.85* 10^(-12)* 0.42)/(4.1* 10^(-5))=0.19\mu F`