Question

Review Part A A parallel plate capacitor is made from two aluminum-foil sheets, each 6.6 cm wide and 5.2 long. Between the sheets is a Teflon strip of the same width and length that is 4.1x102 mm thick What is the capacitance of this capacitor? (The dielectric constant of Teflon is 2.1.) Express your answer using two significant figures AF Submit Previous Answers Request Answer X Incorrect; Try Again; 2 attempts remaining rovide Feedback Next >

Answer 1

Length of the aluminum sheets l= 5.2m

Width of the aluminum sheets w= 6.6 cm = $6.6*10^-2$ m

Area of sheets A= l * w = $5.2*6.6*10^-2$

A=0.3432 $m^2$

Now the thickness of the Teflon d = $4.1*10^-2 mm = 4.1*10^-5 m$

The dielectric constant of Teflon $\zeta$ = 2.1

Now the capacitance of the capacitor

$C=\zeta *\zeta0 * A / d$

$\varepsilon 0 = 8.85* 10^-12$= permittivity of the free space

$C=2.1*8.85*10^-12*0.3432/ 4.1*10^-5$

C= $0.155 * 10^-6 F$

$C= 0.155 \mu F$

Is answer.