Answer to Question #187484 in Electric Circuits for Muzala Seleji

Question

A PHY-102 student at Mulungushi University wants to make a simple 10 𝑛𝐹 capacitor for tuning her crystal radio, using two sheets of aluminum foil as plates, with a few sheets of paper between them as a dielectric. The paper has a dielectric constant of 3.0, and the thickness of one sheet of it is 0.20 mm. If the sheets of paper measure 22×28 𝑐𝑚 and she cuts the aluminum foil to the same dimensions, how many sheets of paper would she use between her plates to get the proper capacitance?

Answer 1

Capacitance, $C=10\space nF=10\times10^{-9}\space F$

Dielectric constant, $K=3$

Thickness of one sheet, $d=0.20\space mm=0.20\times10^{-3}\space m$

Area of plates, $A=22\times28\space cm^2=616\times10^{-4}\space m^2$

$C_o=\dfrac{K\epsilon_o A}{d}$

$C_o=\dfrac{3\times8.85\times10^{-12}\times616\times10^{-4}}{0.20\times10^{-3}}$

$C_o=8.1774\times10^{-9}\space F$

Number of sheets of paper used be $n$

Capacitance needed, $C=10\times10^{-9}\space F$

Equivalent capacitance,

$\dfrac{1}{C}=\bigg[\dfrac{1}{C_o}+\dfrac{1}{C_o}+\dfrac{1}{C_o}+........+\dfrac{1}{C_o}\bigg]$

$C=\dfrac{C_o}{n}$

$n=\dfrac{C_o}{C}=\dfrac{8.1774\times10^{-9}}{10\times10^{-9}}\approx1$