a budding electronics hobbyist wants to make a simple 1.4 nF capcitor for tuning

Question

a budding electronics hobbyist wants to make a simple 1.4 nF capcitor for tuning her cyrstal 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 4.9 and the thickness of one sheet of it is 0.20 mm PART A if the sheets paper measure 27 cm x 36 cm and she cuts the aluminum foil to the same dimensions, how many sheets of paper should she use between her plates to get the proper capacitance? -express as a whole number Suppose for convenience she wants to use a single sheet of posterboard, with teh same dielectric constant but a thickness of 11.0 mm, instead of the paper. what area of aluminum foild will she need for her plates to get her 1.4 nF of capacitance -2 sigfigs

Explanation / Answer

a)C=KoA/d =>d=KoA/C

d =4.9*8.85*10-12*(0.27*0.36)/1.4*10-9

d=3*10-3m =3mm

number of sheets N =3*10-3/0.2*10-3 =15

b)here capacitance and dielectric are constant.

capacitance of paper C1=KA1/d1

capacitance of poster board C2 =KA2/d2

here capacitances are constant i.e C1 =C2

KA1/d1 =KA2/d2

A1/d1 =A2/d2

(0.27*0.36)/(3*10-3)= A2/11*10-3

A=0.3564m2

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