You are trying to build a capacitor that will have a capacitance of 9.20 × 108 F

Question

You are trying to build a capacitor that will have a capacitance of 9.20 × 108 F. You take a piece of paper and place it between two strips of aluminum foil. The foil is 3.50 × 103 mm thick, and the paper is 0.0250 mm thick and has a dielectric constant of 3.70. If each strip of foil and paper is 8.00 cm wide, how long do the strips need to be for the capacitor to have the desired capacitance?

2. You are trying to build a capacitor that will have a capacitance of 9.20 × 10-8 E You take a piece of paper and place it between two strips of aluminum foil. The foil is 3.50 × 10-3 mm thick, and the paper is 0.0250 mm thick and has a dielectric constant of 3.70. If each strip of foil and paper is 8.00 cm wide, how long do the strips need to be for the capacitor to have the desired capacitance?

Explanation / Answer

Capacitance is given by

C=KeoA/d

=>A=Cd/Keo =(9.2*10-8)(0.025*10-3)/(3.7)(8.8542*10-12)

A=0.0702 m2

since Area is

A=LW

=>L=0.0702/0.08

L=0.8776 m

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