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Chapter 17: Problem 71

A capacitor can be made from two sheets of aluminum foil separated by a sheet of waxed paper. If the sheets of aluminum are \(0.30 \mathrm{m}\) by $0.40 \mathrm{m}$ and the waxed paper, of slightly larger dimensions, is of thickness \(0.030 \mathrm{mm}\) and dielectric constant \(\kappa=2.5,\) what is the capacitance of this capacitor?

Short Answer

Expert verified
Question: Calculate the capacitance of a capacitor made up of two rectangular aluminum plates, each with dimensions of 0.30 m x 0.40 m, separated by a 0.030 mm layer of waxed paper with a dielectric constant of 2.5. Answer: The capacitance of the capacitor is approximately \(8.91 \times 10^{-9} \mathrm{F}\).

Step by step solution

01

Calculate the area of the plates

The given dimensions of the sheets of aluminum are \(0.30 \mathrm{m}\) and \(0.40 \mathrm{m}\). To find the area of the plates, we simply multiply these dimensions together: \(A = 0.30 \mathrm{m} \times 0.40 \mathrm{m} = 0.12 \mathrm{m^2}\)
02

Convert the thickness of the waxed paper into meters

The thickness of the waxed paper is given as \(0.030 \mathrm{mm}\). To convert this into meters, we use the following conversion: \(1 \mathrm{mm} = 0.001 \mathrm{m}\) So, \(0.030 \mathrm{mm} = 0.030 \times 0.001 \mathrm{m} = 0.00003 \mathrm{m}\)
03

Calculate the capacitance

Now we have all the necessary values to calculate the capacitance of the capacitor. We plug these values into the capacitance formula: \(C = \kappa \varepsilon_0 \frac{A}{d}\) where \(\kappa = 2.5\), \(\varepsilon_0 = 8.85 \times 10^{-12} \mathrm{F/m}\), \(A = 0.12 \mathrm{m^2}\), and \(d = 0.00003 \mathrm{m}\). \(C = 2.5 \times 8.85 \times 10^{-12} \frac{0.12}{0.00003} = 8.85 \times 10^{-12} \times 2.5 \times 0.12 \div 0.00003\) Calculating the capacitance, we obtain: \(C \approx 8.91 \times 10^{-9} \mathrm{F}\) Thus, the capacitance of this capacitor is approximately \(8.91 \times 10^{-9} \mathrm{F}\).

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Most popular questions from this chapter

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