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Chapter 24: Problem 61

Considering the dielectric strength of air, what is the maximum amount of charge that can be stored on the plates of a capacitor that are a distance of \(15 \mathrm{~mm}\) apart and have an area of \(25 \mathrm{~cm}^{2}\) ?

Short Answer

Expert verified
Answer: The maximum charge that can be stored on the plates of the capacitor is \(6.6375 \times 10^{-7} \mathrm{~C}\).

Step by step solution

01

Convert the given dimensions to meters

First, we need to convert the given dimensions (distance and area) to meters for consistency in calculations: - The distance between the plates is given as \(15 \mathrm{~mm}\). Convert this to meters: \(15 \mathrm{~mm} = 0.015 \mathrm{~m}\). - The area of the plates is given as \(25 \mathrm{~cm}^{2}\). Convert this to square meters: \(25 \mathrm{~cm}^{2} = 2.5 \times 10^{-3} \mathrm{~m}^{2}\).
02

Calculate the maximum electric field

The dielectric strength of air is given as \(3 \times 10^{6} \mathrm{~V/m}\). This is the maximum electric field that air can withstand without breaking down. We can calculate the maximum electric field in air using: \(E = \frac{V}{d}\) where \(E\) is the electric field, \(V\) is the voltage, and \(d\) is the distance between the plates. We can use the dielectric strength of air to find the maximum voltage, \(V_{max}\), that the capacitor can handle: \(V_{max} = E_{max} \times d\) Substituting the values into the formula, we get: \(V_{max} = (3 \times 10^{6} \mathrm{~V/m}) \times (0.015 \mathrm{~m}) = 45,000 \mathrm{~V}\)
03

Calculate the capacitance

Using the capacitance formula: \(C = \epsilon_{0} \frac{A}{d}\) where \(\epsilon_{0}\) is the vacuum permittivity constant, \(A\) is the area of the plates, and \(d\) is the distance between the plates. The value of \(\epsilon_0 = 8.85 \times 10^{-12} \mathrm{~F/m}\). Substitute the values to find the capacitance: \(C = (8.85 \times 10^{-12} \mathrm{~F/m}) \times \frac{2.5 \times 10^{-3} \mathrm{~m}^{2}}{0.015 \mathrm{~m}} = 1.475 \times 10^{-11} \mathrm{~F}\)
04

Calculate the maximum charge stored

Now that we have the capacitance and the maximum voltage that the capacitor can handle, we can find the maximum charge using: \(Q = C \times V\) where \(Q\) is the charge stored, \(C\) is the capacitance, and \(V\) is the voltage across the plates. Substitute the values: \(Q = (1.475 \times 10^{-11} \mathrm{~F}) \times (45,000 \mathrm{~V}) = 6.6375 \times 10^{-7} \mathrm{~C}\) Therefore, the maximum amount of charge that can be stored on the plates of the capacitor is \(6.6375 \times 10^{-7} \mathrm{~C}\).

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Most popular questions from this chapter

Thermocoax is a type of coaxial cable used for high-frequency filtering in cryogenic quantum computing experiments. Its stainless steel shield has an inner diameter of \(0.35 \mathrm{~mm},\) and its Nichrome conductor has a diameter of \(0.17 \mathrm{~mm}\). Nichrome is used because its resistance doesn't change much in going from room temperature to near absolute zero. The insulating dielectric is magnesium oxide \((\mathrm{MgO}),\) which has a dielectric constant of \(9.7 .\) Calculate the capacitance per meter of Thermocoax.

The capacitor in an automatic external defibrillator is charged to \(7.5 \mathrm{kV}\) and stores \(2400 \mathrm{~J}\) of energy. What is its capacitance?

A parallel plate capacitor with a plate area of \(12.0 \mathrm{~cm}^{2}\) and air in the space between the plates, which are separated by \(1.50 \mathrm{~mm},\) is connected to a \(9.00-\mathrm{V}\) battery. If the plates are pulled back so that the separation increases to \(2.75 \mathrm{~mm}\) how much work is done?

Calculate the capacitance of the Earth. Treat the Earth as an isolated spherical conductor of radius \(6371 \mathrm{~km}\).

What is the radius of an isolated spherical conductor that has a capacitance of \(1.00 \mathrm{~F} ?\)
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