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Chapter 24: Problem 58

Design a parallel plate capacitor with a capacitance of \(47.0 \mathrm{pF}\) and a capacity of \(7.50 \mathrm{nC}\). You have available conducting plates, which can be cut to any size, and Plexiglas sheets, which can be cut to any size and machined to any thickness. Plexiglas has a dielectric constant of 3.40 and a dielectric strength of \(4.00 \cdot 10^{7} \mathrm{~V} / \mathrm{m}\). You must make your capacitor as compact as possible. Specify all relevant dimensions. Ignore any fringe field at the edges of the capacitor plates.

Short Answer

Expert verified
The smallest possible size for this parallel plate capacitor has conducting plates with an area of 2.00 x 10^-7 m^2 and a distance of 3.99 x 10^-6 m between them.

Step by step solution

01

Calculate the product of the vacuum and dielectric constants

We can multiply the vacuum permittivity (\(\varepsilon_0\)) with the dielectric constant of Plexiglas (\(\varepsilon_r\)) to get the product of both values: Product = \(\varepsilon_0 \varepsilon_r = (8.854 \times 10^{-12} \mathrm{F} / \mathrm{m}) \times 3.40\)
02

Calculate the distance d between the capacitor plates

We can use the given dielectric strength to calculate the minimum distance between the plates while maintaining safety. Since the dielectric strength is expressed as \(V/m\), we need to consider the capacity of the capacitor (\(7.50 \mathrm{nC}\)) and convert it to the voltage value by dividing the capacity by the given capacitance: \(V = \frac{Q}{C} = \frac{7.50 \times 10^{-9} \mathrm{C}}{47.0 \times 10^{-12} \mathrm{F}} = 159.6 \mathrm{V}\) Now, we can use this voltage value and the given dielectric strength to calculate the minimum distance between the plates: \(d = \frac{V}{E} = \frac{159.6 \mathrm{V}}{4.00 \times 10^7 \mathrm{V} / \mathrm{m}} = 3.99 \times 10^{-6} \mathrm{m}\)
03

Calculate the Area of the Capacitor Plates

We can now use the given capacitance value, calculated product of permittivity and distance (\(d\)) to find the area of the conducting plates: \(A = \frac{C \cdot d}{\varepsilon_0 \varepsilon_r} = \frac{47.0 \times 10^{-12} \mathrm{F} \cdot 3.99 \times 10^{-6} \mathrm{m}}{8.854 \times 10^{-12} \mathrm{F} / \mathrm{m} \cdot 3.40} = 2.00 \times 10^{-7} \mathrm{m^2}\)
04

Final Design

The designed parallel plate capacitor with the smallest possible size has conducting plates with an area of \(2.00 \times 10^{-7} \mathrm{m^2}\) and a distance of \(3.99 \times 10^{-6} \mathrm{m}\) between them. Plexiglas sheets of this area and thickness should be used as the dielectric material.

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Most popular questions from this chapter

The potential difference across two capacitors in series is \(120 . \mathrm{V}\). The capacitances are \(C_{1}=1.00 \cdot 10^{3} \mu \mathrm{F}\) and \(C_{2}=1.50 \cdot 10^{3} \mu \mathrm{F}\) a) What is the total capacitance of this pair of capacitors? b) What is the charge on each capacitor? c) What is the potential difference across each capacitor? d) What is the total energy stored by the capacitors?

A parallel plate capacitor is connected to a battery for charging. After some time, while the battery is still connected to the capacitor, the distance between the capacitor plates is doubled. Which of the following is (are) true? a) The electric field between the plates is halved. b) The potential difference of the battery is halved. c) The capacitance doubles. d) The potential difference across the plates does not change. e) The charge on the plates does not change.

A parallel plate capacitor consists of square plates of edge length \(2.00 \mathrm{~cm}\) separated by a distance of \(1.00 \mathrm{~mm}\). The capacitor is charged with a \(15.0-\mathrm{V}\) battery, and the battery is then removed. A \(1.00-\mathrm{mm}\) -thick sheet of nylon (dielectric constant \(=3.0\) ) is slid between the plates. What is the average force (magnitude and direction) on the nylon sheet as it is inserted into the capacitor?

A dielectric slab with thickness \(d\) and dielectric constant \(\kappa=2.31\) is inserted in a parallel place capacitor that has been charged by a \(110 .-\mathrm{V}\) battery and having area \(A=\) \(100, \mathrm{~cm}^{2},\) and separation distance \(d=2.50 \mathrm{~cm}\). a) Find the capacitance, \(C,\) the potential difference, \(V,\) the electric field, \(E,\) the total charge stored on the capacitor \(Q\), and electric potential energy stored in the capacitor, \(U\), before the dielectric material is inserted. b) Find \(C, V, E, Q,\) and \(U\) when the dielectric slab has been inserted and the battery is still connected. c) Find \(C, V, E, Q\) and \(U\) when the dielectric slab is in place and the battery is disconnected.

A parallel plate capacitor is charged with a battery and then disconnected from the battery, leaving a certain amount of energy stored in the capacitor. The separation between the plates is then increased. What happens to the energy stored in the capacitor? Discuss your answer in terms of energy conservation.
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