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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 Earth can be thought of as a spherical capacitor. If the net charge on the Earth is \(-7.8 \cdot 10^{5} \mathrm{C},\) find \((\) a) the capacitance of the Earth and (b) the electric potential energy stored on the Earth's surface

Two capacitors, with capacitances \(C_{1}\) and \(C_{2},\) are connected in series. A potential difference, \(V_{0}\), is applied across the combination of capacitors. Find the potential differences \(V_{1}\) and \(V_{2}\) across the individual capacitors, in terms of \(V_{0}\), \(C_{1},\) and \(C_{2}\).

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 of capacitance \(C\) has plates of area \(A\) with distance \(d\) between them. When the capacitor is connected to a battery of potential difference \(V\), it has a charge of magnitude \(Q\) on its plates. While the capacitor is connected to the battery, the distance between the plates is decreased by a factor of \(3 .\) The magnitude of the charge on the plates and the capacitance are then a) \(\frac{1}{3} Q\) and \(\frac{1}{3} C\). c) \(3 Q\) and \(3 C\). b) \(\frac{1}{3} Q\) and \(3 C\). d) \(3 Q\) and \(\frac{1}{3} C\).

The Earth is held together by its own gravity. But it is also a charge-bearing conductor. a) The Earth can be regarded as a conducting sphere of radius \(6371 \mathrm{~km},\) with electric field \(\vec{E}=(-150 . \mathrm{V} / \mathrm{m}) \hat{r}\) at its surface, where \(\hat{r}\) is a unit vector directed radially outward. Calculate the total electrostatic potential energy associated with the Earth's electric charge and field. b) The Earth has gravitational potential energy, akin to the electrostatic potential energy. Calculate this energy, treating the Earth as a uniform solid sphere. (Hint: \(d U=-(G m / r) d m\). c) Use the results of parts (a) and (b) to address this question: To what extent do electrostatic forces affect the structure of the Earth?
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