/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 66 An air capacitor is made by usin... [FREE SOLUTION] | 91Ó°ÊÓ

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An air capacitor is made by using two flat plates, each with area \(A,\) separated by a distance \(d\) . Then a a metal slab having thickness a (less than \(d\) ) and the same shape and size as the plates is inserted between them, parallel to the plates and not touching either plate (Fig. P24.66). (a) What is the capacitance of this arrangement? (b) Express the capacitance as a multiple of the capacitance \(C_{0}\) when the metal slab is not present. (c) Discuss what happens to the capacitance in the limits \(a \rightarrow 0\) and \(a \rightarrow d\) .

Short Answer

Expert verified
(a) The capacitance is \(C = \frac{\varepsilon A}{d-a}\). (b) Expressed as a multiple: \(C = \frac{C_0 \times d}{d-a}\). (c) As \(a \rightarrow 0\), \(C\) approaches \(C_0\); as \(a \rightarrow d\), \(C\) approaches infinity.

Step by step solution

01

Understand the Problem and Set Up

We have two flat conducting plates forming a capacitor, with area \(A\) and separated by a distance \(d\). A metal slab of thickness \(a\) (where \(a < d\)) is introduced between the plates. We need to find the capacitance of this arrangement.
02

Concept of Capacitance

Capacitance \(C\) is given by the formula \(C = \frac{\varepsilon A}{d}\), where \(\varepsilon\) is the permittivity of the medium between the plates, \(A\) is the area of the plates, and \(d\) is the separation between the plates. Here, the introduction of the metal slab modifies the effective distance between the plates.
03

Analyze Introduction of Metal Slab

The metal slab divides the capacitor into two capacitors, one with dielectric thickness \(d-a\) and the other also with \(d-a\). The slab itself occupies a thickness \(a\). The effective permittivity for the part without the slab remains \(\varepsilon\), while the slab has permittivity approximation as a conductor (infinite permittivity).
04

Calculate Capacitance with Metal Slab

Consider the metal slab as two capacitors in series. The effective capacitance \(C\) is \[ \frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{(d-a)}{\varepsilon A} + \frac{(d-a)}{\varepsilon A} \].Simplifying, we get the effective capacitance:\[ C = \frac{\varepsilon A}{d-a} \].
05

Express as a Multiple of Initial Capacitance

The capacitance without the metal slab (\(C_0\)) is given by \(C_0 = \frac{\varepsilon A}{d}\). Therefore, the new capacitance as a multiple of \(C_0\) is:\[ C = \frac{C_0 \times d}{d-a} \].
06

Discuss Limits of a

- As \(a \rightarrow 0\), the capacitance \(C\) approaches \(C_0\), since the slab thins out to nothing.- As \(a \rightarrow d\), \(d-a\) approaches 0 making \(C\) approach infinity. This implies the gap is filled entirely by the conductor, reducing the effective separation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Capacitance Formula
Capacitance is a fundamental concept in physics and electronics. It describes the ability of a system to store electric charge. The formula for capacitance, when dealing with a basic parallel plate capacitor, is given by: \[C = \frac{\varepsilon A}{d}\]Here, \(\varepsilon\) represents the permittivity of the medium between the plates. \(A\) is the area of one of the plates, and \(d\) is the distance separating these plates.
Permittivity indicates how easily electric fields can form in a medium. Different materials have different permittivity values, affecting the capacitance.
- In a vacuum or air, we use \(\varepsilon_0\), the permittivity of free space.- If there's a dielectric material instead of air, the permittivity increases, increasing the capacitance.
In the case of the metal slab, though it doesn't change the vacuum permittivity, it effectively alters the distance \(d\), thereby influencing the capacitance. Understanding how different factors modify capacitance helps in designing efficient circuits.
Metal Slab in Capacitors
When a metal slab is introduced between the plates of a capacitor, it alters the effective separation and distribution of electric fields within the device. The metal slab has a significant impact on the capacitance value because it behaves like a dielectric of infinite permittivity.
This creates a unique scenario where the slab divides the space between the plates. Essentially, the capacitor is split into two sections:
  • The space where the metal slab isn't present has a separation \(d - a\).
  • The slab itself, with thickness \(a\), acts as a rearrangement in the electric field distribution.
The effective capacitance of this configuration needs to consider the contributions from these sections, resulting in the calculation:\[C = \frac{\varepsilon A}{d-a}\]It becomes crucial to ensure the metal doesn't touch the plates to prevent a short circuit. Unlike a true dielectric, the metal doesn't polarize but rather, just redirects the field lines. This arrangement demonstrates how inserting conductive materials can efficiently manipulate capacitance.
Series Capacitors Effect
Inserting a metal slab between two plates replicates the effect of capacitors in series. When capacitors are placed in series, the combined capacitance is less than any individual capacitor's capacitance. This is important as it impacts the charge storage ability of the system.
In the configuration with a metal slab:
  • The effective separation is altered by segmenting the capacitor into two regions with separation \(d-a\).
  • You have to sum the effect of these segmented regions, yielding a reduced effective separation, effectively increasing the capacitance.
The expression for combining capacitors in series, here adapted for the slab scenario, is:\[\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2}\]Solving this gives a higher overall capacitance when part of the distance is "replaced" by a conducting slab.
Interestingly, as the slab approaches the full plate distance, the system mimics that of two capacitors shorting, and capacitance tends to infinity. The relationship between series capacitors and metal inserts highlights how components in circuits can fundamentally change function by introducing such elements.

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Most popular questions from this chapter

A capacitor is made from two hollow, coaxial copper cylinders, one inside the other. There is air in the space between the cylinders. The inner cylinder has net positive charge and the outer cylinder has net negative charge. The inner cylinder has radius \(2.50 \mathrm{mm},\) the outer cylinder has radius \(3.10 \mathrm{mm},\) and the length of each cylinder is 36.0 \(\mathrm{cm} .\) If the potential difference between the surfaces of the two cylinders is \(80.0 \mathrm{V},\) what is the magnitude of the electric field at a point between the two cylinders that is a distance of 2.80 \(\mathrm{mm}\) from their common axis and midway between the ends of the cylinders?

The plates of a parallel-plate capacitor are 2.50 \(\mathrm{mm}\) apart, and each carries a charge of magnitude 80.0 \(\mathrm{nC}\) . The plates are in vacuum. The electric field between the plates has a magnitude of \(4.00 \times 10^{6} \mathrm{V} / \mathrm{m}\) . (a) What is the potential difference between the plates? (b) What is the area of each plate? (c) What is the capacitance?

Electronic flash units for cameras contain a capacitor for storing the energy used to produce the flash. In one such unit, the flash lasts for \(\frac{1}{677}\) s with an average light power output of \(2.70 \times 10^{5} \mathrm{W}\) (a) If the conversion of electrical energy to light is 95\(\%\) efficient (the rest of the energy goes to thermal energy), how much energy must be stored in the capacitor for one flash? (b) The capacitor has a potential difference between its plates of 125 \(\mathrm{V}\) when the stored energy equals the value calculated in part (a). What is the capacitance?

A parallel-plate air capacitor is to store charge of magnitude 240.0 \(\mathrm{pC}\) on each plate when the potential difference between the plates is 42.0 \(\mathrm{V}\) (a) If the area of each plate is \(6.80 \mathrm{cm}^{2},\) what is the separation between the plates? (b) If the separation between the two plates is double the value calculated in part (a), what potential difference is required for the capacitor to store charge of magnitude 240.0 pC on each plate?

A \(5.80-\mu \mathrm{F}\), parallel-parallel-plate, air capacitor has a plate separation of 5.00 \(\mathrm{mm}\) and is charged to a potential difference of 400 \(\mathrm{V}\) . Calculate the energy density in the region between the plates, in units of \(\mathrm{J} / \mathrm{m}^{3} .\)

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