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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 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 (\(\textbf{Fig. P24.62}\)). (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) Total capacitance is \(C = \frac{\varepsilon_0 A}{d-a}\). (b) Capacitance as multiple of \(C_0\) is \(\frac{C_0}{1-\frac{a}{d}}\). (c) As \(a\rightarrow 0\), \(C\rightarrow C_0\); as \(a\rightarrow d\), \(C\rightarrow \infty\).

Step by step solution

01

Understanding the Problem

We need to find the capacitance of a capacitor with two parallel plates separated by a distance \(d\) and having a metal slab of thickness \(a\) inserted between them. The challenge is to figure out how the capacitance changes with this insertion.
02

Concept of Series Capacitance

The metal slab divides the capacitor into two capacitors connected in series: one with separation \(d-a\) and one with slab thickness \(a\). We need to calculate the individual capacitances of these two parts and combine them to find the total capacitance.
03

Calculate Capacitance without Metal

The capacitance between the two plates without the metal sheet is given by \(C_0 = \frac{\varepsilon_0 A}{d}\), where \(\varepsilon_0\) is the permittivity of free space and \(A\) is the area of the plates.
04

Capacitance of Each Section

The system can be divided into two capacitors: the first from one plate to one side of the slab with distance \(d-a\) and the second through the thickness of the slab. However, in this scenario, with the slab being conductive, it doesn't introduce a separate capacitance for slab thickness. The first section's capacitance is \(C_1 = \frac{\varepsilon_0 A}{d-a}\) and the section from slab to the second plate is \(C_2 = \frac{\varepsilon_0 A}{d-a}\).
05

Total Capacitance in Series

For capacitors in series, the total capacitance \(C\) is found via \(\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2}\). Using symmetry for \(C_1\) and \(C_2\), this simplifies to \(C = \frac{2C_1C_2}{C_1 + C_2} = \frac{\varepsilon_0 A}{d-a}\).
06

Capacitance Ratio with \(C_0\)

Express the capacitance with the metal slab \(C\) as a multiple of \(C_0\): \(C = \frac{d}{d-a}C_0\).
07

Limit Analysis for Slab Thickness

As \(a\) approaches 0, the metal slab has negligible effect, and \(C \rightarrow C_0\). As \(a\) approaches \(d\), the central slab dominates, resulting in an increase in capacitance, theoretically becoming infinite as the effective gap approaches zero.

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Key Concepts

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

Parallel Plate Capacitor
A parallel plate capacitor is a simple yet powerful device that consists of two conductive plates facing each other, separated by a small distance. The plates are usually flat and are designed to hold opposite electric charges, creating an electric field between them. This field stores electrical energy.
The capacitance of a parallel plate capacitor can be calculated using the formula: \[C_0 = \frac{\varepsilon_0 A}{d}\],where:
  • \(C_0\) is the capacitance,
  • \(\varepsilon_0\) is the permittivity of free space,
  • \(A\) is the area of one of the plates, and
  • \(d\) is the distance between the plates.
Understanding the structure and function of a parallel plate capacitor is fundamental to grasp more complex capacitor systems, such as when modifications such as inserting a metal slab are involved.
Metal Slab Insertion
When a metal slab is inserted between the plates of a parallel plate capacitor, it significantly alters the capacitance. This slab has a thickness \(a\), which is less than the separation distance \(d\) between the plates. Even though the slab doesn’t touch the plates, it divides the space into two regions, essentially forming two capacitors in series.The metal slab increases the capacitance because it shortens the effective distance between oppositely charged regions, enhancing the electric field strength. This alteration can be visualized as two new capacitors, each having a separation of \(d-a\).
Calculating the modified capacitance involves understanding these regions:
  • The effective separation between the two outside faces of the slab itself.
  • The reduced effective distance \(d-a\).
The capacitance increases approximately as a factor of \(\frac{d}{d-a}\) times \(C_0\), as derived from analyzing capacitors in series.
Series Capacitance
Capacitors in series add up differently than those in parallel. When we consider two capacitors \(C_1\) and \(C_2\) in series, the total or equivalent capacitance \(C\) can be calculated with:\[\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2}\]This results in a capacitance lower than that of the smallest individual capacitor. The metal slab's insertion modifies the capacitor into two sections, both having a capacitance \[C_1 = C_2 = \frac{\varepsilon_0 A}{d-a}\]Considering these capacitances as in series, the combined capacitance \(C\) for the entire system becomes:\[C = \frac{2C_1C_2}{C_1 + C_2} = \frac{\varepsilon_0 A}{d-a}\].This analysis shows how the slab effectively reduces the gap between opposing charges, resulting in enhanced capacitance. This knowledge is particularly vital when assessing changes like metal slab insertions or thicker dielectric materials.
Permittivity of Free Space
The permittivity of free space, often denoted \(\varepsilon_0\), is a constant that plays a critical role in the calculation of capacitance. It is the measure of how much capacitance is allowed for a capacitor placed in a vacuum or air. Its value is approximately \(8.85 \times 10^{-12} \mathrm{~F/m}\).In a parallel plate capacitor, the permittivity of free space is a key factor because it affects how much charge can be stored per unit voltage. Together with the plate area and the separation distance, it determines the basic capacitance of the capacitor prior to any insertions or modifications like a metal slab.Understanding \(\varepsilon_0\) is essential when calculating capacitance in various scenarios, as it allows us to compare different capacitor configurations and the effects of dielectric materials or conductive inserts, like the metal slab, on total capacitance.

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Most popular questions from this chapter

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 \({1 \over 675}\) s with an average light power output of 2.70 \(\times\) 10\(^5\) 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 V when the stored energy equals the value calculated in part (a). What is the capacitance?

Your electronics company has several identical capacitors with capacitance \(C_1\) and several others with capacitance \(C_2\). You must determine the values of \(C_1\) and \(C_2\) but don't have access to \(C_1\) and \(C_2\) individually. Instead, you have a network with \(C_1\) and \(C_2\) connected in series and a network with \(C_1\) and \(C_2\) connected in parallel. You have a 200.0-V battery and instrumentation that measures the total energy supplied by the battery when it is connected to the network. When the parallel combination is connected to the battery, 0.180 J of energy is stored in the network. When the series combination is connected, 0.0400 J of energy is stored. You are told that \(C_1\) is greater than \(C_2\). (a) Calculate \(C_1\) and \(C_2\). (b) For the series combination, does \(C_1\) or \(C_2\) store more charge, or are the values equal? Does \(C_1\) or \(C_2\) store more energy, or are the values equal? (c) Repeat part (b) for the parallel combination.

A parallel-plate air capacitor has a capacitance of 920 pF. The charge on each plate is 3.90 \(\mu\)C. (a) What is the potential difference between the plates? (b) If the charge is kept constant, what will be the potential difference if the plate separation is doubled? (c) How much work is required to double the separation?

A parallel-plate air capacitor is to store charge of magnitude 240.0 pC on each plate when the potential difference between the plates is 42.0 V. (a) If the area of each plate is 6.80 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?

Two parallel plates have equal and opposite charges. When the space between the plates is evacuated, the electric field is \(E = 3.20\times10^5\space V/m\). When the space is filled with dielectric, the electric field is \(E = 2.50\times10^5\space V/m\). (a) What is the charge density on each surface of the dielectric? (b) What is the dielectric constant?

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