Chapter 18: Problem 52
In your own words, explain the mechanism by which charge storing capacity is increased by the insertion of a dielectric material within the plates of a capacitor.
Chapter 18: Problem 52
In your own words, explain the mechanism by which charge storing capacity is increased by the insertion of a dielectric material within the plates of a capacitor.
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Get started for freeIn terms of electron energy band structure, discuss reasons for the difference in electrical conductivity between metals, semiconductors, and insulators
At room temperature the electrical conductivity of \(\mathrm{PbS}\) is \(25(\Omega-\mathrm{m})^{-1}\), whereas the electron and hole mobilities are 0.06 and \(0.02 \mathrm{m}^{2} / \mathrm{V}-\mathrm{s},\) respectively. Compute the in trinsic carrier concentration for PbS at room temperature.
A charge of \(2.0 \times 10^{-10} \mathrm{C}\) is to be stored on each plate of a parallel-plate capacitor having an area of \(650 \mathrm{mm}^{2}\left(1.0 \mathrm{in.}^{2}\right)\) and a plate separation of \(4.0 \mathrm{mm}(0.16 \mathrm{in.})\) (a) What voltage is required if a material having a dielectric constant of 3.5 is positioned within the plates?
Germanium to which \(10^{24} \mathrm{m}^{-3}\) As atoms have been added is an extrinsic semiconductor at room temperature, and virtually all the As atoms may be thought of as being ionized (i.e., one charge carrier exists for each As atom). (a) Is this material \(n\) -type or \(p\) -type? (b) Calculate the electrical conductivity of this material, assuming electron and hole mobilities of 0.1 and \(0.05 \mathrm{m}^{2} / \mathrm{V}\) -s, respectively
We noted in Section 12.5 (Figure 12.22 ) that in FeO (wüstite), the iron ions can exist in both \(\mathrm{Fe}^{2+}\) and \(\mathrm{Fe}^{3+}\) states. The number of each of these ion types depends on temperature and the ambient oxygen pressure. Furthermore, we also noted that in order to retain electroneutrality, one \(\mathrm{Fe}^{2+}\) vacancy will be created for every two \(\mathrm{Fe}^{3+}\) ions that are formed; consequently, in order to reflect the existence of these vacancies the formula for wüstite is often represented as \(\mathrm{Fe}_{(1-x)} \mathrm{O}\) where \(x\) is some small fraction less than unity. In this nonstoichiometric \(\mathrm{Fe}_{(1-x)} \mathrm{O}\) material, conduction is electronic, and, in fact, it behaves as a \(p\) -type semiconductor. That is, the \(\mathrm{Fe}^{3+}\) ions act as electron acceptors, and it is relatively easy to excite an electron from the valence band into an \(\mathrm{Fe}^{3+}\) acceptor state, with the formation of a hole. Determine the electrical conductivity of a specimen of wüstite that has a hole mobility of \(1.0 \times 10^{-5} \mathrm{m}^{2} / \mathrm{V}\) -s and for which the value of \(x\) is \(0.040 .\) Assume that the acceptor states are saturated (i.e., one hole exists for every \(\left.\mathrm{Fe}^{3+} \text { ion }\right) .\) Wüstite has the sodium chloride crystal structure with a unit cell edge length of \(0.437 \mathrm{nm}\)
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