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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.

Short Answer

Expert verified
Answer: Inserting a dielectric material between the plates of a capacitor increases its charge-storing capacity by increasing the permittivity (蔚) of the material and reducing the net electric field between the plates due to polarization. This results in a decrease in voltage and an increase in capacitance.

Step by step solution

01

1. Understanding Capacitance

Capacitance is the ability of a capacitor to store electrical energy between two conductive plates separated by an insulating material, typically air or a vacuum. The capacitance is given by the formula C = 蔚A/d, where C is the capacitance, 蔚 is the permittivity of the dielectric material, A is the overlapping area of the plates, and d is the distance between the plates.
02

2. Introducing Dielectric Material

When a dielectric material is inserted between the plates of a capacitor, it replaces the air or vacuum present initially. Dielectric materials have a higher permittivity (蔚) than air or vacuum, which directly affects the capacitance value in the formula C = 蔚A/d.
03

3. Dielectric Polarization

Dielectric materials are made up of molecules with positive and negative charges. When an external electric field is applied across the capacitor plates, the dielectric material gets polarized, causing the positive and negative charges in the material to align with the electric field. This leads to an induced electric field within the material, which opposes the external electric field.
04

4. Increased Capacitance

Due to the induced electric field in the dielectric material, the net electric field between the plates decreases. As a result, the voltage across the capacitor is reduced, which is given by V = Ed, where V is the voltage, E is the electric field, and d is the distance between the plates. As capacitance is directly proportional to the charge (Q) stored and inversely proportional to the voltage (V), according to the formula C = Q/V, the decrease in voltage causes an increase in the capacitance of the capacitor.
05

5. Conclusion

In summary, when a dielectric material is inserted between the plates of a capacitor, it increases the capacitance by increasing the permittivity (蔚) of the material and reducing the net electric field between the plates due to polarization. This results in a decrease in voltage and an increase in charge-storing capacity.

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Most popular questions from this chapter

Briefly describe electron and hole motions in a \(p-n\) junction for forward and reverse biases; then explain how these lead to rectification.

(a) In your own words, explain how donor impurities in semiconductors give rise to free electrons in numbers in excess of those generated by valence band-conduction band excitations. (b) Also explain how acceptor impurities give rise to holes in numbers in excess of those generated by valence band- conduction band excitations.

The polarization \(P\) of a dielectric material positioned within a parallel- plate capacitor is to be \(1.0 \times 10^{-6} \mathrm{C} / \mathrm{m}^{2}\) (a) What must be the dielectric constant if an electric field of \(5 \times 10^{4} \mathrm{~V} / \mathrm{m}\) is applied? (b) What will be the dielectric displacement \(D ?\)

(a) The room-temperature electrical conductivity of a silicon specimen is \(5.93 \times 10^{-3}\) \((\Omega \cdot \mathrm{m})^{-1} .\) The hole concentration is known to be \(7.0 \times 10^{17} \mathrm{~m}^{-3}\). Using the electron and hole mobilities for silicon in Table \(18.3\), compute the electron concentration. (b) On the basis of the result in part (a), is the specimen intrinsic, \(n\)-type extrinsic, or \(p\)-type extrinsic? Why?

Will each of the following elements act as a donor or an acceptor when added to the indicated semiconducting material? Assume that the impurity elements are substitutional. \begin{tabular}{ll} \hline Impurity & Semiconductor \\ \hline \(\mathrm{P}\) & \(\mathrm{Ge}\) \\ \(\mathrm{S}\) & \(\mathrm{AIP}\) \\ \(\mathrm{In}\) & \(\mathrm{CdTe}\) \\ \(\mathrm{Al}\) & \(\mathrm{Si}\) \\ \(\mathrm{Cd}\) & \(\mathrm{GaAs}\) \\ \(\mathrm{Sb}\) & \(\mathrm{ZnSe}\) \\ \hline \end{tabular}
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