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An air capacitor is made by using two flat plates, each

University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman ISBN: 9780321675460 31

Solution for problem 66P Chapter 24

University Physics | 13th Edition

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University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman

University Physics | 13th Edition

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Problem 66P

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 (Fig. P24.62). (a) What is the capacitance of this arrangement? (b) Express the capacitance as a multiple of the capacitance C0 when the metal slab is not present. (c) Discuss what happens to the capacitance in the limits

Step-by-Step Solution:

Solution 66P Step 1 of 9: The capacitor is a device used to store electrical energy in a confined region and also which is capable of releasing the electrical energy instantly. The capacitance of the capacitor is the capability of the capacitor to store the charge or electrical energy. Step 2 of 9: Before inserting the dielectric medium(initial) Consider two parallel plates at d distance apart. with air between them, having capacitance C. Under a potential difference V as in figure below, Capacitance is, 0 A C = d .............1 Where is the permittivity of free space , A is the area of the parallel plates and d is the 0 distance between them. Step 3 of 9: After inserting dielectric medium, For the same above configuration, we are inserting a dielectric medium which is having a dielectric constant K as shown below, Capacitance is, KA C di = 0 .....................2 d Where i0 the permittivity of free space , K = dielectric constant of medium, A is the area of the parallel plates and d is the distance between them. Step 4 of 9: (a) What is the capacitance of this arrangement In the given problem, the two plates with area A each at distance d apart. A metal slab having the thickness a is inserted between them as shown in the figure below. Here we need to calculate the net capacitance of the arrangement. Step 5 of 9: From above figure, we can see that; there are two capacitor in series. One being the capacitor with thickness (d-a) without dielectric and the other capacitor with thickness a with dielectric medium. In order to calculate the net capacitance, we need to first calculate the individual capacitance. Capacitance of capacitor without dielectric, Using equation 1, C = 0A without dielectrda Capacitance of capacitor with dielectric medium, KA Using equation 2, C = 0 dielectric a

Step 6 of 9

Chapter 24, Problem 66P is Solved
Step 7 of 9

Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

The answer to “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 (Fig. P24.62). (a) What is the capacitance of this arrangement? (b) Express the capacitance as a multiple of the capacitance C0 when the metal slab is not present. (c) Discuss what happens to the capacitance in the limits” is broken down into a number of easy to follow steps, and 89 words. The full step-by-step solution to problem: 66P from chapter: 24 was answered by , our top Physics solution expert on 05/06/17, 06:07PM. This textbook survival guide was created for the textbook: University Physics, edition: 13. Since the solution to 66P from 24 chapter was answered, more than 326 students have viewed the full step-by-step answer. University Physics was written by and is associated to the ISBN: 9780321675460. This full solution covers the following key subjects: capacitance, plates, slab, metal, Limits. This expansive textbook survival guide covers 26 chapters, and 2929 solutions.

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