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A parallel-plate capacitor has plates with area 0.0225 m2

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

Solution for problem 46E 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 46E

A parallel-plate capacitor has plates with area 0.0225 m2 separated by 1.00 mm of Teflon. (a) Calculate the charge on the plates when they are charged to a potential difference of 12.0 V. (b) Use Gauss’s law (Eq. 24.23) to calculate the electric field inside the Teflon. (c) Use Gauss’s law to calculate the electric field if the voltage source is disconnected and the Teflon is removed.

Step-by-Step Solution:

Solution 46E Step 1 of 4: Gauss’s law in dielectrics has the same form as in vacuum except that the electric field is multiplied by a factor of K and the charge enclosed by the Gaussian surface is the free charge. The capacitance of an object depends on its shape and size . When the space between the conductors is filled with a dielectric material, the capacitance increases by a factor K called the dielectric constant of the material. The quantity = K 0 C = KC 0 Step 2 of 4: a)The capacitance of a parallel plate capacitor is 12 2 2 C = K0A = 2.1×8.85×10 C /N.m ×(0.0225)= 4.18 × 10 10F d 1×10 m The charge on its plates is Q = CV = 4.18 × 10 10F × 12 V

Step 3 of 4

Chapter 24, Problem 46E is Solved
Step 4 of 4

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

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A parallel-plate capacitor has plates with area 0.0225 m2

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