Giancoli 7th Edition textbook cover
Giancoli's Physics: Principles with Applications, 7th Edition
17
Electric Potential
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17-1 to 17-4: Electric Potential
17-5: Potential Due to Point Charges
17-6: Electric Dipoles
17-7: Capacitance
17-8: Dielectrics
17-9: Electric Energy Storage
17-10: Digital
17-11: TV and Computer Monitors

Question by Giancoli, Douglas C., Physics: Principles with Applications, 7th Ed., ©2014, Reprinted by permission of Pearson Education Inc., New York.
Problem 48
Q

What is the capacitance of a pair of circular plates with a radius of 5.0 cm separated by 2.8 mm of mica?

A
1.7×1010 F1.7 \times 10^{-10} \textrm{ F}
Giancoli 7th Edition, Chapter 17, Problem 48 solution video poster
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VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. The capacitance between two parallel plates is the dielectric constant times the permittivity of free space times the area of the plates divided by their separation. And in this case, its the circular plates, their area will be pi times the radius squared. So the dielectric constant for Mica is seven and multiply that by 8.85 times ten to the minus 12 coulomb square per newton meter squared times pi times 5.0 times ten to the minus two meters and then square that converting the centimeters into meters there, and then divide by 2.8 times ten to the minus three meters separation and you get 1.7 times ten to the minus ten farads for the capacitance.

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