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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
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17-5: Potential Due to Point Charges
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17-6: Electric Dipoles
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17-7: Capacitance
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17-8: Dielectrics
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17-9: Electric Energy Storage
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Question by Giancoli, Douglas C., Physics: Principles with Applications, 7th Ed., ©2014, Reprinted by permission of Pearson Education Inc., New York.
Problem 36
Q

An 8500-pF capacitor holds plus and minus charges of 16.5×10−8 C16.5 \times 10^{-8} \textrm{ C}. What is the voltage across the capacitor?

A
19 V19 \textrm{ V}
Giancoli 7th Edition, Chapter 17, Problem 36 solution video poster
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VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. The charge in a capacitor is equal to its capacitance times the voltage and we can solve for voltage by dividing both sides by C and so the V is going to be the charge in each plate divided by the capacitance. So 16.5 times ten to the minus eight coulombs divided by 8500 picofarads and Pico is ten to the minus 12 and that gives us about 19 volts.

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Giancoli Answers, including solutions and videos, is copyright © 2009-2024 Shaun Dychko, Vancouver, BC, Canada. Giancoli Answers is not affiliated with the textbook publisher. Book covers, titles, and author names appear for reference purposes only and are the property of their respective owners. Giancoli Answers is your best source for the 7th and 6th edition Giancoli physics solutions.