SProblem 22: An insulating material with a dielectric constant of 4.37 and a dielectric strength of 6.3 MV/m will be used to fill the gapbetween the plates of a 6.25 nC parallel-plate capacitor.What is the minimum plate area, in squared meters, if the capacitor must withstand applied voltages up to 4.1 kV/m?Amin =m,2

Capacitance is defined as the property of an electric instrument to store charge in it. Its unit is…

Answered: Problem 22: An insulating material with…

University Physics Volume 2

18th Edition

ISBN:9781938168161

Author:OpenStax

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Chapter8: Capacitance

Section: Chapter Questions

Problem 53P: A parallel-plate capacitor has charge of magnitude 9.00F on each plate and capacitance 3.00F when...

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(a) Will the electric field strength between two parallel conducting plates exceed the breakdown strength of dry air, which is 3.00106 V/m, if the plates are separated by 2.00 mm and a potential difference of 5.010V is applied? (b) How close together can the plates be with this applied voltage?
A 6.50-F capacitor is connected to a battery. What is the charge on each plate of the capacitor if the voltage of the battery is a. 10.0 V and b. 2.00 V?
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(a) Will the electric field strength between two parallel conducting plates exceed the breakdown strength for air ( 3.0 106 V/m) if the plates are separated by 2.00 mm and a potential difference of 5.0 103 V is applied? (b) How close together can the plates be with this applied voltage?
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The dielectric to be used in a parallel-plate capacitor has a dielectric constant of 3.60 and a dielectric strength of 1.60107 V/m. The capacitor has to have a capacitance of 1.25 nF and must be able to withstand a maximum potential difference 5.5 kV. What is the minimum area the plates of the capacitor may have?
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