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A parallel-plate capacitor is charged by a 17.0V battery, then the battery is removed (Question in details).?

A.) What is the potential difference between the plates after the battery is disconnected?

B.) What is the potential difference between the plates after a sheet of Teflon is inserted between them?

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Since this is a physics class question, we can assume that everything is ideal.

The 17.0 volt battery deposits enough charge on the plates to create a back voltage of 17.0 volts. When the battery is removed, that voltage does not go away.

We assume that there was air or vaccum between the plates before, as the problem doesn't mention it. Teflon has a higher dielectric constant, than air, so if you go by the parallel plate capacitor formula, the charge has remained constant, the plate area and spacing has remained constant, so the voltage must go down. You can look up the dielectric constant of Teflon to calculate what the new voltage should be, assuming that the thickness of the Teflon completely fills the air gap in the capacitor.

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