A parallel-plate capacitor is charged by a 13.0Vbattery, then
the battery is removed.What is the potential difference between the
plates after the battery is disconnected? Answer: 13V. I got that
part. Now it's asking... What is the potential difference between
the plates after a sheet of Teflon is inserted between them? My
anwers thus far have been: 
Can you please give me the
equation and work and why I know teflon is 2.1 value for k and that
I believe the voltage decreases. Thank you!
Answer
Write the formula to calculate the charge of a parallel-plate capacitor, when it is charged by a battery Substitute 13.0 V for Vin the above equation. 0 (13.0 V)xC Here, Q is he charge of the parallel-plate capacitor after removing the battery The potential difference between the parallel-plates after removing the battery is, Substitute equation (1) in equation (2). (13.0 V)xC 13.0 V Hence, the potential difference between the parallel plates when the battery is removed is 13.0 V Calculate the potential difference between the plates, when the sheet of Teflon is inserted. _(13.0 V)x C Teion Calculate the charge of the parallel-plates, when the battery is not removed and the sheet of the
and the sheet of the Teflon is inserted Q (13.0 V)x Teflon Teflon Teion (13.0 V)xCTelon Tefion - 13.0 V Hence, the potential difference between the parallel-plates, when the battery is not removed and the sheet of the Teflon is inserted is 13.0V Calculate the capacitance of the parallel-plate capacitor, when the battery is removed and the Teflon is inserted Tefion Tefon Teflon Tefion 13.0 V 2 Hence, the potential difference between the plates, when the Teflon is inserted and the battery is removed is 6.5 V