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Teflon Sheet
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Frequently Asked Questions...
Desperate for homework help! A parallel plate capacitor has an area A = 20.0 cm^2 and a plate separation of d?
A parallel plate capacitor has an area A = 20.0 cm^2 and a plate separation of d = 4.00 mm.
(a) Find the capacitance in air and the maximum voltage and charge the capacitor can hold.
(b) A Teflon sheet is slid between the plates, filling the entire volume, Find the new capacitance
and maximum charge.
(c) Before the insertion of the Teflon, the plates are connected to a 24.0 V battery and then
disconnected. What are the energies in the capacitor before and after the Teflon is inserted?
(d) Was work done in inserting the Teflon?
Suppose now, that the Teflon sheet that is inserted between the plates is only 2 mm thick and
fills only half the volume. Before the Teflon is inserted, the disconnected capacitor carries a
charge of 1.00 nC.
(e) Find the electric field everywhere.
(f) Find the new capacitance.
Given:
Dielectric strength (air) = 3.00×10^6 V.m^-1
Dielectric strength (Teflon) = 6.00×10^7 V.m^-1
Dielectric constant (Teflon) = 2.1
Answer:
27 nF
14 nF
24 volts
pi^2-4X^Y^2-2 nF









































































