Question

A parallel plate capacitor has an air filled gap. It is attached an EMF and charged it to charge Q_o and voltage difference ΔV_co. Two different experiments are done.

Case A: The capacitor is disconnected from the EMF first before the gap in between the plates is then filled with Teflon.

Case B: The gap between the capacitor plates is then filled with Teflon with the capacitor still connected directly across the terminals of the EMF.

Which statement is true after the Teflon is inserted into the gap?

In case A, ΔV_c is unchanged but Q decreases to keep the net E field between the plates unchanged. In case B, Q is unchanged but ΔV_c increases due to a larger net E field between the plates.

In case A, Q is unchanged but ΔV_c increases due to a larger net E field between the plates. In case B, ΔV_c is unchanged but Q decreases to keep the net E field between the plates unchanged.

In case A, ΔV_c is unchanged but Q increases to keep the net E field between the plates unchanged. In case B, Q is unchanged but ΔV_c decreases due to a smaller net E field between the plates.

In case A, Q is unchanged but ΔV_c decreases due to a smaller net E field between the plates. In case B, ΔV_c is unchanged but Q increases to keep the net E field between the plates unchanged.

Answer 1

Case-A :: capacitor is disconnected to EMF at the time of the Teflon is inserted into the gap . Then

1) Capacitance will increase as Teflon has greater permittivity than air.

2) charge Q will remain same as there is no way to pass the charge ( EMF is disconnected) . So to hold law of conservation of charge , the amount of charge will remain same.

3) since Q=CV then Voltage del(Vc) will decrease as C increases.(to keep Q constant)

Case-B:: in this case capacitor is not disconnected while filling the gap with Teflon .then

1) capacitor will increase as above case-A.

2) here Voltage remains same as constant EMF is applied i.e EMF is not disconnected.

3) as Q=CV , then Q will increase i.e. more charge will come from source as C increases. ( to keep V constant)

Conclusion:: So, In case-A , Q will remain constant and V decreases. In case-B, V remains the same and Q increases.