A parallel plate capacitor has an air-filled gap. It is attached an EMF and charged it to charge Qo and voltage difference ΔVco. Two different experiments are done. Case A: The gap between the capacitor plates is then filled with Teflon with the capacitor still connected directly across the terminals of the EMF. Case B: The capacitor is disconnected from the EMF first before the gap in between the plates is then filled with Teflon. Which statement is true after the Teflon is inserted into the gap? 1.) In case A, ΔVc is unchanged but Q increases to keep the net E field between the plates unchanged. In case B, Q is unchanged but ΔVc decreases due to a smaller net E field between the plates. 2.) In case A, Q is unchanged but ΔVc decreases due to a smaller net E field between the plates. In case B, ΔVc is unchanged but Q increases to keep the net E field between the plates unchanged. 3.) In case A, Q is unchanged but ΔVc increases due to a larger net E field between the plates. In case B, ΔVc is unchanged but Q decreases to keep the net E field between the plates unchanged. 4.) In case A, ΔVc is unchanged but Q decreases to keep the net E field between the plates unchanged. In case B, Q is unchanged but ΔVc increases due to a larger net E field between the plates. Now the parallel plate capacitor with an air-filled gap (having a capacitance of 5 μF, connected to a 6 Volt EMF and allowed to charge) where the EMF still connected the gap between plates is completely filled with Teflon. What is the electric energy stored in the capacitor now? Use the Dielectric constant k=1.0006 and Dielectric strength 3*10^6 V/m for your calculations.
A parallel plate capacitor has an air-filled gap. It is attached an EMF and charged it to charge Qo and voltage difference ΔVco. Two different experiments are done.
Case A: The gap between the capacitor plates is then filled with Teflon with the capacitor still connected directly across the terminals of the EMF.
Case B: The capacitor is disconnected from the EMF first before the gap in between the plates is then filled with Teflon.
Which statement is true after the Teflon is inserted into the gap?
1.) In case A, ΔVc is unchanged but Q increases to keep the net E field between the plates unchanged. In case B, Q is unchanged but ΔVc decreases due to a smaller net E field between the plates.
2.) In case A, Q is unchanged but ΔVc decreases due to a smaller net E field between the plates. In case B, ΔVc is unchanged but Q increases to keep the net E field between the plates unchanged.
3.) In case A, Q is unchanged but ΔVc increases due to a larger net E field between the plates. In case B, ΔVc is unchanged but Q decreases to keep the net E field between the plates unchanged.
4.) In case A, ΔVc is unchanged but Q decreases to keep the net E field between the plates unchanged. In case B, Q is unchanged but ΔVc increases due to a larger net E field between the plates.
Now the parallel plate capacitor with an air-filled gap (having a capacitance of 5 μF, connected to a 6 Volt EMF and allowed to charge) where the EMF still connected the gap between plates is completely filled with Teflon. What is the electric energy stored in the capacitor now? Use the Dielectric constant k=1.0006 and Dielectric strength 3*10^6 V/m for your calculations.
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