Dielectric problem

[inaccensa]

If a dielectric was inserted between the plates of the
capacitor in the defibrillator when the switch is in
position A (connected to a battery) :
A. the energy stored in the capacitor would
increase.---correct ans
B. the energy stored in the capacitor would
decrease.
C. the electric field between the plates would
increase.
D. the electric field between the plates would
decrease.

Why isn't the answer D?
Dielectric decreases the electric field when it is inserted between the capacitor plates. Although the switch is on,which implies that the voltage is same, but the charge and the capacitance increase. I feel like both A and D are correct.

Second q- If the battery was no longer connected. The charge would remain the same, the voltage will decrease. What happens to the capacitance in this case?

Thanks!!

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[eric51]

The electric field is only a property of the difference in potential between the plates and the distance. Since you are not changing either of those, the field will not change. You are right that a dielectric would change the field strength if the capacitor was not connected to a battery, but that is because the voltage would change as well. As long as the battery is connected, the voltage will not change, and thus the field stays the same.

 

[inaccensa]

The electric field is only a property of the difference in potential between the plates and the distance. Since you are not changing either of those, the field will not change. You are right that a dielectric would change the field strength if the capacitor was not connected to a battery, but that is because the voltage would change as well. As long as the battery is connected, the voltage will not change, and thus the field stays the same.

thanks. But what happens to the capacitance when you insert a dielectric when its discharging? Will it still increase? The voltage will decrease, since there is no external battery, the charge will remain the same and the electric field will decrease. Thus the capacitance should increase?

 

[zmatrix]

thanks. But what happens to the capacitance when you insert a dielectric when its discharging? Will it still increase? The voltage will decrease, since there is no external battery, the charge will remain the same and the electric field will decrease. Thus the capacitance should increase?

Suppose the capacitor has been charged for a long period of time (remember it takes an infinite amount of time to fully charge a capacitor). Then you flip the switch to bypass the battery to let it discharge, and you immediately stick a dielectric into the capacitor.

Since C = Q/V, you increase C by inserting the dielectric, Q remains the same because it is dependent on the charge separation between the plates, therefore V must decrease inversely proportional to the increase in C.

Energy is stored within a dielectric by doing work to polarize it. A dielectric is an insulator so no electron travel (current).

An analogous example would be isothermally increasing the volume of a compressed ideal gas. Although the total kinetic energy (charge separation) of the system is still the same since particles are traveling at the same speed, we see that potential energy was lost because of gas expansion. Because of the reduced pressure (voltage), each particle (electron) does less work for a certain unit volume expansion.

So the dielectric in the example is like the extra volume you add. After adding the dielectric, for a certain voltage (pressure) you can store more charge (gas molecules) than if you did not add the dielectric.

 

[inaccensa]

thanks

 

[drechie]

To clarify ----


does the potential of a capacitor change by adding a dielectric if plugged into a battery? (C = Q / Potential)