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 chiddler 04-30-2012 11:13 PM

Dielectric insertion

2 µF capacitor, 12 V battery. The capacitor is fully charged and with the battery still connected, an insulator with a dielectric constant equal to 4 is inserted into the capacitor.

How much work is required to insert the dielectric?

The book has the mathematical answer, but I don't understand how they did what they did.

thanks.

 milski 05-01-2012 12:05 AM

Potential energy stored in the capacitor is U=1/2CV^2. Dielectric with d=4 will increase the capacitance by 4. Calculate the energy stored in the capacitor before and after the dielectric is inserted - the difference will be the work done to insert the dielectric.

The justification why ΔU is the work done in this case is somewhat fishy, at least from what I can think of at the moment. There is another example, where you have two connected capacitors with no battery and a dielectric is inserted in one of them causing change of charge. Calculating the total energy in both capacitors before and after the insertion shows that there is a change in energy. In that case it's easy to argue that the change of energy comes from work because there are no other sources - the two capacitors are a closed system. In your problem there is a transfer of energy from the battery, so...

What's the source of the problem?

 pm1 05-01-2012 07:42 AM

Quote:
 Originally Posted by milski (Post 12460689) Potential energy stored in the capacitor is U=1/2CV^2. Dielectric with d=4 will increase the capacitance by 4. Calculate the energy stored in the capacitor before and after the dielectric is inserted - the difference will be the work done to insert the dielectric. The justification why ΔU is the work done in this case is somewhat fishy, at least from what I can think of at the moment. There is another example, where you have two connected capacitors with no battery and a dielectric is inserted in one of them causing change of charge. Calculating the total energy in both capacitors before and after the insertion shows that there is a change in energy. In that case it's easy to argue that the change of energy comes from work because there are no other sources - the two capacitors are a closed system. In your problem there is a transfer of energy from the battery, so... What's the source of the problem?
I agree with your reasoning, but I can't get to the right answer (4.3*10^-4). I keep getting 2.8*10-4. Am I just messing up on my calculations?

thanks

 milski 05-01-2012 07:55 AM

Quote:
 Originally Posted by pm1 (Post 12461192) I agree with your reasoning, but I can't get to the right answer (4.3*10^-4). I keep getting 2.8*10-4. Am I just messing up on my calculations? thanks
Probably.

Ui=1/2cv^2=1/2*2e-6*12^2=1.44e-4
Uf=1/2*8e-6*12^2=5.76e-4
ΔU=Uf-Ui=5.76e-4 - 1.44e-4 = 4.32e-4

 pm1 05-01-2012 07:58 AM

Quote:
 Originally Posted by milski (Post 12461252) Probably. Ui=1/2cv^2=1/2*2e-6*12^2=1.44e-4 Uf=1/2*8e-6*12^2=5.76e-4 ΔU=Uf-Ui=5.76e-4 - 1.44e-4 = 4.32e-4
ha! okay, thanks! :)

 chiddler 05-01-2012 10:27 AM

Quote:
 Originally Posted by milski (Post 12460689) Potential energy stored in the capacitor is U=1/2CV^2. Dielectric with d=4 will increase the capacitance by 4. Calculate the energy stored in the capacitor before and after the dielectric is inserted - the difference will be the work done to insert the dielectric. The justification why ΔU is the work done in this case is somewhat fishy, at least from what I can think of at the moment. There is another example, where you have two connected capacitors with no battery and a dielectric is inserted in one of them causing change of charge. Calculating the total energy in both capacitors before and after the insertion shows that there is a change in energy. In that case it's easy to argue that the change of energy comes from work because there are no other sources - the two capacitors are a closed system. In your problem there is a transfer of energy from the battery, so... What's the source of the problem?
it's a princeton review question.

It takes energy to insert the dielectric because it would cause a change in potential energy in the capacitor. Energy conserved, and thus the energy gained must be the energy spent to insert the dielectric. And with the experiment you mentioned, it must be from the insertion not from the battery.

this question isn't very good. without knowing what you wrote, i would have thought that work would come from the battery. is there another way of figuring this out without knowing the prior example?

 milski 05-01-2012 11:08 AM

Quote:
 Originally Posted by chiddler (Post 12461820) it's a princeton review question. It takes energy to insert the dielectric because it would cause a change in potential energy in the capacitor. Energy conserved, and thus the energy gained must be the energy spent to insert the dielectric. And with the experiment you mentioned, it must be from the insertion not from the battery. this question isn't very good. without knowing what you wrote, i would have thought that work would come from the battery. is there another way of figuring this out without knowing the prior example?
The problem with the question is that some of the energy can and will come from the battery. The only way to make a practical setup that can be evaluated is the two capacitors setup that I mentioned above. You would be correct that some part of the work will be coming from the battery.

 chiddler 05-01-2012 11:21 AM

Quote:
 Originally Posted by milski (Post 12462006) The problem with the question is that some of the energy can and will come from the battery. The only way to make a practical setup that can be evaluated is the two capacitors setup that I mentioned above. You would be correct that some part of the work will be coming from the battery.
thanks very much.

 milski 05-01-2012 11:37 AM

Quote:
 Originally Posted by chiddler (Post 12462083) thanks very much.
Ok, a bit of correction, since "the only way" is a bit too strong. Any setup where you can evaluate the total energy will work. What makes it impractical with a battery is that you have no good way to measure what happened to the battery.

 chiddler 05-01-2012 11:47 AM

Quote:
 Originally Posted by milski (Post 12462144) Ok, a bit of correction, since "the only way" is a bit too strong. Any setup where you can evaluate the total energy will work. What makes it impractical with a battery is that you have no good way to measure what happened to the battery.
okie dokie i understand.

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