Newton's law and energy of conservation

[A Dhingra]

hi everyone...

i was recently thinking about a generator that would work as per the principle of electrostatics, the capacitor in specific...
the system i had thought of had two parallel charged capacitor plates with air as the dielectric medium and with minimum leakage current... between the two plates two more metallic parallel plates were placed such that they have no potential difference on their surface ( so equipotential surface)... on such a surface if we apply force to move the charges ( with the help of a battery) ,then the charge should move forever until restricted by nay external force (the Newton's first law)...

in the system the two added metal plates are connected by a wire and a circuit with battery along the two way key ...and the other end of the two way key is connected to the normal wire.

when the key with the battery is attached to circuit ,the charges of the two metal plates will start flowing (unrestricted) and as soon as the key is changed... the result should be that the current should still flow through te circuit, but this is against the law of conservation of energy...

i am not going to make any such generator, was just thinking...

so the problem that arises here is that either the 1st law of Newton is correct or the law of conservation of energy ....
i know that in this case where there is no external resistive force on the charge, law of conservation of energy should not be applied... but if this type of a system works even for a month and gives electricity with only a few minutes of supply from the battery , then this can solve the problem of energy crisis... i am sure there must be something to not allow this to happen... can you tell me what is that resistive factor...

or which law is more authentic...

 

[n.karthick]

You can think of charging a capacitor as equivalent to compressing a spring. Definitely you need energy to compress a spring. Assuming no friction, if you apply an external force to a spring, will the compression of spring goes on for ever? Won't you reach a point where anymore compression is impossible, irrespective of presence of external force?

 

[JaredJames]

so the problem that arises here is that either the 1st law of Newton is correct or the law of conservation of energy

I direct you to my signature.

Anyhow, I can't quite work out your design, it just doesn't make sense and it appears you are neglecting losses.

 

[A Dhingra]

OK. i will try to put a diagram of the system i am talking about ... and by that time can you please tell me what losses are possible in the system...

 

[A Dhingra]

You can think of charging a capacitor as equivalent to compressing a spring. Definitely you need energy to compress a spring. Assuming no friction, if you apply an external force to a spring, will the compression of spring goes on for ever? Won't you reach a point where anymore compression is impossible, irrespective of presence of external force?

i agree to your point ... i know the system that i have explained was not clear but the external force is to be applied to the two other plates inserted between the charged capacitor plate...
and with the application of Newton's 1st law, the charge when given some unbalanced force will continue to move until there comes any resistive force in the picture...

 

[sophiecentaur]

In any problem in which you connect capacitors, there is always the question of apparent energy loss - or continual oscillation of energy between the capacitors. I have a feeling that this may be a similar, apparent paradox - although I haven't seen a diagram yet and don't get exactly what the arrangement would be.

 

[A Dhingra]

here is the diagram...in attachment...

i hope it is clear...now...

 

[sophiecentaur]

If the two plates are connected as shown then they will be at the same potential all the time. The battery will be shorted out when the switch is connects. You may want to redraw the diagram.

 

[sophiecentaur]

If the movable plates are not isolated from each other and inserted into the space between the charged plates, the effective Capacitance between the charged plates will increase significantly. This means that the PD between them will reduce and work will have been done on the inserted plates (they will be pulled into the gap). This is because the unbalanced charge is the same but the capacity will be less and energy has to go somewhere. The plates could then be rotated with no change in the energy situation - charge will flow from one side to the other of the pair with no energy used when the resistance between them is zero.

If, however, you disconnect the plates and put a resistive path between them, there will be work done (energy expended) as current flows through the resistor to equalise the PD between them. Rotating the plates, so the charge flows the other way, will involve work, every time their positions are reversed. You would be generating AC power, which would be dissipated in the resistor - but you would need to provide energy to do this.

I am not sure what function you are suggesting the battery would have in your circuit. I imagine it could charge and discharge a tiny bit as the plates were reversed.

 

[A Dhingra]

...
sorry but the plates are not movable( at least i did not consider them like that.)... but what i was focusing, was that, the battery would give out electrons from its negative pole... and these electrons would cause a drift to the electrons of the two uncharged plates ( which are at same potential on the surface of the plate... and any external force will make them change their state and cause them to move... as per Newton's law) to move due to the potential difference generated due to the capacitor plates...
and even after the external force of the battery is removed it should continue in its state of motion...

have i been able to make it clear now?

as you were saying it was about rotation and the production of normal ac current...
which did not think of... but thanks for the idea...

 

[A Dhingra]

help me with this as well...

https://www.physicsforums.com/showthread.php?t=488158

thanks

 

[Dale]

here is the diagram...in attachment...

i hope it is clear...now...

Hi A Dhingra, when the switch is connected to the battery the bulb will briefly light up as current flows to charge up the inner capacitor. When the switch is turned to the path without the battery the bulb will briefly light up as the current discharges in the capacitor. The amount of energy lost by the battery will be greater than the amount of energy provided to the light bulb. The large plates don't do much as drawn, except to add an offset voltage for the small plates. This is not a generator.

 

[sophiecentaur]

The middle plates appear connected in the diagram. That means that there is no PD across them so they take no part in the circuit at all - they are just a connecting wire in the battery circuit.

Could you go over, again, what you actually want this thing to do? :-)
I think Dalespan has said it all - unless you had something more in mind.

 

[Dale]

The middle plates appear connected in the diagram. That means that there is no PD across them so they take no part in the circuit at all - they are just a connecting wire in the battery circuit.

Oh, you are right. I mentally edited out that line since it didn't make sense. So with that you simply get the light turning on when the battery is connected and off when the other line is connected. Both sets of capacitor plates are irrelevant.

 

[A Dhingra]

well now i agree with you... it won't work...

but still i was thinking that the presence of the two charged capacitor plates, will make sure that the plates confronting them should have some electric induction and hence potential difference... as the two metal plates are connected their potential difference should cause the current to flow... even if the battery was removed... this was the main thing that i was thinking about the system...

(but due to various losses, i guess it won't work for long.)

have you any ideas of a generator using a capacitor?

 

[A Dhingra]

DaleSpam and sophiecentaur...
thanks to both of you...

 

[Dale]

the presence of the two charged capacitor plates, will make sure that the plates confronting them should have some electric induction and hence potential difference... as the two metal plates are connected their potential difference should cause the current to flow

The connecting wire prevents any potential difference from forming. All points connected by a good conductor are at the same potential. That is what it means for something to be a good conductor.

If you want a potential difference then you have to remove the wire, but then the charge can't flow that way. If you want the charge to flow that way then you have to add the wire, but then you don't get a potential difference.

 

[sophiecentaur]

But if you put a resistor in circuit, energy is transferred and work is done when things are moved.

 

[Dale]

Are you suggesting to replace the connecting wire between the plates with a resistor?

 

[sophiecentaur]

Well, the diagram has a bulb in it so I thought it was relevant.
Although 'which plates' might be a question.

 

[Dale]

OK, if you take the circuit drawn previously and replace the wire between the two inner plates with a resistor then you get the following two situations. When the battery is connected you get a voltage divider between the bulb and the resistor. When the other wire is connected you get no current flow.

The outer plates charge up a little stray capacitance when they first turn on, but otherwise they do not affect the circuit in either switch position.

 

[A Dhingra]

The connecting wire prevents any potential difference from forming. All points connected by a good conductor are at the same potential. That is what it means for something to be a good conductor.

If you want a potential difference then you have to remove the wire, but then the charge can't flow that way. If you want the charge to flow that way then you have to add the wire, but then you don't get a potential difference.

i am getting more confused...
but it won't work... that's it...right?

and the outside plate won't help maintain the PD...?

 

[A Dhingra]

OK, if you take the circuit drawn previously and replace the wire between the two inner plates with a resistor then you get the following two situations. When the battery is connected you get a voltage divider between the bulb and the resistor. When the other wire is connected you get no current flow.

The outer plates charge up a little stray capacitance when they first turn on, but otherwise they do not affect the circuit in either switch position.

what do you mean by the voltage divider...?

 

[Dale]

what do you mean by the voltage divider...?

http://en.wikipedia.org/wiki/Voltage_divider

 

[Dale]

and the outside plate won't help maintain the PD...?

Correct. Not if there is a wire connecting the inside plates.

 

[A Dhingra]

i guess i got it , thanks...