Physical paradox

[kavlas]

I would like to hear your opinion, if this possible, in the following paradox:

 

 

 

here

Edited August 20, 2011 by swansont
fix link

[swansont]

If you want to discuss it, you really should post the question here.

 

Short answer: an electric field has energy.

[kavlas]

If you want to discuss it, you really should post the question here.

 

Short answer: an electric field has energy.

 

Sorry by "here" i meant the forum in concern. Certainly the discussion must be here.

 

Where did the charge acquired the energy to form the electric field??

[swansont]

Sorry by "here" i meant the forum in concern. Certainly the discussion must be here.

 

Where did the charge acquired the energy to form the electric field??

It has always had it.

[kavlas]

It has always had it.

 

Where??

[swansont]

Where??

Charge is a conserved quantity. It doesn't just pop into existence. If you have a charge, it was either already there, or you separated a neutral system into a positive and negative charge, and in doing so, you did work on that system.

 

 

This is a manufactured "paradox"

[kavlas]

Charge is a conserved quantity. It doesn't just pop into existence. If you have a charge, it was either already there, or you separated a neutral system into a positive and negative charge, and in doing so, you did work on that system.

 

 

This is a manufactured "paradox"

 

 

Then this work must of nearly infinite amount because you can let fall that charge from the top of a Cliff as many times as you want .

 

Every time it will produce the same energy ,will it not ??

[swansont]

Then this work must of nearly infinite amount because you can let fall that charge from the top of a Cliff as many times as you want .

 

Every time it will produce the same energy ,will it not ??

 

You have to do work to get it to the tip of the cliff, but yes. The problem is that the claim "On the creation of this kinetic energy of the charge we did not spend any kinetic energy of the body mvv / 2, because if this was so, it would be contrary to the law of gravity, where all the bodies have the same acceleration." is wrong.

 

They both have the same acceleration from gravity, but the acceleration of a body depends on the total force, and there is an additional force on the accelerating charged particle. The "analysis" is simply ignoring it. Accelerating charges radiate. That causes (on average) an acceleration in the opposite direction.

[kavlas]

You have to do work to get it to the tip of the cliff, but yes. The problem is that the claim "On the creation of this kinetic energy of the charge we did not spend any kinetic energy of the body mvv / 2, because if this was so, it would be contrary to the law of gravity, where all the bodies have the same acceleration." is wrong.

 

They both have the same acceleration from gravity, but the acceleration of a body depends on the total force, and there is an additional force on the accelerating charged particle. The "analysis" is simply ignoring it. Accelerating charges radiate. That causes (on average) an acceleration in the opposite direction.

 

You mean that the charge will hit the ground with an acceleration less than 9,8 meters/sec^2 ??

[swansont]

You mean that the charge will hit the ground with an acceleration less than 9,8 meters/sec^2 ??

 

Yes.

[kavlas]

Yes.

 

Then the rule : all bodies fall in vacuum with the same acceleration 9,8m/s^2 ,should be altered to:

 

All not charged bodies fall in vacuum with the same acceleration

[swansont]

Then the rule : all bodies fall in vacuum with the same acceleration 9,8m/s^2 ,should be altered to:

 

All not charged bodies fall in vacuum with the same acceleration

 

Or one can simply recognize that a law that describes mechanical systems applies to mechanical systems.

[kavlas]

Or one can simply recognize that a law that describes mechanical systems applies to mechanical systems.

 

Then highly charged airplanes will never fall ??

 

Or ,in general ,highly charged bodies never fall.

 

 

Suggest a practical experiment ,if this is possible.

Edited August 23, 2011 by kavlas

[swansont]

Then highly charged airplanes will never fall ??

 

Or ,in general ,highly charged bodies never fall.

 

 

Suggest a practical experiment ,if this is possible.

 

No, the force from being charged in this situation will only be present while there is acceleration, and will not exceed the gravitational force. The statement "all bodies fall in vacuum with the same acceleration 9,8m/s^2" assumes that the only force present is gravity. It is an application of F=ma, but is only true when no other forces are present.

[kavlas]

No, the force from being charged in this situation will only be present while there is acceleration, and will not exceed the gravitational force. The statement "all bodies fall in vacuum with the same acceleration 9,8m/s^2" assumes that the only force present is gravity. It is an application of F=ma, but is only true when no other forces are present.

 

When i say "never fall" ,i mean the following:

 

Suppose a body is so highly charged that after 1/10 sec there is no forces acting on the body ,then the speed with which the body will hit the ground will be 3,528km/hour.

 

Suppose now the body is charged more so after 1/100sec there is no forces acting on the body ,then it will hit the ground with 0,3528 km/hour

 

So we can highly charge the body up to the point that it will hit the ground with an infinitesimal speed.

 

Probably with a speed : 3,528x 10^-1000Km/h

[swansont]

If you have a single charge, the radiation that will slow it down depends on the acceleration. There can never be a time when there is no net force. This is not a speed-dependent force, like air resistance.

[kavlas]

If you have a single charge, the radiation that will slow it down depends on the acceleration. There can never be a time when there is no net force. This is not a speed-dependent force, like air resistance.

 

You mean that,if a charge Q produces an ,x deceleration,then a 2Q charge will produce the same deceleration??

[swansont]

You mean that,if a charge Q produces an ,x deceleration,then a 2Q charge will produce the same deceleration??

 

No. The acceleration from the effect will always be less than g. If there is no net acceleration, there is no radiation.

[kavlas]

No. The acceleration from the effect will always be less than g. If there is no net acceleration, there is no radiation.

 

Anyway ,do the charges Q and 2Q produce the same deceleration on the falling body??

Edited August 28, 2011 by kavlas

[DrRocket]

Then the rule : all bodies fall in vacuum with the same acceleration 9,8m/s^2 ,should be altered to:

 

All not charged bodies fall in vacuum with the same acceleration

 

The law is that the gravitational force is [math]F=G \dfrac {m_1m_2}{r^2}[/math]

 

It is not that "all bodies fall in a vacuum at the same acceleration 9,8m/s^2".

 

Other forces also apply to an accelerating charged body. http://en.wikipedia....93Lorentz_force

Edited August 28, 2011 by DrRocket

[swansont]

Anyway ,do the charges Q and 2Q produce the same deceleration on the falling body??

 

No. The radiated power varies as Q^2