kavlas Posted August 20, 2011 Share Posted August 20, 2011 (edited) I would like to hear your opinion, if this possible, in the following paradox: here Edited August 20, 2011 by swansont fix link Link to comment Share on other sites More sharing options...
swansont Posted August 20, 2011 Share Posted August 20, 2011 If you want to discuss it, you really should post the question here. Short answer: an electric field has energy. Link to comment Share on other sites More sharing options...
kavlas Posted August 20, 2011 Author Share Posted August 20, 2011 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?? Link to comment Share on other sites More sharing options...
swansont Posted August 20, 2011 Share Posted August 20, 2011 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. Link to comment Share on other sites More sharing options...
kavlas Posted August 20, 2011 Author Share Posted August 20, 2011 It has always had it. Where?? Link to comment Share on other sites More sharing options...
swansont Posted August 20, 2011 Share Posted August 20, 2011 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" Link to comment Share on other sites More sharing options...
kavlas Posted August 20, 2011 Author Share Posted August 20, 2011 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 ?? Link to comment Share on other sites More sharing options...
swansont Posted August 21, 2011 Share Posted August 21, 2011 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. Link to comment Share on other sites More sharing options...
kavlas Posted August 21, 2011 Author Share Posted August 21, 2011 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 ?? Link to comment Share on other sites More sharing options...
swansont Posted August 21, 2011 Share Posted August 21, 2011 You mean that the charge will hit the ground with an acceleration less than 9,8 meters/sec^2 ?? Yes. Link to comment Share on other sites More sharing options...
kavlas Posted August 21, 2011 Author Share Posted August 21, 2011 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 Link to comment Share on other sites More sharing options...
swansont Posted August 21, 2011 Share Posted August 21, 2011 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. Link to comment Share on other sites More sharing options...
kavlas Posted August 23, 2011 Author Share Posted August 23, 2011 (edited) 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 Link to comment Share on other sites More sharing options...
swansont Posted August 23, 2011 Share Posted August 23, 2011 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. Link to comment Share on other sites More sharing options...
kavlas Posted August 27, 2011 Author Share Posted August 27, 2011 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 Link to comment Share on other sites More sharing options...
swansont Posted August 27, 2011 Share Posted August 27, 2011 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. Link to comment Share on other sites More sharing options...
kavlas Posted August 27, 2011 Author Share Posted August 27, 2011 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?? Link to comment Share on other sites More sharing options...
swansont Posted August 27, 2011 Share Posted August 27, 2011 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. Link to comment Share on other sites More sharing options...
kavlas Posted August 28, 2011 Author Share Posted August 28, 2011 (edited) 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 Link to comment Share on other sites More sharing options...
DrRocket Posted August 28, 2011 Share Posted August 28, 2011 (edited) 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 Link to comment Share on other sites More sharing options...
swansont Posted August 28, 2011 Share Posted August 28, 2011 Anyway ,do the charges Q and 2Q produce the same deceleration on the falling body?? No. The radiated power varies as Q^2 Link to comment Share on other sites More sharing options...
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