This is an improved mode to understand energy creation or energy increase. All the positive potentials have been changed to zero. The plates of vertical displacement have move towards the anode as closely as possible. Then the passing electron is only repelled and is always accelerating. HOWEVER THE ELECTRON SPENT ENERGY UP TO THE ANODE, AFTER PASSING THE ONLY GAINS ENERGY FOR FREE AS THERE NO CONSUMPTION AFTER THIS POINT!!!!!!!!!!!!
IMPORTANT NOTICE: The anode is at zero potential as we said and it is very well known from electronics applications it acts also as an Faraday shield for the fields behind it.(For those who are mean I MIGHT MAKE THIS ANODE TO BE A COMPLETE CLOSED SURFACE BY CLOSING IT FROM THE LEFT END, MAKING IT A COMPLETE CLOSED FARADAY CAGE WITH A LITTLE HOLE ON THE RIGHT, ENCLOSING THE CATHODE . NOTHING WILL BE CHANGED !). Unless the little hole makes all the creation of energy !. Let me know by contacting me ppappas@papimi.com etc phones, and FAX.. I may publish your point unanimously with my answer together.
Notice also i1 = 0, i2 < i3 i3 - i2 corresponds to the passing electrons. only. Typical value for this current 200 mA. It may go up to 2A or more. This current may charge or discharge the capacitor plates in less than 1 nanosecond. These values are given to show to an invited and objecting University professor, that a relative small energy system such as the horizontal capacitor's 1/2CV2 may divert the huge relative energy of the passing electrons VIt= 5000voltx2AXt
IMPORTANT NOTICE: The anode is at zero potential as we said and it is very well known from electronics applications it acts also as an Faraday shield for the fields behind it.(For those who are mean I MIGHT MAKE THIS ANODE TO BE A COMPLETE CLOSED SURFACE BY CLOSING IT FROM THE LEFT END, MAKING IT A COMPLETE CLOSED FARADAY CAGE WITH A LITTLE HOLE ON THE RIGHT, ENCLOSING THE CATHODE . NOTHING WILL BE CHANGED !). Unless the little hole makes all the creation of energy !. Let me know by contacting me ppappas@papimi.com etc phones, and FAX.. I may publish your point unanimously with my answer together.
Notice also i1 = 0, i2 < i3 i3 - i2 corresponds to the passing electrons. only. Typical value for this current 200 mA. It may go up to 2A or more. This current may charge or discharge the capacitor plates in less than 1 nanosecond. These values are given to show to an invited and objecting University professor, that a relative small energy system such as the horizontal capacitor's 1/2CV2 may divert the huge relative energy of the passing electrons VIt= 5000voltx2AXt
NUMERICAL EXAMPLE with typical (reasonable) values.
In the following we shall use only definitions and Mathematical identities that are always and under any circumstances valid
In the following we shall use only definitions and Mathematical identities that are always and under any circumstances valid
Such as
Energy = mechanical work by definition = int(Fds)=int(medu/dt)(ds)=int{meudu}= 1/2meu2
Energy = mechanical work by definition = int(Fds)=int(medu/dt)(ds)=int{meudu}= 1/2meu2
Energy = electrical work by definition = eE, E = Electromotive Force (Voltage), using the better terminology of older Electrodynamics
(KOSTA) Remember that q on a conductive surface or a conductor is a function -distribution, not a constant. Also that, the E.M.F. on a conductor or a surface is zero, so no matter whatever the charge distribution is on it, the conductor or the metallic surface remains always equi-potential that determines the electric field outside and near it!
therefore eV = 1/2mev2 therefore
v = (2eV/me)1/2 = (2x 5x103x1,602 X 10-12 10-8Joules / 9,109 x 10-28 x 10-3 Kgr)1/2 = (1,7587x1014)1/2 = 1,326X107 meters/ seconds in non relativistic mechanics. So for relativistic mechanics, the passing velocity is close to the velocity of light. Approximately: v= 98% C VELOCITY OF LIGHT
(KOSTA) Remember that q on a conductive surface or a conductor is a function -distribution, not a constant. Also that, the E.M.F. on a conductor or a surface is zero, so no matter whatever the charge distribution is on it, the conductor or the metallic surface remains always equi-potential that determines the electric field outside and near it!
therefore eV = 1/2mev2 therefore
v = (2eV/me)1/2 = (2x 5x103x1,602 X 10-12 10-8Joules / 9,109 x 10-28 x 10-3 Kgr)1/2 = (1,7587x1014)1/2 = 1,326X107 meters/ seconds in non relativistic mechanics. So for relativistic mechanics, the passing velocity is close to the velocity of light. Approximately: v= 98% C VELOCITY OF LIGHT
mass to electron charge me/ e = 1/(5.2727x 1017 ESU x gr-1)
5000eV =5000eE = 5x103x1,602 X 10-12 erg= 5x103x1,602 X 10-12 10-8Joules
5000eV =5000eE = 5x103x1,602 X 10-12 erg= 5x103x1,602 X 10-12 10-8Joules
me =9,109 x 10-28 g r= 9,e109 x 10-28 x 10-3 Kgr
distance cathode to the anode = irrelevant however with higher order corrections due to the small hole, which tends to zero, the bigger this distance, we choose.
distance cathode to the anode = irrelevant however with higher order corrections due to the small hole, which tends to zero, the bigger this distance, we choose.
Homogeneous Intensity of field between the plates of big capacitor = V/L= -200Volts/1cm=-20000Volts/ meter
L = 1 cm separation = 0,01meters
V = -200 VOLT
E = Electromotive force of the cathode = -5000 Volts
Length of the capacitor S= 300 cm= 3 meters
Width of the capacitor = 40 cm= 0,4 meter
time travel between the plates = t=S /v= 3 Meters/30000000 m/sec =3/3x10-8seconds = approximately 10 nanoseconds.
t time passing through the capacitor = approximately 10 nanoseconds
The passing short time is short and it will correspond to a practical small increase in the vertical velocity and due, to the electron's high rapidness of descending from the capacitor with a great velocity, the electron results also to a practical small increase in the horizontal velocity. Therefore for this realistic and reasonable example we have a small but definite increase in the over all velocity and energy.
Therefore this definite but in practice small increase of velocity and corresponding energy explains why experimentally energy was not known to increase.
Therefore this definite but in practice small increase of velocity and corresponding energy explains why experimentally energy was not known to increase.
Improving factors for making the energy increase bigger are to minimize the cathode| to the anode potential difference down to less even than 50 Volts, even down to -15volts (cut off negative anode potential), letting even thermal electronss through, to increase the capacitor length to 30 meters or more, to increase the potential of the capacitor plate to even 5000 or more and finally to decrease the diameter of the hole which what ever small it will be, still (thermal) electrons will go through it even without any cathode to anode potential,. Q.E.D.
Matters like the self degreasing system's entropy will be published on another occasion elsewhere
Source : http://www.papimi.gr/IMPROVED.htm
No comments:
Post a Comment