INTRODUCTION
ALL 6,7 AND 8TH CONFIGURATION HAVE THE SAME CHARGED NEGATIVELY CATHODE. THE SAME GROUNDED ANODE WITH THE SAME HOLE TO ALLOW PASSAGE OF THE REPELLED ELECTRONS.
CONFIGURATION 6 IS EQUIPPED WITH A NEGATIVELY AND ELECTROSTATICALLY CHARGED RING AS CLOSELY AS POSSIBLY TO THE ANODE AND ON THE RIGHT HAND SIDE.
ALL 6,7 AND 8TH CONFIGURATION HAVE THE SAME CHARGED NEGATIVELY CATHODE. THE SAME GROUNDED ANODE WITH THE SAME HOLE TO ALLOW PASSAGE OF THE REPELLED ELECTRONS.
CONFIGURATION 6 IS EQUIPPED WITH A NEGATIVELY AND ELECTROSTATICALLY CHARGED RING AS CLOSELY AS POSSIBLY TO THE ANODE AND ON THE RIGHT HAND SIDE.
THE ELECTRONS IN CONFIGURATION 6 AS SOON AS, THEY PASS THE HOLE OF THE ANODE , THEY ARE BEING REPELLED BY THE ELECTROSTATICALLY CHARGED RING, AWAY FROM IT. THUS INCREASING THEIR HORIZONTAL VELOCITY AND TOTAL ENERGY WITHOUT TO HAVE TO PAY ANY EXTRA POTENTIAL ENERGY TO HAVE TO CROSS THE HOLE.
CALCULATION OF GAINED FOR FREE ENERGY TO AVERY AWAY POINT, WHERE THE FIELD IS ZERO = to the first approximation, ignoring induced charges on the anode (second order approximation)=
= E = Q/R
WHERE R IS THE RADIUS OF THE RING, AND Q IS THE TOTAL CHARGE ON THE RING.
= E = Q/R
WHERE R IS THE RADIUS OF THE RING, AND Q IS THE TOTAL CHARGE ON THE RING.
CONFIGURATION 7 PRESUMABLY HAS A LITTLE CHARGED NEGATIVELY SPHERE ON THE RIGHT OF THE ANODE, REPRESENTING A NEGATIVE POINT CHARGE. THE POINT NEGATIVE CHARGE ACTS THE SAME WAY AS THE NEGATIVE RING OF CONFIGURATION 6, THUS INCREASING THE HORIZONTAL AND TOTAL ENERGY WITHOUT THE LOSS OF ANY CHARGE ON THE RIGHT OR ENERGY
QUANTITATIVE CALCULATION OF INCREASED ENERGY.
SUPPOSE THE POTENTIAL OF THE POINT CHARGE AT THE ENTRY POINT IS q/r1
WHERE q IS THE CHARGE ON THE SMALL SPHERE-POINT CHARGE AND r1 IS THE DISTANCE OF POINT CHARGE TO THE HOLE.
THE POTENTIAL AT A FAR AWAY POINT IS. q/r2. THEN THE GAINED ENERGY IS:
THE GAINED ENERGY PER PASSING ELECTRON AT A FAR AWAY POINT OF DISTANCE r2
= to the first approximation, ignoring induced charges on the anode (second order approximation)=
E = (q/r1 - q/r2)xe
= to the first approximation, ignoring induced charges on the anode (second order approximation)=
E = (q/r1 - q/r2)xe
WHERE e IS THE CHARGE OF THE PASSING ELECTRON.
CONFIGURATION 8 CONSISTS OF THE SAME ANODE AND CATHODE. ON THE RIGHT OF THE ANODE IT IS EQUIPPED WITH A NUTRAL ISOLATED PLATE. AS SOON THE ELECTRON PASSES OVER THE NUTRAL PLATE OPPOSITE ELECTROSTATIC CHARGES ARE INDUCED THERE, WHICH ATTRACT THE ELECTRON. THUS THE ELECTRON GAIN VERTICAL VELOCITY BY THIS ATTRACTION AS SMALL NEUTRAL PARTICLES ARE ATTRACTED BY A CHARGED BODY AS LONG AS THE ELECTRON COMES AWAY FROM THE PLATE THE ECTROSTATIC INDUCTION BECOMES WEAKER. THUS REDUCING THE HOLDING BACK VELOCITY.
THUS THE GAINED OVER ALL VELOCITY AND ENERGY IS BIGGER.
THUS THE GAINED OVER ALL VELOCITY AND ENERGY IS BIGGER.
CONCLUSION
WHAT IS CHARACTERISTIC IN THE 3 EXAMPLE CONSISTING OF THREE DIFFERENT OBJECTS:
ONE CHARGED NEGATIVELY RING,
ONE CHARGED NEGATIVELY SMALL BALL, POINT CHARGE,
ONE CHARGED NEGATIVELY PLATE.
IT IS THE NON SYMMETRY ON THE LEFT AND RIGHT HAND SIDE.
ON THE LEFT THERE IS A GROUNDED ANODE WITH A SMALL HOLE, WHICH ANODE ACTS AS A FARADAY SHIELD FOR THE COMING ELECTRON. SO THE NEEDED APPROACHING SPENT ENERGY TO THESE OBJECTS IS MUCH LESS THAN THE DESCENDING GAINED ENERGY.
Cathode and anode are both grounded to zero potential. An electron with kinetic energy received by the heated cathode approaches the neutral anode. There on the anode, by induction, opposite positive charges, are accumulated, which attract and accelerate the electron. The electron passes the new hole again, which a negative concetretic negative ring is placed, as shown. The negative ring repells and farther accelerates the electron to the right towards the next anode. The whole process is repeated, increasing the velocity of the electron, farther and farther,..., making U0, <U1., < U2,< U3 .....
Source: http://www.papimi.gr/678.htm
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