Approach this question from the position limit values. What can be the maximum enegiya secluded world? If you look at the classical formula: \[ W_{c}=\frac {Q^{2}} {2C} \qquad (2.1) \] it turns out that less than \(C\), i.e. the capacitance, the higher the potential energy. Then what can be the minimum capacity? To do this, recall the formula for capacitance secluded Shar: \( C=4\pi \varepsilon \varepsilon _{0}r\), where \(r\) is the radius of the ball. What is the minimum radius can be from globe charge? Yes, that's right — the electron radius \(r_{e}\) [2]. Where do we find his own capacity: \[ C_{e}=4\pi \varepsilon \varepsilon _{0}r_{e} \qquad (2.2) \] it is Clear that the charge of the ball would be exactly equal to the charge of an electron \(e\). Relative permittivity \(\varepsilon\) is taken equal to unity (for a vacuum) and get the maximum energy for minimum capacity — the potential energy of the electron charge: \[ W_{ce}=\frac {e^{2}} {8\pi \varepsilon _{0}r_{e}} \qquad (2.3) \]

Recall that \( e=1.6\cdot 10^{-19} \;TC \) and \( r_{e}=2.82\cdot 10^{-15} \;m \).

But the resulting formula is exactly equal to half of the Einstein mass-energy: \[ W_{ce}=\frac {e^{2}} {8\pi \varepsilon _{0}r_{e}}\;=\;\frac {m_{e}c^{2}} {2}, \qquad (2.4) \] where: \(m_{e}\) is the mass of the electron is \(9.1\cdot 10^{-31} \;kg\), \(c\) is the speed of light is \(3\cdot 10^{8} \;\frac {m} {it}\). So we got the bundle: **the charge-mass-energy** and answer the question — where does the energy come from.

The potential energy of a system of electrons will be maximum if the capacitance of the capacitor, in which they are located, will tend to zero. Apparently, such critical state of the electron system is the electron gas or plasma in the vacuum.

This is the answer to the original question — where does the energy come from, if the capacity of the system of electrons increase? The electrons simply contact capacity and cease to be free, and the more capacity, the more they are related.

In this paper we will not delve into the wilds of electrodynamics and quantum physics, and we will consider free charges from the point of view of electrical engineering and electronics.

Once the electron is a kind of elementary capacity, then why can't he be a simple inductance? And indeed, we find such a justification in [3] and present a formula for the inductance of the electron: \[ L_{e}=\frac {4m_{e}r_{e}^{2}} {e^{2}}\;=\;\frac {\mu_{0}r_{e}} {2\pi}, \qquad (2.5) \] where: \(\mu_{0}\) — magnetic constant equal to \(1.26\cdot 10^{-6} \;\frac {GN} {m}\).

For the full picture we have to do the last assumption is that the electron is a perfect resonant circuit with its resonant frequency, characteristic impedance and infinite quality factor. As is known, energy in an ideal resonant circuit can circulate forever, or for as long as the circuit will not be connected to a radiating antenna, for example.

Another interesting conclusion may be: once the electron — oscillation circuit, so while he's part — **his entire potential energy of the jet**. It becomes active when an electron becomes a wave, and the manifestations of this energy we can feel in the form of light, heat, etc.

If all our assumptions are correct, then the task of extracting energy from the electron is reduced to one simple rule: we must create the conditions for electron, in which case it's reactive energy can be converted to active. Next, we consider the following conditions.

__For reference __

- self-capacitance of the electron: \(C_{e}=1.57\cdot 10^{-25} \;f\)
- self-inductance of the electron: \(L_{e}=2.82\cdot 10^{-22} \;GN\)
- impedance of electron: \(Z_{e}=\sqrt {\frac {L_{e}} {C_{e}}} = 60 \;Ohms\)
- resonant frequency: \(\nu_{e}=\frac {1} {2\pi\sqrt {L_{e}C_{e}}} = 1.69\cdot 10^{22} \;Hz\)
- wavelength: \(\lambda_{e}=\frac {c} {\nu_{e}} = 1.77\cdot 10^{-14} \;m\).