KR20000073781A - Induction Generator - Google Patents

Abstract

PURPOSE: A capacitive induction generator is provided to make the direction of electric line of force become 90 deg. so as for electricity to flow into an external circuit without electric resistance. CONSTITUTION: In a capacitive induction generator, a plurality of electric magnet and generating magnet groups are formed so as to make a rate of the pole number of the electric magnet to the generating magnet become 1 to 2. The induction generator is constructed by forming the electric magnet with a multiple system so that electricity flows into an external circuit without electric resistance.

Description

Electrostatic Induction Generator

The present invention relates to an electrostatic induction generator which complements the inventions of Applicants' Patents 8006 and Publication Nos. 91-13655 to more effectively obtain a larger output than the input.

According to Patent No. 8606 and Publication No. 91-13655 invented by the applicant, the ratio of the number of poles of the electric field and the generator is 1: 2 (in the case of Patent No. 8606 and Publication No. 91-13655, stator and rotor are described). In the present application, this is called an electric field or a generator. If the direction of the electric line is set to 90 ° with respect to the rotational movement of the generator, this electric force can be developed without supplying energy because it does not prevent the generator from rotating ( For further details, see Patent No. 8606 and Publication Nos. 91-13655. Subsequently, as a result of further studies and experiments by the Applicant, as shown in FIG. It is confirmed that the input increases and the output decreases by remaining in (b) and causing the electricity to deviate significantly from the direction of the electric field line by 90 °.

Therefore, in order to generate electricity without supplying energy as claimed in Patent Nos. 8606 and Published Publication Nos. 91-13655, the electric power lines should be 90 ° in an easy way to flow electricity to an external circuit regardless of electric resistance (actually at 90 °). You need to figure out how to get close to it.

1 to 3 are explanatory diagrams for explaining the present invention.

4 is a model diagram of an electric field.

5 is a model diagram of a generator.

6 to 10 are explanatory diagrams of a basic circuit and an output calculation.

11 is an explanatory diagram for generating sine waves.

12 is an area display diagram that changes in the case of a single pole.

Fig. 13 is a model diagram of an electric field and a generator in the case of multiple poles.

14 is an area display diagram that changes in the case of a multipole;

15 is a model and a circuit diagram of an electric field in the case of multiple poles.

16 and 17 are models and circuit diagrams of the generator in the case of multiple poles.

18 is a cross-sectional view taken along line II of FIG. 17.

19 is an assembly view of the generator.

20 is an exploded view of an electric field and a generator when the electric field is doubles;

21 is a partial cutaway view showing the assembled state of the electric field and the generator.

22 is a layout and connection display diagram of groups A and B of electric fields.

* Explanation of Signs of Major Parts of Drawings *

1. Electrode of electric field 2. Non-electrode part 3,4,5,6.Connector of electric field 12,13,14a, 14a ', 23,24,25b, 25b' ... Connector of generator 7,8 is the connector of the electric field

9,10 ... groove of electric field 15, 16, 21, 22 ... circular circuit of generator 17 ... hole 19a, 19a ', 20b, 20b' ... hole through which wire can penetrate 26. Small round plate

27 Bearing shaft 28, 29, 31 Nut 33, 34 Switching 38 Stand

39,39 '... support 40 ... brush 41 ... lower acrylic plate 42 ... upper acrylic plate

44A, 45A, 44B, 45B ... Conductor wire connecting electric field 43 ... Bearing

2 and 3 is a basic principle diagram of the present invention, the pole plates (a) and (b) on both sides of the insulator (Y) and (a ') and (b') are connected by resistance wires (R1) and (R2), and the pole plates. (+) And (-) electricity are stored in (A) and (a), and (-) and (+) electricity are stored in electrode plates (B) and (b). And since (A) and (B) are completely insulated, the (+) and (-) electricity of (A) and (B) once stored are retained for a long time, and the (+), It is assumed that the negative electric force acts on (a, a ') and (b, b') only by the lengths of (A) and (B), as indicated by the dotted line.

As shown in FIG. 3, when (A) and (B) are pushed or pulled in the → direction, (a) has a portion deviating from the (+) electric force of (A), and thus (-) electricity appears there. In (b) there is a part of (B) which is out of the (-) electric force, so there is (+) electricity, but these (-) and (+) electricity are (+) of (A), (B). , (-) a (R ₁₎ by a deviation and at the same time force the action of its own in an electrical force, where the plate (a) of the (+) electric-and the dotted line outside of the (a) electricity which have not left () There is no attraction, and the (-) electricity in (B) is not capable of attraction with the outside of the dotted line in (b), where no (+) electricity remains.

At the same time, (a ') and (b') have a part of (A) and (B) that receive (+) and (-) electric force, so that electrostatic induction action occurs at (a ') and (-), (+) and (b ') generate (+) and (-) electricity. At this time, the (-) electricity of (a ') and the (+) electricity of (b') remain attracted to the (+) and (-) electricity of (A) and (B), but the (+) of (a ') The electricity and the negative electricity of (b ') flow to (R ₂ ) by its attractive force. In this case, the (+) electricity of (A) is outside the dashed line of (a ') without (+) electricity remaining, and (-) electricity of (B) is (b') without (-) electricity remaining. There is no repulsion with the outside of the dotted line. Therefore, the direction of the electric force lines with respect to the horizontal movement of (A) and (B) is in the 90 ° direction, not in the (a), (b) side, nor in the (a '), (b') side.

Fx = Fcos 90 ° = 0, Fy = Fsin 90 ° = F

(cos 90 ° = 0, sin 90 ° = 1)

Equation of In this way, it becomes clear that the electric force does not interfere with the horizontal movement of (A) and (B). Since the current flowing through (R ₁ ) and (R ₂ ) does not occur in the magnetic field, the current does not prevent horizontal movement of (A) and (B). Therefore, the electric and magnetic forces can be developed without supplying energy from the outside because they do not interfere with the horizontal movement of (A) and (B).

In reality, however, the electric force is weakened due to the thickness and dielectric constant of the insulating material, so that electricity that cannot be turned into current remains in (a) and (b). For this reason, the direction of the electric field lines also deviates from 90 degrees. Therefore, in order to make the direction of the electric field lines as close to 90 ° as possible, the thickness of the insulator (Y) is thin, and the lower the dielectric constant is, the better. In this regard, as the insulating material, a high vacuum of 10 ^-4 [mmHg] or less is best.

The following derives the formula for calculating the output:

1. Capacity calculation formula of electrostatic induction generator

In the case of using the electric field of FIG. 1B as a single electric field, and the electric field of the double electric field in the case of FIGS. 2 and 3 (A) and (B) as an electric field, the capacity is the same regardless of whether the electric field is single or double. The calculation describes the case where the electric field is fasting. Single field and double field will be described later. FIG. 1B shows an electric field (which plays a role of causing an electrostatic induction to the generator while maintaining a high voltage), and this is shown as in FIG. 4, and (b, b ') is represented by the generator (high voltage of the electric field). The static electricity is induced by a strong electric field and serves to flow the electricity to an external circuit.

Wherever (B) is in FIG. 1, this can be considered as a capacitor. Therefore, the capacitance of the electrostatic induction generator can be cited by the formula for calculating the capacitance of the capacitor. First, the effective area must be calculated first.

The effective area of the generator can be subtracted from the area of πr ₁ ² and the insulated wire from πr ₂ ² , but as shown in FIGS. 4 and 5, the area of the electric field is 1/2 of the area of the generator, so it should be 1/2 of the effective area of the generator. do. In other words, the effective area of the electric field may be calculated. Therefore, if the number of generators is calculated as n, the capacity of the capacitor should be calculated as 2n considering both sides of the generator, but the effective area of the electric field is 1/2 of the effective area of the generator. Calculate If the number of poles of the electric field (the number of poles will be described later) is E, the number of insulated wires is 2E (the number of insulated wires is equal to the number of poles of the generator). Therefore, ① the distance from the center of the generator to the inner circumference r ₁ [cm], the distance from the center of the generator to the outer circumference r ₂ [cm], ③ the distance between the electric field and the generator d [cm], ④ The formula is to calculate the capacitance of a capacitor when the number of rulers is n, the number of insulated wires is 2E, and the width of the insulated wire is α [cm].

The capacity can be calculated by citing. Therefore, the effective area must be obtained first. If the total area of one generator is S ₂ and the invalid area is S ₁ , the effective area S is

S = S ₂ -S ₁

to be. And

S ₁ = πr ₁ ² + 2E (r ₂ -r ₁ ) α

ego,

S ₂ = πr ₂ ²

Since the effective area S of one generator is

S = πr ₂ ^2- {πr ₁ ² + 2E (r ₂ -r ₁ 1) α}

_{^{= Πr 2 2 - πr 1 2}} - 2E (r 2 - r 1) α

= π (r ₂ ² -r ₁ ² )-2E (r ₂ -r ₁ ) α

= (r ₂ -r ₁ ) {π (r ₂ + r ₁ )-2Eα}

to be. And since the number of generators is n, the total effective area S of generators is

nS = n (r ₂ -r ₁ ) {π (r ₂ + r ₁ )-2Eα}

to be. Therefore, the formula for calculating the capacity C of an electrostatic induction generator is

Becomes.

2. Generator's rotational movement and electric energy charging and discharging

For the convenience of explanation, the principle is explained by the horizontal movements of FIGS. 2 and 3 (A) and (B), but the rotational movement is actually performed, so that (A) and (B) are electric fields, (a, b) and (a). ', b') is represented by the generator as shown in Figs. And for the sake of simplicity, the house transition is represented by four concentric circles of different sizes, but in FIGS. 7 to 10, the house transition and the brush are omitted, so that (R, R ₂₎ is (a, b), (a ', b') Marked as directly connected. Also, for convenience of description, the capacity when (A), (B) and (a, b) coincides with each other as shown in FIG. 6 is Cab, and (A), (B) and (a ', b) as shown in FIG. The capacity when ') is matched is expressed as Ca'b'.

As shown in Fig. 6, when (A) and (a) store (+), (-), (B) and (b) (-) and (+) electricity, the formula,

W = 1/2 CV ² [J]

6, in the case of

Wab = 1/2 CabV ² [J]

Of electrical energy is stored. Since (A) and (B) are completely insulated, it is assumed that the (+) and (-) electricity stored once in (A) and (B) are maintained for a long time.

Next, as shown in FIG. 7, when the generator rotates in the → direction, (a) is out of the (+) electrical attraction of (A) is (-) electricity, (b) is out of (-) +) Electricity appears, but the electricity flows out of (A), (B) positive (+) and (-) electric force and flows to (R ¹ ) by its own attraction. At the same time, (a ') has a part receiving (+) electric force of (A), so electrostatic induction action occurs there and (-), (+) electricity is generated, and (b') of (B) There is a part that receives the electric manpower, so there is also (+), (-) electricity. At this time, the (-) electricity of (a ') and the (+) electricity of (b') remain attracted to the (+) and (-) electricity of (A) and (B), and the (+) of (a ') The electricity and the negative electricity of (b ') flow to (R ² ) by its attractive force. Thus, as shown in Figure 8 (a ', b') was initially stored during the rotation until it matches (A), (B),

Wab = 1/2 CabV ² [J]

Newly generated by electrical energy and electrostatic induction

Wa'b '= 1/2 Ca'b'V ² [J]

Of electrical energy flows into (R ¹ ) and (R ² ), this time because (A), (B) and (a ', b') coincide.

Wa'b '= 1/2 Ca'b'V ² [J]

Of electrical energy is stored. Therefore, the amount of electrical energy flowing into (R ¹ ) and (R ² ) during the generator's half turn is twice that of the original stored energy.

Subsequently, when the generator rotates as shown in FIG. 9, this time, in contrast to FIG. 7, (a ') which is out of the (+) electric force of (A), (-) electricity is out of the (-) electric force of (B), and (b) ') Shows positive (+) electricity, but it flows to (R ₂ ) by its attractive force as it escapes the (+) and (-) electric force of (A), (B). At the same time, (a) has a part receiving (+) electric force of (A), so electrostatic induction action occurs there and (-), (+) electricity is generated, and (b) (-) electricity of (B) There is a part that receives the attraction, so there is also (+) (-) electricity. At this time, the (-) electricity of (a) and the (+) electricity of (b) remain attracted to the (+) and (-) electricity of (A) and (B), but the (+) electricity of (a) The negative electricity in (b) flows to (R ₁ ) by its attractive force. However, since the pole is changed at this time, the direction of the current is opposite to that of FIG.

If the generator continues to rotate until the state as shown in Figure 10, the stored of Figure 8,

Wa'b '= 1 / 2Ca'b'V ² [J]

Newly generated by the electrical energy and also electrostatic induction,

Wab = 1 / 2Cab V ² [J]

Of the electrical energy flows into (R ₂ ), (R ₁ ), where (A), (B) and (a), (b) coincided with each other as in the first FIG.

Wab = 1 / 2CabV ² [J]

Of electrical energy is stored. Therefore, the amount of electrical energy that flows into (R ₁ ), (R ₂ ) while the generator rotates once,

2 (Wab + Wa'b ') = 2 (1 / 2CabV ² + 1 / 2Ca'b'V ² ) [J]

to be. By the way, for convenience of explanation, Cab and Ca'b 'is described separately, but since it is actually the same capacity, if Cab and Ca'b' is called C,

2 (Wab + Wa'b ') = 2 (1 / 2CabV ² + 1 / 2Ca'b'V ² )

= 2 (1 / 2CV ² + 1 / 2CV ² )

= 2CV ² [J]

to be. Since the amount of electrical energy flowing through (R ₁ ) and (R ₂ ) is proportional to the rotational speed, assuming that the rotational speed for t seconds is N,

W = 2CV ² Nt [J]

It becomes

In this way, each time the generator rotates halfway, the newly generated electrical energy flows to the external circuit and is also stored and stored at the same time due to the stored electrical energy and the electrostatic induction. That is, each time the generator rotates half a turn, the direction of the current is changed, and until the current is changed again, the power of the current is developed into a square wave alternating current. But this is not a good waveform. Because in order to get a practically large output, you must increase the voltage to develop it and lower it back to a voltage suitable for use with a transformer, in which case the sine wave is the most efficient. Therefore, it must be developed into a sine wave, which is closer to the sine wave as the number of poles increases, or the pole of the electric field is spherical (dotted line) as shown in FIG. This can be done by producing a sinusoidal shape. In practice, however, the poles of the generator are not necessarily sinusoidal because the poles are as high as possible to increase the output.

3. Relationship between number of poles, output and current

The combined area of the electric field of FIG. 4 and the generator of FIG. 5 forms a generator as shown in FIG. 12, and when the generator is rotated for t seconds at a speed in the direction →, the changed area is a [cm ² ], a '[cm ² ] ([cm ² ] is omitted below).

Next, as shown in Figs. 4 and 5, when the generator is formed as shown in Fig. 14 as the electric field of the two poles and the generator of the four poles as shown in Fig. 13, and the generator is also rotated in the t direction for t seconds at the same speed as in Fig. The changed area becomes b, b ', c, c'. therefore,

a = a '= b = b' = c = c '

Because of,

2 (a + a ') = (b + b') + (c + c ')

to be.

In this way, the same size and the number of poles changed to 2 poles is 2 times, the changed area is 3 times, and if 3 poles is n times. This means that the changing area is proportional to the poles (the poles are the poles of the electric field). This also means that the capacity of the electrostatic induction generator increases in proportion to the poles. Therefore, the amount of electric energy flowing to the external circuit is proportional to the capacity, so

W = 2CV ² Nt [J]

If you add E, the number of poles, the amount of electrical energy flowing to the external circuit while the generator rotates for t seconds,

W = 2CV ² NEt [J]

It becomes If you mark this as power,

[W]

Because of,

[W]

This is the basic formula for calculating the output of a static induction generator.

Next, we break down this output equation into voltage and current.

P = Vi [W]

Because of

P = 2 CV ² NE [W]

To

P = Vi [W]

Can be decomposed into therefore

P = 2 CV ² NE [W]

To

P = V · 2CVNE [W]

If we decompose into, V is the voltage,

2CVNE

Is the current. As can be seen from the above equation, the output and the current are proportional to the number of poles. And because the frequency f is synergistic between the rotational speed N and the pole number E,

f = NE / _Sec

The frequency of the electrostatic induction generator is much higher than the frequency of commercial alternating current (50/60 [Hz]) because power is generated by increasing the number of revolutions and the number of poles to increase the output. (It is disclosed in Publication No. 91-13655 that the output and current are proportional to the poles.)

But the number of poles is not always high. This is because increasing the number of poles increases the number of insulated wires proportional to the number of poles, thereby reducing the effective area. Therefore, the pole number should be set properly.

The following explains how to make electric fields and generators.

The electric field maintains a strong electric field due to the high voltage of the DC high voltage generator and causes an electrostatic induction to occur in the generator. A single electric field is generated when only one of the positive and negative poles of the DC high voltage is generated. As shown in (B) of FIG. 1 and FIG. 2, 3, the case of developing both positive and negative poles is called a double electric field (A, B of FIGS. 2 and 3), and hereinafter, a double electric field means a double electric field. .

A plurality of acrylic plates (insulation plates) having a thickness of 2 [mm] are made into squares as shown in FIG. 15, and then cut in half and cut the opposite edges of the connecting holes (7, 8) to allow disassembly and assembly at will. . Next, after attaching the aluminum foil on the electrode 1, the connecting line (3, 4) connecting the electrode and the electrode and the connecting line (5, 6) connecting all the electrodes to the connection hole (7, 8) and Figure 15 and Use a razor blade to cut the gold to form the same shape, and remove the aluminum foil from other parts except the electrode (1), connecting lines (3, 4, 5, 6), and connecting holes (7, 8). Next, make the electrode, the connecting line, and the hole on the back in the same way as the front side, where both electrodes (1) should be paired with each other on both sides without matching. And the insulation problem is very important because the electric field must maintain high voltage at all times. Therefore, the insulation tape was affixed four times and the insulation tape was cut and insulated with a strong adhesive.

However, this is only a method used by the present inventors as convenient, and in reality, since there is no perfect insulator, this degree of insulation method is not sufficient when the electric field is practically manufactured. This is because, unless a perfect insulator is present, when a high voltage is applied, electricity is also conducted in the non-electrode 1, so that the voltage of the electrode 1 and the non-electrode 2 become almost the same. This is the same as developing at a lower voltage, which causes the output to drop significantly. Therefore, there is a need for a method of preventing electricity from conducting to the non-electrode. This may be done by removing or recessing the non-electrode 2 (dashed line) as shown in FIG. 15.

The generator generates electrostatic induction due to the strong electric force of the electric field and makes the electricity flow to the external circuit while rotating.

However, since the electric field is subjected to DC high voltage but no electricity flows, the material of the electric field only needs to have high insulation capability. However, since high voltage, high-frequency alternating current flows through the generator, the generator's material must have a very low dielectric loss as well as insulation ability. In particular, when the electric field is doubled, high frequency alternating current flows on both sides of the generator. Considering the previous point, polystyrene (Polystylene), which has a very low dielectric loss, was selected as the material of the generator.

A thin polystyrene plate having a thickness of 1.5 [mm] is formed in a circle to drill a hole 17 through which the generator shaft can penetrate and a hole 18 through which a long, long walk can pass. Next, the aluminum foil is put on this, and the razor blade is used to attach gold having the shape as shown in FIG. 16 to the electrodes (d, e, f, g, h, i, j, k), connecting lines (12, 13, 14a, 14a '), The portions other than the circular circuits 15 and 16 and the collecting electrodes a and a 'and the aluminum foil attached to the insulating line 11 are removed. Next, in the same manner as on the front surface of the back of Figure 16, the electrodes (d ', e', f ', g', h ', i', j ', k'), the connecting line (23, 24, 25b, 25b ') ), Circular circuits 21 and 22, collector electrodes b and b ', and the like, as shown in FIG. 17, wherein electrodes d and d', e and e 'on both sides, .. K and K 'Should be paired with each other on both sides of the field as if they were electric fields. In addition, small holes 19a, 19a ', 20b, and 20b' are drilled in all of the collecting electrodes a, a ', b, and b' to make it easier to connect the collecting electrode and the collecting electrode with a conductor. In addition, since the electrostatic induction generator generates high voltage, there is a fear that electricity may leak into the atmosphere. Therefore, a thin layer of transparent lacquer (paint) was applied to the surface of the generator so that electricity was not shorted in the atmosphere.

In this way, a small circular plate (26) (hereinafter referred to as 'small circular plate') for making a large number of generators and controlling the distance between electric field and generator, is also drilled through the holes through which the generator shaft, long walkways, and conductors can penetrate. Alternately, the generator shaft 27 and the long walkway 30 penetrate the generator shaft ball 17 and the sidewalk hole 18 drilled in the center of the generator and the small circular plate. When the same collector electrode of the generator is connected to each other and connected to the collector switching (33, 34) and then tighten the nuts (28, 29, 31) on both sides of the generator shaft 27 and the long walkway 30 as shown in Figure 19 The generator is completed.

As described above, the assembly method of the generator is briefly described, but it is not so simple in practice. In particular, when assembling the generator, the most important thing is to connect the same collector electrodes incorrectly. If any one of them is connected incorrectly, the connection problem is very important because not only one of them is connected incorrectly. Therefore, a method in which the same collector electrodes can be conveniently connected without being wrong will be described using a simple development diagram of the electric field and the generator of FIG. 20. In addition, (a, a ', b, b') shown in the developed view means a collector electrode of the generator shown in FIGS. 16 and 17, and the (+) and (-) poles indicate that the same poles are connected.

In order to connect the same collector electrodes of the generators, it is convenient to use the conductors of different colors by dividing the generators in half and half. In other words, half of the same generators are marked with C and the other half with D, and the conductors with blue color of the coating are the collecting electrode a of group C, red is b, black is a ', white is b', Yellow is a group D, green is b, blue is a ', and pink is b'.

After this preparation is completed, as shown in Figs. 19 and 20, generator C, then D, and then C,... If the same color wires are connected, the generators can be connected to the same collection electrodes in the C and D groups. Connect the generators C and D in this way, then solder the same poles, green and blue, yellow and red, black and pink, white and blue, and then connect the same poles again. And connect these two wires to the current switch (33, 34). However, this is only a description of the basic method, and there are many other ways. After all the collecting electrodes are connected, the electrodes or insulated wires of the generator should be lined up.

In this way, no matter how many generators, the collection electrodes of the same group can be connected easily.

In this way, assembling the generator and assembling the electric field again and combining them, an electrostatic induction generator is described. Next, the method of assembling the electric field and assembling the electric generator will be described. First, half of the electric fields made as shown in FIG. 15 are marked with (A) and the other half with (B) so that the lower electric field (A) is (A ₁ ), the upper electric field (A) is (A ₂ ), The electric field B is represented by (B ₁ ), and the upper electric field B is represented by (B ₂ ).

As shown in Fig. 21, an acrylic plate 41 having a thickness of about 3 [mm] is mounted on both sides of the pedestal 38 and fixed on both sides, and A ₁ , next to B ₁ , and next to the inner wall of the acrylic plate 41. In addition, A ₁ , in this order to fix the electric field to the inner wall of the acrylic plate (41) with an adhesive (chloro groove), the spacing should be adjusted so that the generator does not hit when rotating.

Next, the generator of Fig. 19 is placed on this electric field just assembled, so that the bearing 43 of the generator shaft (another bearing is not hidden by the support 39 ') is placed on the supports 39, 39'. Tighten the nut in order to be able to develop self smooth rotation and grooves (9, 10) is also of a _2, B ₁ grooves 9 and 10 of the a ₁ is knocked to push the B ₂ is also established on either side of the acrylic plate (42 ) [Acrylic plates (41, 41 ', 42, 42') on both sides are completely covered with four sides of the electric field. Hidden and invisible] should also be well spaced so that the generator does not hit the electric field as it rotates. And then the wire into the wire (44A _2), the connecting hole (8) of the B ₁ in the lead (44A _1), the connecting hole (7) of A ₂ in the connecting hole (8) of the A ₁ as shown in Fig. 22 (44B ₁ ), connect the conductors 44B ₂ to the connection hole 7 of B ₂ in an appropriate manner, and connect the four conductors again to 44A ₁ and 44A ₂ , 44B ₁ and 44B ₂ , The result is a double connection. When power is generated, connect these two wires one by one to the (+) and (-) poles of the DC high voltage generator (connect all the electric fields to one wire and connect the DC high voltage generator If you connect only one of the (+) and (-) pole of is called a single electric field.) Finally, the brush (40) is fixed to the support (39) so that it is in good contact with the switch (33, 34). When fixed, the electrostatic induction generator is completed.

As described above, the fabrication of the electrostatic induction generator has been described, but this is only intended for the experimenter's experiment and is not practical. Therefore, it is a matter of course that the material selection and manufacturing method should be taken differently, and it is obvious that this also cannot be deviated from the scope of the present invention.

According to the present invention, the effect of obtaining a larger output than the input is obtained.

Claims (2)

When the number of poles is 1: 2, a large number of multi-pole electric fields and generators are formed, and when a generator is formed by double electric fields, the generator's electricity flows to the external circuit more easily than when the electric field is single. Electrostatic induction generator. The electrostatic induction generator according to claim 1, wherein portions other than the electrode of the electric field are removed or recessed so that no electricity is conducted to the portions other than the electrode.