MPEMathematical Problems in Engineering1563-51471024-123XHindawi10.1155/2020/72306407230640Research ArticleTheoretical Calculation and Application Test of Lift Force for Ideal Electric Asymmetric CapacitorHuangJian-Guo1https://orcid.org/0000-0003-2655-1153ChengXiang-Yu23LiuJia-Bao1Anhui Technical College of Water Resources and Hydroelectric PowerHefei 231603China2Hefei Institute of Physical ScienceChinese Academy of ScienceHefei 231603Chinacas.cn3No. 38 Research Institute of China Electronics Technology Group CorporationHefei 230088China2020582020202009062020030720205820202020Copyright © 2020 Jian-Guo Huang and Xiang-Yu Cheng.This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

The asymmetric capacitor’s lift force formula can be obtained on the basis of literature review, which can almost cover all practical forms of asymmetric capacity forms. But there are still some problems we should solve. The first and foremost one is whether the formulas are correct and can they be verified in engineering practices? On the contrary, the parameter q in the formulas is normally unknown in the beginning of calculations, how can we get or reckon up it so as to use the formulas smoothly? In this paper, we set out to solve these questions.

Chinese Academy of SciencesY86CT21051Research Activity Funding of Postdoctoral Fellow of Anhui Province2018B250Natural Science Research Project of Anhui Education DepartmentKJ2018A0725High Magnetic Field Laboratory of Anhui Province
1. Introduction

How can we solve lift force produced by a lifter formed of an asymmetric capacitor? Based on some hypothetical conditions, a formula was obtained through three methods in an ideal scenario [1, 2]. But an unknown parameter q is still contained so that numerical calculations are difficult to carry out. This paper intends to solve this problem by eliminating the unknown factor in the hope that the formula can be effortlessly put into practical application and engineering. Following that, experimental tests and practical estimations are provided to verify its validity.

In former papers [1, 2], the same result is acquired through three ways using the following equation of electric lift force of asymmetric capacitor loaded by high voltage in ideal condition:(1)f=q2ε1S11S2.

An unknown variable q is still included in the above formula. In order to solve this problem thoroughly, the carried charge q should be figured out. Normally, carried charge q of capacitor is relevant to the voltage U and the capacitance C. The voltage U can be known. But the capacitance C is difficult to calculate when the capacitor is in irregular shape.

Nevertheless, the analysis of hypothetical predetermined conditions verifies that the capacitance C of the asymmetric capacity is calculable. When the small plate of asymmetric capacitor is in a slender cylinder form, its capacitance could be estimated at a cylindrical way. When the small plate of asymmetric capacitor  is in sphere form, its capacitance could be estimated at a spherical way. The result might not be ideal in the case of precision. It can still be applied to estimation in engineering assessment [6, 7]. Furthermore, the subsequent test data verified that the estimate result was fairly accurate unexpectedly.

2. Theoretical Derivation

Regarding the reason why the experimental result is more precise than expected, the analysis of the unique characteristics of the asymmetric capacitor has presented several objective reasons: (1) the distance d between two plates is more larger than the dimension of surface area S1 of plate 1 (small plate), that is, dS1/l or dS1; (2) the area of plate 2 (large plate) is larger than that of plate 1, that is, S2S1; and (3) the voltage loaded between two plates is below the breakdown voltage that is relevant to the gap distance.

Under the initial condition, we begin to deduce capacitance of the asymmetric capacitor  and then to estimate its lift force [11, 12]. Deducing processes are as follows:

For S2S1, when high voltage is loaded on two plates, the electric field intensity on plate 1, E1=q/εS1, is larger than that on plate 2, E2=q/εS2, i.e., E1E2. It leads to the voltage drop ΔU=ΔdE, which mainly centralizes around plate 1. So when calculating the capacitance C=q/ΔU=q/U1U2dU, the field intensity near plate 1 should be taken into major consideration. That is to say, the capacitance calculation can be carried out by combining the following equations (2), (3), (4), (5), and (6):(2)E=qεS,(3)du=Edr,(4)C=qΔU,(5)rSlfor thin wire and board plates,(6)rSfor sphere point and board plates.

Equations (5) or (6) can be also written as(7)r=ksh1Sl,r=ksh2S,

where r is the nominal dimensional size of plate 1, l is the length of plate 1, and ksh1 and ksh2 are the shape coefficient relevant to the plates’ structure size.

Because the distance d between two plates is far larger than the nominal size of plate 1 r, to simplify the calculation, we assume that surface charge of plate 1 is uniformly distributed, and voltage drop of thin wire plate or spherical capacitor plate is integrated for estimating the capacitance in magnitudes. The details are shown as follows.

For a thin wire small plate capacitor, we can take(8)S1=2πR1l,S2=2H2l,

where H2 is the width of the board plate.

Because

(9)E=qεS,

referring to equation (3), we get

(10)du=qεSdr.

Mainly considering the electric field variation beside the thin wire, we have

(11)Scir=2πrl.

Considering the effective fan-shaped part, we have

(12)Sfan=ScirR1+dR1+d+H22dr2πr=Scir1πlnR1+d+H2R1+d=2rlln1+H2R1+d, dS=dSfan=2lln1+H2R1+ddr, dr=12lln1+H2/R1+ddS.

Integrating both sides of equation (10), we obtain

(13)ΔU=U1U2du=r1r2qεSdr=qεS1S212Slln1+H2/R1+ddS=qlnS2/S12εlln1+H2/R1+d=qln2πH2l/2πR1l1/πlnR1+d+H2/R1+d2εlln1+H2/R1+d1,=qlnH2/R1ln1+H2/R1+d2εlln1+H2/R1+d1.

So we get the capacitance

(14)C=qΔU=2εlln1+H2/R1+dlnH2/R1ln1+H2/R1+d.

For a spherical small plate capacitor, we can take

(15)S1=4πr2.

Combining equation (2), we get

(16)E=qε4πr2.

Referring to equation (3), we get

(17)du=qε4πr2dr.

Integrating both sides, we obtain

(18)ΔU=U1U2du=R1R1+dqε4πr2dr.

For the distance dR1, we have

(19)ΔU=U2U1=R1R1+dq4πr2εdr=q4πε1R11R1+dq4πεR1.

So we get the capacitance

(20)C=qΔU=4πεR1.

We can calculate the electric lift force of asymmetric capacitor loaded by high voltage with the capacitance C.

For thin wire small plate capacitor, using equation (14),(21)f=q2ε1S11S2=UC2ε1S11S2=1εU2εlln1+H2/R1+dlnH2/R1ln1+H2/R1+d212πR1l12H2l2εlU2ln21+H2/R1+dln2H2/R1ln1+H2/R1+d1πR11H2.

This is the lift force formula about a normal lifter in thin wire asymmetric capacitor form under high voltage loaded.

For spherical small plate capacitor, using equation (20),(22)f=q2ε1S11S2=UC2ε1S11S2=U4πRε2ε1S11S2.

Considering the condition S2S1, we have(23)f=U4πRε2ε1S1=U24πε4πR21S1.

If simplifying calculation as a spherical plate, the surface area of plate 1 S1=4πR2, we can get(24)f=4πεU2.

This is the concised formula that finally turned out, from which we can tell the maximum lift force produced by spherical asymmetric capacitor under high voltage loaded.

3. Formula Application

The formulas are exerted on two applications to test their validity. They are, respectively, lift force estimation of a electricity lifter [13, 14] and a high-voltage rising hair experiment [15, 16].

3.1. Lift Force Estimation of a Lifter

When a lifter loads with high voltage (Figure 1) U=30kV, what is the lift force produced by the lifter? The physical sizes (Figure 2) of the lifter are, respectively, length of thin wire l=15cm×3=0.45m, radius of thin wire R1=0.08mm=8×105m, width of large board H2=30cm=0.3m, and gap between 2 plates d=40cm=0.4m. We take permittivity of atmosphere ε=8.85×1012C2N1m2.

A lifter flying up loaded with high-voltage power.

An asymmetric capacitor in lifter form.

On the initial conditions, using equation (21), we have(25)f=2εlU2ln21+H2/R1+dln2H2/R1ln1+H2/R1+d11πR11H2=2×8.85×1012×0.45×300002×ln21+0.3/8×105+0.4ln20.3/8×105×ln1+0.3/8×105+0.411π×8×10510.3=0.115 N11.7 gf.

That is to say, a lifter loaded with 30 kV voltage can produce a largest lift force of 11.7 gf.

3.2. Lift Force Estimation of High-Voltage Charged Conducting Sphere

As we know, when a high voltage loads on human body, our hair may be lifted up by the static electricity . But there is no precise data or concrete calculating method of the length of the lifted hair by the high voltage. By equation (24), the mentioned problem can be solved. We can use the formula to quantitatively calculate the hair length lifted by the static electric field. The details are shown as follows.

3.2.1. Initial Conditions

Voltage loaded on the head U=100kV, diameter of the human head D=10cm=1×102m, average diameter of the hair Dh=70μm=70×106m, density of the hair ρ=1.25g/cm3=1.25×103kg/m3, and permittivity of atmosphere ε=8.85×1012C2N1m2.

3.2.2. Target Problem

When the voltage is loaded on the hair under the above initial conditions, what is the maximum length of the hair (lh=?) that can be lifted up?

3.2.3. Solving Process

In this case, the head and ground can be considered as the two plates of asymmetric, where the head may be regarded as a small plate and its area of sphere surface is S1 and the distant ground as a large plate and its area of flat surface S2. However, S2S1, and the distance between the two plates dS1. The voltage loaded between small plate (head) and large plate (ground) should not reach to breakdown threshold. Under this condition, we apply equation (24) to calculate the electrostatic lift force acted on hairs.

The sum of the electrostatic lift force acted on the hair is(26)f=4πεU2=4×3.14×8.85×1012×100×1032=1.11N.

The surface area of the head is(27)S1=πD2=3.14×1022=3.14×104m2.

The intensity of pressure supported by electrostatic force is(28)p=fS1=1.11×1063.14×104=3.54×103Pa.

The average transverse area of a hair is(29)Sh=πDh24     =3.14×70×10624     =3.8×109m2.

Head surface area occupied by a hair can support a mass by the electrostatic force:(30)mh=Ghg=Shpg=3.8×109×3.54×1039.8=1.4×106kg.

The mass is converted into length of a hair:(31)lh=VhSh=mh/ρhSh=1.4×1061.25×103×3.8×109m=0.29m=29cm.

Therefore, the final result is obtained: when human being’s hair is loaded by high-voltage static electricity of DC 100 kV through a conducting metallic ball, approximately 29 cm length hair floats up into air.

4. Explanation and Conclusion

Based on some assumptions with simplified calculation, we derived lift force formula produced by an asymmetric capacitor in different conditions, with which the assess in certain survey and qualitative research can be undertaken in spite of unsatisfying precision. The method also provides a convenient way to calculate static electricity lift capacity produced by an asymmetric capacitor or lift force of lifters. It also contributes to the parameter optimization in designing  a larger load force of lifter formed by an asymmetric capacitor.

Data Availability

The data used to support the findings of this study are included within the article.

Conflicts of Interest

The authors declare that the research was conducted in the absence of any commercial or financial relationships that could be construed as a potential conflict of interest.

Acknowledgments

The authors gratefully acknowledge the support of the Thirteenth Five-Year Plan of Hefei Institute of Physical Science of Chinese Academy of Science (Grant no. Y86CT21051, “Electric and Magnetic Propulsion System”), Research Activity Funding of Postdoctoral Fellow of Anhui Province (Grant no. 2018B250, “High-Energy Ions Accelerated Thruster”), and Natural Science Research Project of Anhui Education Department (Grant no. KJ2018A0725, “The Uniformity Optimization and Software Development for MRI Magnets”). A portion of this work was supported by the High Magnetic Field Laboratory of Anhui Province.

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