/mo> E 2 = σ 2 ε = q ε S 2 (3)

Because of actual electric charge q is n = q q 1 h = S 2 S 1 times of the hypothetical electric charge q 1 h , it needs n electric fields E 1 h to superpose each other to compose the actual field strength on plate 1.

The residual electric charge q 1 r removed the assuming charge q 1 h from actual charge q is

q 1 r = q q 1 h = q S 1 σ 2 = q ( 1 S 1 S 2 ) (4)

We need n residual electric charges q 1 r to superpose each other to compose the actual field strength on plate 1 E 1 r = q ( 1 S 1 S 2 ) n . So we can get the electric field force f of the n residual electric charge q 1 r in assuming electric field E 1 h = E 2

f = q ( 1 S 1 S 2 ) E 2 n = q ( 1 S 1 S 2 ) q ε S 2 S 2 S 1 = q 2 ε ( 1 S 1 1 S 2 ) (5)

Therefore we get the right result of the thrust force of practical asymmetric capacitor.

3.2. The Second Way

We assume the air medium between the two plates of the asymmetric capacitor as a whole block in physical shape, and electrical non-conduction at the inner (refer Figure 4). The gaps between the air block surfaces near to the plates are infinitesimal. So the molecular collision between them causes to charge exchange, and then leads to the upper surface positively charged and the lower surface negative charged. As the reason of like charges repelling, there is a pair of reaction forces produced on the surfaces of air block and the plates. You can further find the force loaded on the upper plate is larger than the lower plate for reason of different plate areas causing to different electric densities. That is to say their sum is positive. It is exactly the reason that an asymmetric capacitor like as lifter loaded by high voltage DC power can produce an upward lift force. Then, we can attempt to figure out the strength of the levitation force.

The forces of the top and bottom surfaces of the air block exerted by the asymmetric capacitor are different. Because of different areas S lead to different

charge densities σ , combining E = q σ and f = q E , we can find that the sum

force of the air block is not equal to zero. It gets a total trend to downward by the electric field force exerted, and makes the asymmetric capacitor to get the trend to upward which is from the large plate pointing to the small plate. So, the lift force produced by the asymmetric capacitor is

Figure 4. Air block analytical approach of asymmetric capacitor.

f = f 1 u f 2 d = f 1 d f 2 u = q E 1 q E 2 = q q ε S 1 q q ε S 2 = q 2 ε ( 1 S 1 1 S 2 ) (6)

4. Brief Analysis of Result

The results by above methods are the same. Except q, the other three variables ε , S 1 and S 2 are known. So, before we can calculate the final lift force, we must to figure out the carried charge firstly. If when

ε = 8 .854187817 × 10 12 F / m ,

S 1 [ 1.8 9 × 10 4 , 240 × 10 4 ] m 2 ,

S 2 [ 1.8 9 × 10 4 , 240 × 10 4 ] m 2 ,

q = 1.8 × 10 8 C ,

and using above formula f = q 2 ε ( 1 S 1 1 S 2 ) , we can get the lift force produced by the asymmetric capacitor that is between the maximum 0.2 N and the minimum −0.2 N. The trend of f ( S 1 , S 2 ) is drawn in Figure 5.

We can see that smaller small-plate than the large-plate means stronger force output. But too smaller dimension may lead to structural strength problem. In other works, by reference to kinds of lifters, we found that thinner copper wire can produce more lift force. The trend reflected by the formula is verified to be correct.

From Equation (1) and Figure 5, we can find the lift force f is sensitive to amount of the charges carried on asymmetric capacitor. As q = U C , the amount of the charges q is proportional to the load voltage U. Through an asymmetric capacitor in lifter form (refer Figure 6), we test the relation between load voltage U and force f, and achieve a data table (refer Table 1). This trend of force change reflects (refer Figure 7) that f is assuredly sensitive to U. As the voltage is increased, the lift force increases quickly with charge growing on the plates.

Figure 5. A lifter composed by asymmetric capacitors.

Figure 6. Test lifter of asymmetric capacitor.

Table 1. The relation table between voltage and lift force of a lifter.

Figure 7. Relative changing trend between voltage and lift force of a lifter.

5. Conclusion

Through two methods, we achieved lift force formula of asymmetric capacitor successfully. But how to obtain the value q of the asymmetric capacitor, which needs much more content to present the derivation details? So including the experiment verification of the calculating method for the lift force of asymmetric capacitor, we plan to write them in the follow-up papers. In this one, we mainly completed the lift force calculation established on some theoretical assumptions by two ways, the results by both ways point to the same form formula. It hints us that the calculation may be right, as for the exact verification of the formula, we will continue to achieve it at next papers to thoroughly resolve the lift force problem of asymmetric capacitor loaded by high voltage.


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”), the Research Activity Funding of Postdoctoral Fellow of Anhui Province (Grant No. 2018B250 “High-energy Ions Accelerated Thruster”), the Natural Science Research Project of Anhui Education Department (Grant No. KJ2018A0725 “the Uniformity Optimization and Software Development for MRI Magnets”), and a portion of this work was supported by the High Magnetic Field Laboratory of Anhui Province.

Cite this paper
Cheng, X. , Kuang, G. , Zhang, Y. , Huang, P. , Jiang, D. , Jiang, S. , Qian, X. , Ding, H. and Chen, W. (2020) Theoretical Calculation of Lift Force for Ideal Electric Asymmetric Capacitor Loaded by High Voltage. Engineering, 12, 33-40. doi: 10.4236/eng.2020.121003.

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