Theoretical Calculation of Lift Force for Ideal Electric Asymmetric Capacitor Loaded by High Voltage

Asymmetric capacitor like the so-called lifter can fly up from the ground. Some common characteristics exist in the asymmetric capacitor: high-voltage, capacitor, lift force. What are the accurate quantitative relations among them and how can we figure the lift force out? It’s a thorny problem so far. In this article, we attempt to establish a model that can match the actual experimental data and theory derivation result. After checking, the calculation result is verified to be correct.


Introduction
was performed for a kind of ideal asymmetric capacitor [1] in former paper. It looks simple and brief. But there are still a series of problems that need us to solve like the quantity of electric charges and what's the surface distribution. The formula is suitable for an ideal asymmetric capacitor and its surface electric charges are uniformly distributed. But for an irregular surface of a general asymmetric capacitor of which electric charges are nonuniformly distributed, how can we figure out the lifter force is the main problem we Engineering will further analyze. That is to say, we will make the lifter force calculation formula of asymmetric capacitor generalizations.
We derived the lift force formula of asymmetric capacitor [2] in ideal case. But in normal conditions, asymmetric capacitor isn't in an ideal regular shape (refer Figure 1). So we must expend the equation into general case suitable for practical engineering. In the next content below, we will show the lift force formula of asymmetric capacitor in general firstly, and then, the derivation process is given out.
For a normal asymmetric capacitor, after loading with a DC high-voltage on its two plates, the lift force produced on the capacitor is where f is lift force of asymmetric capacitor, q is carried electric charge of the is charge density distribution function relevant to geometric shape and location.

Derivation Process
The lift force calculation formula is shown in Equation (1 [7]. The detail derivation process is expanded as follows. Before concrete derivation, we will list some preconditions and steps.
The derivation starting point is established on these assumption conditions: 1) Electric field line is always closed. All of the Electric field lines emit from positive electrode patches, return to negative electrode patches, but can never be blocked by others neutral objects. Total charge carried on the positive electrode is ever equal to that on the negative electrode 2) For a different element area on a negative electrode, the more charge it carries, the more electric field strength it owns. In other words, the electric field strength is distributed in proportion to the negative charge it carried.
Based on these conditions, we have following corollaries.
For plate 1, there are ( ) For plate 2, there are similarly ( ) So, the electric field strength nearby plate 2 is From the first and the third assumption conditions, we known that: the total electric field lines 2 dS E emitted from plate 2, not only distribute to the area differential element 1 ds on plate 1 (which is charged with electric quantity 1 dq ), but also distribute to the rest of area (the total area corresponding to 2 ds ) on plate 1 (which is charged with total electric quantity q). Hence the area differential element 1 ds actually achieves a electric field strength The assumption of equal electric density causes the remaining charge produced on one differential element of plate 1 which electric density is actually higher than plate 2. The remaining charge So we can get the force produced by remaining charge q ∆ in the assumption electric field Combining the initial condition , we can further simplify above equation So we get the lift force formula in general form.

Affiliated Verification
A brief verification is performed by simplifying Equation (8) from general to special case. If the both plates of an ideal asymmetric capacitor are distributed with uniform charge, there are following relations.
Firstly, we can take a part area from plate 1 1 p S whose electric quantity is  ( ) So we get the distribution function of electric charge density that is Secondly, we can take a part area from plate 2 2 p S whose electric quantity is ( ) Therefore, we can finally find that the lastly corollary is as the same as the

Conclusion
From above derivation, we get the general calculation formula of lift force of asymmetric capacitor. And by further simplification, we finally find the formula in ideal case is the same as previous deduction. The general calculation method can help us to get the lift force of asymmetric capacitor precisely in any shape. It is really an exciting method which can universally solve the hard thrust calculation problem completely.