Computing capacitance in the general case

[coquelicot]

This may well be a textbook question, or even a duplicate in this forum, but I have no idea of how to compute the capacitance between two conductors in the general case: all the computations I saw in textbooks make heavy use of symmetries to compute the capacitance, e.g. capacitance between two infinite planes separated by a fixed distance d, two concentric spheres etc. But in the general case, even if there is no solution in terms of elementary functions, how to attack this problem whenever symmetries are not available ? if only a simulation is possible, what is the exact method to obtain the simulation ?
As a practical example, I would like to compute (or simulate) the exact capacitance between two rectangular planes at fixed distance d, separated by a dielectric ε (but without approximating the planes by infinite planes).

Note: The following points are know and need not be explained here:
* The potential inside and at the surfaces of the the conductors is constant
* The field inside the conductors is null
* The field at the surfaces of the conductors is normal to the surfaces
* Gauss law

EDIT: In the following link, the author says that it has searched for a formula unsuccessfully, and that he finally derived it himself (but does not explain how):
http://chemandy.com/calculators/rectangular-capacitor-calculator.htm
Please, see the comments of the author there.

 

[davenn]

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html

 

[coquelicot]

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html

I think this formula is based on the approximation by infinite planes. Even if it is not, this does not explain the general case.

 

[davenn]

I think this formula is based on the approximation by infinite planes. Even if it is not, this does not explain the general case.

well if it doesn't, then maybe you could try google yourself

 

[coquelicot]

well if it doesn't, then maybe you could try google yourself

davenn, I think you are missing the point. I've added an "edit" to my question.

 

[Carrock]

I think this formula is based on the approximation by infinite planes. Even if it is not, this does not explain the general case.

http://srjcstaff.santarosa.edu/~yat.../Fringe Field of Parallel Plate Capacitor.pdf looks useful.
I used "Fringing field of finite parallel-plate capacitors" on DuckDuckGo to find that and similar papers.

 

[coquelicot]

http://srjcstaff.santarosa.edu/~yataiiya/UNDER_GRAD_RESEARCH/Fringe Field of Parallel Plate Capacitor.pdf looks useful.
I used "Fringing field of finite parallel-plate capacitors" on DuckDuckGo to find that and similar papers.

Thank you a lot: this is rather astonishing: there seems to be few literature on this topic.

 

[Delta Kilo]

I'm afraid for arbitrary shaped conductors you would have to find the exact surface charge distribution and electric field everywhere as a byproduct of computing C.
Here is one way of doing it numerically (first that comes to mind, haven't tried it myself but there is no reason why it shouldn't work):
1. define evenly distributed points x_i on a surface of each conductor. Assume point charge q_i at each point x_i.
2. define arbitrary reference points x_j inside the conductors, at sufficient distance from the surface.
3. Compute contribution k_ij of each surface point i to the Coulomb potential at each reference point j: $k_{ij}=\frac{1}{4 \pi \epsilon_0} \frac{1}{|x_i-x_j|}$, $U_j = \sum_i k_{ij} q_i + C$
4. Add boundary conditions: Set potential of reference points to 0V and 1V, set sum of surface charges to -Q and Q respectively.
6. Solve the resulting linear system for Q

 

[coquelicot]

I'm afraid for arbitrary shaped conductors you would have to find the exact surface charge distribution and electric field everywhere as a byproduct of computing C.
Here is one way of doing it numerically (first that comes to mind, haven't tried it myself but there is no reason why it shouldn't work):
1. define evenly distributed points x_i on a surface of each conductor. Assume point charge q_i at each point x_i.
2. define arbitrary reference points x_j inside the conductors, at sufficient distance from the surface.
3. Compute contribution k_ij of each surface point i to the Coulomb potential at each reference point j: $k_{ij}=\frac{1}{4 \pi \epsilon_0} \frac{1}{|x_i-x_j|}$, $U_j = \sum_i k_{ij} q_i + C$
4. Add boundary conditions: Set potential of reference points to 0V and 1V, set sum of surface charges to -Q and Q respectively.
6. Solve the resulting linear system for Q

Thank you Delta kilo: a finite element method is what I expected. It is indeed simpler than what I thought. Nevertheless, from the theoretical point of view, I'm not yet entirely satisfied:
1) there lacks a theoretical proof that there is a unique solution, for given potentials U1 and U2 at the surfaces of each conductor, and
2) By what theoretical mean can you assert that the sum of the surface charges are respectively Q and -Q ? Furthermore, are you sure that arbitrary charge Q is consistent with arbitrary potentials U1 and U2: (suppose for example that Q is very very small, ant that the potentials are 0V and 1V).
Note: I suggest you edit your answer, because some things are irrelevant: it is known that the charges are all located at the surface of the conductors, so the points x_j are unnecessary. I would would also replace the term distributed points by "division of the surface centred at points x_i", and U_i by U1 and U2, since the potential at the surface of each conductor is constant (as you did in step 4). Finally, there are as many unknown q_i as equations, so it is conceivable that it is only necessary to fix arbitrary potentials U1 and U2 to obtain the density of charge q_i.

 

[Delta Kilo]

1) there lacks a theoretical proof that there is a unique solution, for given potentials U1 and U2 at the surfaces of each conductor

Well, since we are approximating smooth surface charge distribution with a finite number of point charges, we cannot expect the potential to be exactly the same everywhere inside the conductor, but with sufficient number of points it should be possible to get it reasonably flat. For a given number of point charges on the surface we can always add more and more points inside the conductor until the system is well and truly overdetermined and then solve it in least squares sense. Since it is a linear system, there will be a unique solution minimizing the residuals.

2) By what theoretical mean can you assert that the sum of the surface charges are respectively Q and -Q ?

It is an arbitrary choice. I just want the capacitor to be electrically neutral when seen from a distance. There is a (mutual) capacitance between conductors and there is also self-capacitance of individual conductors. Under normal circumstances, the former is orders of magnitude larger, so it shouldn't matter.

Furthermore, are you sure that arbitrary charge Q is consistent with arbitrary potentials U1 and U2: (suppose for example that Q is very very small, ant that the potentials are 0V and 1V).

Q is the unknown variable we want to find (so we can divide it by voltage to get the capacitance). Sorry I didn't make it clear. Also, when I mentioned 0V and 1V, these are defined up to a constant C. A better way of doing it would be to introduce two more variables U1 and U2 and then set U2 = U1 + 1V.

Note: I suggest you edit your answer, because some things are irrelevant: it is known that the charges are all located at the surface of the conductors, so the points x_j are unnecessary. I would would also replace the term distributed points by "division of the surface centred at points x_i", and U_i by U1 and U2, since the potential at the surface of each conductor is constant (as you did in step 4). Finally, there are as many unknown q_i as equations, so it is conceivable that it is only necessary to fix arbitrary potentials U1 and U2 to obtain the density of charge q_i.

As I said above, we are trying to approximate a potential induced by a smooth distribution of surface charges by that of a cloud of points. It works fine when the distance from the point of interest to a surface is at least several times greater than distance between charges (which is what I meant by "sufficient distance"). As you get closer, the potential becomes more and more bumpy, until it completely breaks down on the surface itself (it shoots up to infinity at x_i). That is the reason to choose reference points inside the conductor and not on a surface.

 

[coquelicot]

Well, since we are approximating smooth surface charge etc.

Thank you again for answering me Delta Kilo. For the convenience of the future reader, I reformulate here your ideas (tell me if this is OK):

At the macroscopic scale, it is well known that the charges are distributed on the surfaces of the conductors. The distribution of the charges at the surfaces of the two conductors can be approximated by continuous surface densities $\rho_1$ and $\rho_2$, that we have to determine.
To this end,
1) divide the surfaces of the conductors into a large number of small patches $P_i$ , with central point $x_i$, say $n$ patches;
2) Assume the density is approximately constant inside each patch $P_i$ , and denote it by $q_i$;
3) Compute $\alpha_{i,j} = {1\over 4\pi\varepsilon_0||x_i-x_j ||}$ for every $i,j$. At equilibrium, the potentials at each surface must be constant because if it were not, there would be a current flow. Denote the potentials at surfaces 1 and 2 by $U_1$ and $U_2$ resp. Fix them to arbitrarily values, like 0V and 1V;
3) for every $i$ such that $x_i$ belongs to surface 1, there holds $U_1 = \sum_{j\not= i} \alpha_{i,j} q_j$, and for every $i$ such that $x_i$ belongs to surface 2, $U_2 = \sum_{j\not = i} \alpha_{i,j} q_j$. This gives $n$ linear equation in $n$ unknowns $q_i$, and solving this system leads to the values of $q_i$ for the chosen potentials $U_1$ and $U_2$;
4) finally, let $Q_1 = \sum_{surface~1} q_i$ and $Q_2 = \sum_{surface~2} q_i$. There should hold (approximately, since the surface density has been approximated by patches) $Q_1 = - Q_2$ (is this exact? why?), and the capacitance between the two conductors is $C = |Q_1/(U_2-U_1)|$. It is known that the capacitance is independent of the choice of the potentials $U_1$ and $U_2$ (the electrostatic field depend only on the ddp $U_2-U_1$, hence also the distribution of charges, and multiplying the ddp by a scalar multiplies the distribution of charges by the same scalar, hence leaves C invariant).