# I Question about capacitance matrix

1. Dec 18, 2016

### mnb96

Hello,

suppose I have four conductors (1,2,3,4) and I know their mutual capacitances cij where i,j∈{1,2,3,4}. Note that the quantities cij are essentially the elements of the capacitance matrix of this system.

Now, if I apply a voltage to two conductors and leave the other two grounded (e.g. V1=V2=1 and V3=V4=0), I would like to know the capacitance between the two pairs of conductors (1,2) and (3,4).

Is it possible to derive this quantity from the sole knowledge of the capacitance matrix? If so, how?

2. Dec 18, 2016

### jasonRF

Assuming the medium between the conductors is linear, the capacitance is only a function of the geometry and the material properties (dielectric constant) not the applied voltages. The capacitance matrix simply relates the applied voltages to the charges on the conductors.

EDIT: my answer assumes you are simply setting voltages of the conductors. Is this all you are doing, or are you adding conductors (such as wires) in order to electrically connect the conductors?

Jason

3. Dec 19, 2016

### Svein

The physicist answer: Go measure it!

4. Dec 19, 2016

### mnb96

You correctly understood that all I am doing is applying voltages to the conductors: I am not adding wires to connect them. In this case, as you said, assuming the medium is linear then the capacitance is only a function of the geometry and the material. Fine, but is seems to me that the question remains: if we know the full capacitance matrix of 1,2,3,4, is it possible to directly deduce the capacitance between the pairs (1,2) and (3,4)?

From the last part of your answer, I am inclined to guess that, if the medium is not linear, then the answer is no.

5. Dec 19, 2016

### jasonRF

I guess I don't understand what you mean by
Could you explain?

6. Dec 20, 2016

### mnb96

Yes. I would like to consider the conductors 1,2 as a single entity (analogous to a capacitor plate) and the conductors 3,4 as another entity (analogous to the other capacitor plate) and calculate the capacitance between them.

The only thing I have information about is the mutual capacitances between each of the four conductors.

7. Dec 21, 2016

### Staff: Mentor

It's tempting to consider that the capacitance could be the sum of the obvious capacitances.

However, if you draw a parallel plate capacitor and insert a much larger plate between them, the tentative theory of summing two capacitances looks like not holding here because of the shielding effect of that middle plate. So I'd generalise this to say there is no general method you can apply to that limited data set.

https://www.physicsforums.com/attachments/110502.gif

8. Dec 22, 2016

### jasonRF

If you want to consider 1,2 as a single entity, what does it mean for 1,2 to have charge +Q and 3,4 to have charge -Q? Are 1 and 2 somehow free to exchange charge (without being connected!)? I still don't understand the question ... I will hang back and let others continue to answer.

9. Jan 4, 2017

### mnb96

Maybe I did not manage in first place to formulate my question properly.

Let's try with one example which slightly differs from the original post: Suppose we have a very simple circuit: one battery whose terminals are connected to the plates 1,2 of a capacitor (see figure at this link). Now, imagine that I take a pair of scissors and I make a straight cut on both plates. Plate 1 has now been split into two smaller plates (let's call them 1A and 1B). Analogously, plate 2 has been split into 2A and 2B. Note that we have now four plates.
I also assume that the two wires that connect the terminals of the battery to the plates bifurcate into 1A-1B and 2A-2B.

Now, do we agree that it makes sense to calculate the capacitance between the pairs (1A,1B) and (2A,2B). And do we agree that this capacitance will be similar (but not equal to) the capacitance between 1-2 (the original plates without the cuts)?

If yes, then we can continue the discussion, otherwise my intuition is wrong.

10. Jan 4, 2017

### jasonRF

Yep, I agree. Now I understand - since your battery wires are bifurcated the 1A and 1B are electrically connected. Perhaps it didn't matter, but I couldn't wrap my brain around the question any other way.

Anyway, if 1 and 2 in your original question are electrically connected (so can exchange charges) to form (1,2), and likewise for (3,4), then I would approach the problem as follows. First consider the 4 conductor case, for which the charges are given by,
$$\begin{eqnarray*} Q_1 & = & C_{12}(V_1-V_2) + C_{13}(V_1-V_3) + C_{14}(V_1-V_4) \\ Q_2 & = & C_{12}(V_2-V_1) + C_{23}(V_2-V_3) + C_{24}(V_2-V_4) \\ Q_3 & = & C_{13}(V_3-V_1) + C_{23}(V_3-V_2) + C_{34}(V_3-V_4) \\ Q_4 & = & C_{14}(V_4-V_1) + C_{24}(V_4-V_2) + C_{34}(V_4-V_3) \end{eqnarray*}$$
Here $C_{12}$ is the capacitance between 1 and 2, etc. Note that I have used the fact that $C_{12}=C_{21}$, etc. This formulation follows that in "Field and Wave Electromagnetics" by Cheng, but many many books cover this stuff.

Setting $V_1=V_2=V^\prime$ and $V_3=V_4=0$ I get,
$$\begin{eqnarray*} Q_1 & = & (C_{12}+ C_{13})V^\prime \\ Q_2 & = & (C_{23}+ C_{24})V^\prime \\ Q_3 & = & -(C_{13}+ C_{23})V^\prime \\ Q_4 & = & -(C_{14}+ C_{24})V^\prime \end{eqnarray*}$$
Since the charge on (1,2) is simply $Q^\prime=Q_1+Q_2$ I get
$$\begin{eqnarray*} Q^\prime & = & (C_{12}+ C_{13}+C_{23}+ C_{24})V^\prime \\ & \equiv & C^\prime V^\prime \end{eqnarray*}$$
So I get that the capacitance of the new system is $C^\prime = C_{12}+ C_{13}+C_{23}+ C_{24}$.

I also see that that $Q_3+Q_4 = -Q^\prime$, as we expect for a capacitor.

Does my explanation make sense, or did I do something screwy?

EDIT: if you draw a picture of 4 "blobs" with capacitors between them and add shorts to connect (1,2) and (3,4) the picture will show you 4 capacitors in parallel, which should add just like the formula I derived. So this seems plausible.

Jason

Last edited: Jan 4, 2017
11. Jan 8, 2017

### jasonRF

I just took another look at my post and see a typo: it should be
$$\begin{eqnarray} Q_1 & = & (C_{13}+C_{14}) V^\prime \\ C^\prime & = &C_{13}+C_{14} + C_{23}+C_{24} \end{eqnarray}$$

Also, I didn't specify that I was considering a scenario where the 4 conductors were the only things in the universe. Often these kinds of problems are modeled by essentially assuming our 4 conductors are inside a grounded spherical shell of infinite radius; if we do this then the starting equations are
$$\begin{eqnarray*} Q_1 & = & C_{01} V_1 + C_{12}(V_1-V_2) + C_{13}(V_1-V_3) + C_{14}(V_1-V_4) \\ Q_2 & = & C_{02} V_2 +C_{12}(V_2-V_1) + C_{23}(V_2-V_3) + C_{24}(V_2-V_4) \\ Q_3 & = & C_{03} V_3 +C_{13}(V_3-V_1) + C_{23}(V_3-V_2) + C_{34}(V_3-V_4) \\ Q_4 & = & C_{04} V_4 +C_{14}(V_4-V_1) + C_{24}(V_4-V_2) + C_{34}(V_4-V_3) \end{eqnarray*}$$
where $C_{01}$ is the capacitance of conductor 1 with respect to ground (which as at infinity) .

Jason

Last edited: Jan 9, 2017
12. Jan 11, 2017

### mnb96

Thanks jasonRF!

Your explanation makes sense to me and it seems to be what I was looking for.
Now I just need to spend some time to make sure I fully understood the reasoning. I am now reading the chapter on capacitors of "Field and Wave Electromagnetics" by Cheng.
I will post again in the next days.

By the way, are those four additional terms that you introduced in the last post the self-capacitances of the four conductors?

13. Jan 15, 2017

### jasonRF

Yes, the addition terms are self-capacitances - or the capacitances with respect to ground. I should probably have used the notation $C_{11}$ instead of $C_{01}$.

jason

14. Mar 1, 2017

### mnb96

Hello,

after working on this problem, I would have an additional question related to it.

Let's consider a similar scenario where we have four conductors such that V1=V2=1 and V3=0 (as in the original post), but now the 4th conductors is "floating" instead of being grounded (i.e. it is physically disconnected from the rest of the circuit).

If we consider the following equations:

How do we treat floating conductors?
Should we simply set Ci4=C4i=0 in the above matrix?