# Potential in capacitance network

1. Sep 23, 2009

### hectoryx

1. The problem statement, all variables and given/known data

A capacitance matrix represents the charge coupling within a group of conductors — that is, the relationship between charges and voltages for the conductors. Given the three conductors shown in the following link, with the outside boundary taken as a reference,

http://i1021.photobucket.com/albums/af335/hectoryx/capacitancenetwork.jpg

2. Relevant equations

the net charge on each object will be:

$$\begin{array}{l} {Q_1} = {C_{10}}{V_1} + {C_{12}}({V_1} - {V_2}) + {C_{13}}({V_1} - {V_3}) \\ {Q_2} = {C_{20}}{V_2} + {C_{12}}({V_2} - {V_1}) + {C_{23}}({V_2} - {V_3}) \\ {Q_3} = {C_{30}}{V_3} + {C_{13}}({V_3} - {V_1}) + {C_{23}}({V_3} - {V_2}) \\ \end{array}$$

3. The attempt at a solution

The problem is, if all of the capacitances are already known, and $${V_1} - {V_2} = U$$ is also known, but $${V_1}$$ and $${V_2}$$ is not known.

then how to calculate $${V_3}$$ or $${Q_3}$$?

Regards

Hector

2. Sep 24, 2009

### turin

I don't believe that it is possible. You have six unknowns, but only four equations. Do you know any other values besides just V1-V2? Two more independent values should do it.

3. Sep 25, 2009

### hectoryx

However, in the equivalent circuit of the capacitor network, the voltage of conductor 3 to the reference ground can be caculated.......so...

4. Sep 25, 2009

### turin

Are you suggesting that this is possible if you only know V1-V2? I don't believe so. If you simply mean that, yes, there is this additional input information, then the solution is trivial: V3=V3, right? I must not be understanding your problem.