How to Calculate V3 or Q3 in a Capacitance Network?

hectoryx
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Homework Statement



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


Homework Equations



the net charge on each object will be:

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


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}?

Thanks in advance.

Regards

Hector
 
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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.
 
Thanks for your reply.

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