2
\$\begingroup\$

I would like to have a rough estimate of the capacitance per unit length of a system composed by N parallel identical straight wires. I tried to sketch the capacitance network but I'm able to solve it only up to 4 wires... Any idea how to smplify the problem?

enter image description here

\$\endgroup\$
  • \$\begingroup\$ Is the distance far enough that we can approximate the wires as 'flat'? \$\endgroup\$ – Oldfart Jun 16 '18 at 8:35
  • \$\begingroup\$ @Oldfart: if this can help, sure. At this stage I would like just to know the equivalent capacitance; the physical implementation would be a second step, but you are more than welcome to already take this aspect in consideration . \$\endgroup\$ – Andrea Jun 16 '18 at 8:44
  • \$\begingroup\$ Sorry, second question which I should have asked at the same time. The LHS is V, RHS is GND, the middle wires are connected to what? \$\endgroup\$ – Oldfart Jun 16 '18 at 8:48
  • \$\begingroup\$ @Oldfart: still, the whole picture is more complicated than this... If you are interested in the voltage at each point we can say that it is 2V/3 on the first wire, V/3 on the second one. But to solve the capacitance network you don't really need to know all these details right? (I really appreciate your effort) \$\endgroup\$ – Andrea Jun 16 '18 at 8:54
  • \$\begingroup\$ @Oldfart: if you really want you can see the entire system as a meander line \$\endgroup\$ – Andrea Jun 16 '18 at 9:07
0
\$\begingroup\$

It has been a loong time since I did basic electricity and fields.

The simplest representation of a capacitor is two flat electric plate at a distance. There will be an electric field between these plates:
enter image description here

The electric field starts at potential V and, with a constant dielectric, will linear with distance go to 0V (Gnd). This means that in the middle you find a potential of V/2.

If we place a very thin electric plate at 1/3d from V, the potential of the plate will then be 2/3V:

enter image description here

If you would externally set a potential of 2/3 V on that plate the electric field, and thus the capacitance would be the same.

This means that all your wires in the middle do not make a difference and can be ignored.

But to solve the capacitance network you don't really need to know all these details right?

As you can see, the voltage on those line places a crucial role.

\$\endgroup\$
  • \$\begingroup\$ Let us call C1 the capacitance between the first neighbours, C2 the capacitance between the second ones and so one, what is the value of the total capacitanc? Cn? \$\endgroup\$ – Andrea Jun 16 '18 at 10:57
  • \$\begingroup\$ Please see my answer: This means that all your wires in the middle do not make a difference and can be ignored. \$\endgroup\$ – Oldfart Jun 16 '18 at 10:59
  • \$\begingroup\$ sorry but why solving the lumped element circuit Ctot=C3+2(C1C2/(C1+C2))? \$\endgroup\$ – Andrea Jun 16 '18 at 11:04
  • \$\begingroup\$ Sorry, but there is nothing more I can do. For some reason you keep going into a direction which is wrong. There is no lumped capacitance, only the capacitance between the outer wires counts. (As @Andyaka also stated) \$\endgroup\$ – Oldfart Jun 16 '18 at 11:15
  • \$\begingroup\$ does the same (i.e. the field doesn't change) happen even if you substitute the "parallel plate capacitor" approach with wires with circular cross section? \$\endgroup\$ – Andrea Jun 16 '18 at 12:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.