System of multiple conductors: image method

  • #1
Granger
168
7

Homework Statement


The following figure represent the traversal cut of a system with two cylindrical equal conductors of radius r0 length l at a distance d from one another and at the same distance h of a plane conductor (conductor zero). The dielectric that surrounds the conductors is the air which maximum electric field 20 kV/cm. Consider the approximation of thin conductors. By applying the image method determine the capacitance matrix of the system
qayoF.png


Homework Equations


3. The Attempt at a Solution [/B]

So I have no idea where to start...

I now how to apply the image method to a single cylindrical conductor and a plane which would leave us to obtain:

$$ \frac{2 \pi \epsilon_0 l}{\ln (\frac{h}{r_0} + \sqrt{(\frac{h}{r_0})^2 -1})} =\frac{2 \pi \epsilon_0 l}{\ln (\frac{2h}{r_0})} $$

Where I applied the fact that the conductor is thin in the last equality.
Now how do I relate this with the second conductor.
I read that this is the solution for the capacitance between each conductor and the ground plane (so the main diagonal of the matrix). But how is it valid that we "ignore" the second conductor when we calculate the capacitance between one conductor and the ground plane? Is it because the conductors are thin?

Now what about the conductance between conductor 1 and conductor 2 (C12 and C21 on the matrix)? How do we apply the image method? Do I need to consider 4 conductors? What is the form of the equipotentials? Are they still circumferences as in the case of a single conductor and a plane?

I am really confused... Can someone give hints about how to attack the problem?
 

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Answers and Replies

  • #2
Baluncore
Science Advisor
2021 Award
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You have not specified the relative potential of conductors 1 and 2, or the relative direction of current flow.

The ground plane is a mirror. So you have a 4 wire system, two real conductors above ground, with the inverted image of those two conductors “underground”.

You might consider expanding the problem to four wires by eliminating the ground, or by placing another mirror plane half way between the two conductors which reduces it to a one wire problem.
 

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