# Electromagnetism - co-axial

Discussion in 'Homework Help' started by smarch, Jun 30, 2010.

1. ### smarch Thread Starter Active Member

Mar 14, 2009
52
0

A co-axial conductor comprises a high voltage central wire, a layer of oil with
a relative dielectric constant of 3, an insulator layer with a relative dielectric
constant of 5 and an outer earthed sheath (see below for the various
dimensions). The breakdown field for the oil is 2.8 MVm​
-1 and the
breakdown field for the insulator is 4.2 MVm-1. Calculate the maximum​

voltage that can be applied between the central wire and the outer sheath.

r1=2mm
r2=5mm
r3=10mm

Region 1 (
r=0-r1)
Region 2 (r=r1-r2)​

Region 3 (r=r2-r3)

Am I right to say that I have to use gauss's law?

2. ### steveb Senior Member

Jul 3, 2008
2,433
469
Gauss' Law is useful in this problem. It could also be useful to think of deriving the capacitance of this structure.

If you have trouble, start with a simpler version of the problem in which there is only one dielectric material inside. The capacitance of a coaxial line with one dielectric is well known, and the derivation can be looked up in your book or online. Once you know the approach, the real problem will be easier.

3. ### smarch Thread Starter Active Member

Mar 14, 2009
52
0

So far I have come up with what is attached to this post.
Is it correct? If not can you point me in the correct direction?

thanks

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4. ### steveb Senior Member

Jul 3, 2008
2,433
469
I could be misinterpreting what you are doing here, but it doesn't look correct to me. I'm not sure why you have different sigmas (1 and 2 for example). I assume these sigmas are charge per unit length along the coaxial line. There will only be one value of charge per unit length.

The approach I would take is to derive the capacitance per unit length which relates voltage to charge per unit length. Then apply Gauss's law which relates charge per unit length to electric field. This way you can derive the field versus voltage in each region. Then you need to think about which region is going to break down at the lower voltage.

EDIT: By the way, if you can, try to work out the final numerical answer. This allows others to check more easily. We might use different approaches that are both correct and give the same answer.

Last edited: Jun 30, 2010
5. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
The sigma values will be the charge density per unit length and these decrease with increasing radii.

It seems to me that the most likely position for maximum electric field strength would be at the inner radius - r1. Also given the oil has the lower breakdown strength, the point which is the weakest and therefore the limiting condition for the cable voltage rating, is reasonably obvious.

It's an interesting problem.

6. ### steveb Senior Member

Jul 3, 2008
2,433
469
I'm not sure I follow you here, and I'm still confused on his solution. Sigma isn't charge density per unit length, but is charge per unit length (i.e. total charge per unit axial length). It is basically "linear charge density". This terminology gets confusing for sure. The charge only exists on the surface of the conductors and has cylindrical symmetry. So, the only radii (a very thin cylindrical shell) that have charge are r1 and r3, and the linear charge density is the same on both conductors. In his solution, he has calculated charge per unit length at r2, which doesn't seem meaningful to me. (EDIT: acutally looking again, maybe he's not doing that, but I still can't follow his logic)

For the OP's benefit, one way to avoid this confusion about charge density is to just assume a 1 meter length (assuming you are using SI units) with a fixed charge Q. This way charge Q and linear charge density (sigma) are basically the same.

The basic thing here is that Gauss' Law only cares about the total net charge within a surface, and the charge per unit (axial) length on the inner conductor allows calculation of the field inside the shield. As you know, outside the shield, the equal and opposite charge on the outer conductor cancels that on the inner conductor and the field is zero outside.

I think I agree with you here. I haven't solved the problem myself, but your logic seems sound.

Last edited: Jul 2, 2010
7. ### smarch Thread Starter Active Member

Mar 14, 2009
52
0
To be hoenst I'm not sure where I was going with it.

I have attached a JPEG of some help I recieved from someone else, though I am still quite confused by it, it does not show the full question, but apparently it should be easy to calculate the rest of the question from what is there.

Can anyone shed some light on it?

File size:
82.1 KB
Views:
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8. ### steveb Senior Member

Jul 3, 2008
2,433
469
That looks correct to me. So, now you have the capacitance per unit length for the structure. Remembering that Q=CV, you can use Gauss' Law again to find electric field in terms of applied voltage. Than use T_N_K's advice from post #5, and you should arrive at the correct answer.

9. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
As a check I worked an equation for Vmax based on the maximum expected field strength occurring at radius r1 - as follows ...

$Vmax=\frac{r_1}{\epsilon_3}$\epsilon_2 ln(\frac{r_3}{r_2})+\epsilon_3 ln(\frac{r_2}{r_1})$E_{2max}$

Given ....

$\epsilon_2=3$

$\epsilon_3=5$

$E_{2max}=2.8MV \ per \ metre$

$\epsilon_0 \ term \ cancels \ out$

with r1, r2, & r3 as given but expressed in metres

10. ### smarch Thread Starter Active Member

Mar 14, 2009
52
0
I can understand up to there, though where do I get the length from to calculate the capacitance?

11. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
As Steveb suggested treat everything with respect to unit length. So just assume a cable length of 1 metre and the length term in any particular equation is taken care of. Whether the cable is 1 metre or 100metres in length will have no bearing on the maximum safe operating voltage.

12. ### smarch Thread Starter Active Member

Mar 14, 2009
52
0
Oh I get you, cheers!