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Hi,

actually I would expect the COMSOL result and not your analytical calculation. The potential is obviously continuous at the internal boundary, but with a kink,which is to be expected, because the conductivity changes. And thus the field is discontinuous.

Cheers
Edgar

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Edgar J. Kaiser
emPhys Physical Technology
www.emphys.com

Dear Edgar,

Thank you for your reply.

However, I can not agree with you on this matter, considering the following:
- the conductivity, as its constant in regards to space and time (does not depend on "r" or on "t"), does not interfere in the expression of potential V, nor in the expression of field intensity E.

I may be wrong, but can you please correct me?
The regime I am using (electric currents, stationary) isn't governed by these expressions? :
1. E = - grad V
2. div J = 0
3. J = sigma * E

And that my 2D geometry has symmetry and can be considered as a 1D geometry with variations of just "r", my expression: [ div ( sigma * ( - grad V ) ) = 0 ] turns into ( in cylindrical coordinates, with variations just on the "r" axis ):

( 1 / r ) * [ (d / dr) * r * sigma * [ ( d/dr) * - V ] ] = 0.


if I am considering that expression and the expressions I've mentioned above in the post regarding properties of material and values of expressions at geometric borders, more specifically:
i) V_domain1 ( r = r_min ) = V_potential
ii) V_domain2 ( r = r_max ) = V_ground
iii) V_domain1 (r = r_med ) = V_domain2 ( r = r_med)
iv) E_domain1 (r = r_med ) = E_domain2 ( r = r_med)
with r_med being the radii of the border separating the two domains within the ring
( from 0.02 m to 0.06m the first domain, and from 0.06 to 0.10 m the 2nd domain )

if I integrate considering all of the above I will have:

( 1 / r ) * [ (d / dr) * r * sigma * [ ( d/dr) * - V ] ] = 0
=> [ d/dr * r * sigma * [ (d/dr) - V(r) ] ] = 0 ( simplifications by 1/r as the expression in the right equals to 0 )
=> [ d/dr * r * sigma * [ (d/dr) V(r) ] ] = 0 ( simplifications by a minus sign)
=> r * sigma * [ d/dr V(r) ] = - ConstantA ( integrating the previous expression, on the left removing the d/dr and on the right integrating leaves a constant )
=> [ d/dr V(r) ] = ConstantA / ( r*sigma ) ( defining d/dr V(r) as whatever else remains)
And considering that E(r) equals to minus d/dr V(r) then we'll have E(r) equals to minus the expression on the right.
Then, if we integrate another time, we'll have V(r) as
=> V(r) = ConstantA * log(r) /sigma + ConstantB

Thus, I'll be left with:
E_domain1(r) = - Constant1 / (sigma_domain1 * r)
V_domain1(r) = Constant1 * log(r) / sigma_domain1 + Constant2

E_domain2(r) = - Constant3 / (sigma_domain2 * r)
V_domain2(r) = Constant3 * log(r) / sigma_domain2 + Constant4

With the expressions of the constants as following: (deduced from the system of equations i)- iv) )
Constant1 = (V_potential - V_ground) * sigma_domain1 / log(r_min / r_max)
Constant2 = V_potential - (V_potential - V_ground) * log(r_min) / log (r_min / r_max)
Constant3 = (V_potential - V_ground) * sigma_domain2 / log (r_min / r_max )
Constant4 = V_ground - (V_potential - V_ground) * log(r_max) / log(r_min / r_max)

And introducing the constants in the expressions of E and V I will be left with:

V_domain1(r) = V_potential + (V_potential - V_ground) * log(r / r_min) / log(r_min / r_max)
V_domain2(r) = V_ground + (V_potential - V_ground) * log(r / r_max) / log(r_min / r_max)

E_domain1(r) = E_domain2(r) = - (V_potential - V_ground) / ( r * log (r_min / r_max) )


Thus, as the material properties aren't varying with coordinates they will not interfere in this regime and I should have a smooth potential curve from one side to the other of the ring, governed only by the potentials of the two extremities, the fraction between the r_min and r_max and by the "r" of the point where the potential is to be found.

Are you saying that Comsol is using other expressions, and not these that I have mentioned above?
Or are you saying that I may be mistaking the regime that I am using for another one, and that my analytically expressions above aren't the correct expressions for this regime?

Thank you for your answer again, and hopefully you (or someone else) can provide an additional answer for what I wrote in this message.

But the conductivity is different in the two domains, so it is dependent on a spatial coordinate. A simple analogy would be a serial circuit of 2 resistors. Say you have 1 Ohm and 9 Ohm in series and 1V attached to the terminals. This will result in 0.1 V and 0.9 V voltage drop.

Or maybe I misunderstand you setup completely,

Cheers
Edgar

--
Edgar J. Kaiser
emPhys Physical Technology
www.emphys.com

Your equation iv should read
iv) J_domain1 (r = r_med ) = J_domain2 ( r = r_med)
not
iv) E_domain1 (r = r_med ) = E_domain2 ( r = r_med)
Jeff

Ahh, that explains some things.

Thank you very much for the answers!