Capacitance Calculation for Dielectric-Filled Coaxial Cable

[CAF123]

Homework Statement

A long cable consists of two coaxial conducting cylindrical shells of radius b and 3b. The region with radius $\delta$ between b and 2b is filled with a material of relative permittivity $\epsilon_r \neq 1$ and relative permeability $\mu_r = 1$; the remaining space between the cylinders is empty.
(a) Suppose the cable carries charge per unit length $\pm \gamma$ on the inner and outer cylinders. Find $\underline{D}$ and $\underline{E}$ everywhere within it, where $\underline{D}$ is the electric displacement vector.
(ii) Hence calculate the potential difference between the inner and outer shell, and obtain an expression for the capacitance per unit length of the cable.

Homework Equations

Gauss' Law, $\underline{D} = \epsilon_o \epsilon_r \underline{E}$, potential on surface of inner cylinder is $-\int_{\infty}^{b} \underline{E} \cdot$d$\underline{r}$. Similarly for other cylinder.

The Attempt at a Solution

(a)Because of the dielectric in the space $b < r < 2b$, the E field in the material will go down by a factor of $\epsilon_r$. In the space $2b < r < 3b$, the E field is unchanged. So the E field for the region [b,2b] is $$\underline{E_2} = \frac{1}{\epsilon_r} \frac{\gamma}{2 \pi \epsilon_o r_1} \underline{e}_r $$ and that in [2b,3b] is $$\underline{E_1} = \frac{\gamma}{2 \pi \epsilon_o r_2}\underline{e}_r,$$where r1 between b and 2b and r2 between 2b and 3b. $\underline{D}$ is then these expressions multiplied by $\epsilon_r$.
(b) Potential diff = potential at inner cylinder - potential at outer cylinder:
Consider potential at inner cylinder first: $$V_{inner} = -\int_{\infty}^{b} = -\int_{3b}^{\delta}\underline{E} \cdot d\underline{r} - \int_{\delta}^{b}\underline{E} \cdot d\underline{r}$$ where E1 and E2 are the electric fields in the non-dielectric and dielectric areas respectively. Subbing in, I get $$V_{inner} = \frac{\gamma}{2\pi \epsilon_0} \left(\ln\left(\delta^{1/\epsilon_r -1} \right) + \ln(3) \right)$$ The potential of the outer cylinder is $$V_{outer} = -\int_{\infty}^{3b} \underline{E} \cdot d\underline{r} = 0$$ since E is zero outside the cylinder. Hence $\Delta V = V_{inner}$

The capacitance/length = γ/ΔV. Is it okay? Many thanks.

 

[rude man]

(a)Because of the dielectric in the space $b < r < 2b$, the E field in the material will go down by a factor of $\epsilon_r$. In the space $2b < r < 3b$, the E field is unchanged. So the E field for the region [b,2b] is $$\underline{E_2} = \frac{1}{\epsilon_r} \frac{\gamma}{2 \pi \epsilon_o r_1} \underline{e}_r $$ and that in [2b,3b] is $$\underline{E_1} = \frac{\gamma}{2 \pi \epsilon_o r_2}\underline{e}_r,$$where r1 between b and 2b and r2 between 2b and 3b. $\underline{D}$ is then these expressions multiplied by $\epsilon_r$.

Look again at your equation for D. Start with Gauss' theorem.

Your answer for D (or εE) is dimensionally incorrect, so you know it can't be right.

 

[CAF123]

Look again at your equation for D. Start with Gauss' theorem.

Your answer for D (or εE) is dimensionally incorrect, so you know it can't be right.

Yes, I meant to say D is E multiplied by $\epsilon_r \epsilon_0$. In the region [b,2b], I obtain $\underline{D} = \frac{\gamma}{2 \pi r_1}\underline{e}_r$ for b<r1<2b. I am not really sure how to obtain the D vector in the region [2b,3b] using Gauss (i.e how to define the Gaussian surface appropriately). Using D=ε_rε_oE, I get that D = ε_rγ/2πr₂.

 

[rude man]

Sorry, I thought gamma was a surface charge. Your expressions for E are all correct.

Oops, your D is not.

The nice thing about using D is that D is continuous across the vacuum-dielectric layer. So I suggest you always use D, then convert to E at the last minute, with E = D/ε.

 

[CAF123]

What is Gauss' theorem? Use D but you can use E if you insist, just make sure you know which epsilon to use in that case.

In media, it states $$\oint_S \underline{D} \cdot d\underline{S} = (Q_f)_{enc}$$ with $Q_f$ the free charge (in this case $\gamma l$, l the length of the cylinder.)

 

[CAF123]

Sorry, I thought gamma was a surface charge. Your expressions for E are all correct.

Oops, your D is not.

The nice thing about using D is that D is continuous across the vacuum-dielectric layer. So I suggest you always use D, then convert to E at the last minute, with E = D/ε.

Should that not be E = D/ε_oε_r (**)? In the region [b,2b], I get D = γ/2πr₁ and in [2b,3b] I get D = ε_rγ/2πr₂. This is just by applying the formula (**). I managed to get the eqn for D in the region [b,2b] via Gauss thm as well, but not the D vector in [2b,3b]. I am not sure how to define my Gauss surface in this case.

 

[rude man]

In media, it states $$\oint_S \underline{D} \cdot d\underline{S} = (Q_f)_{enc}$$ with $Q_f$ the free charge (in this case $\gamma l$, l the length of the cylinder.)

Right. Sorry again.

 

[rude man]

Should that not be E = D/ε_oε_r (**)?

Same thing. ε = ε_rε₀.

... and in [2b,3b] I get D = ε_rγ/2πr₂. This is just by applying the formula (**).

You are misapplying the formula. Take out the ε_r.

Again: use D and you won't get into this kind of trouble. Change to E at the end:
E = D/ε where ε = ε_rε₀
and where ε_r = 1 in vacuo and >1 in a dielectric.

I notice you assumed a unit length (1m) for your Gaussian cylinders. That is OK but I prefer to assume a length L, then you can do dimensional checks on all your terms.

 

[CAF123]

Again: use D and you won't get into this kind of trouble. Change to E at the end:
E = D/ε where ε = ε_rε₀
and where ε_r = 1 in vacuo and >1 in a dielectric.

I don't think I see it yet - given that I have already calculated E let's find D. $D = εE,$ where $E = \frac{\gamma}{2 \pi \epsilon_o r_2}\,\,, 2b<r_2 < 3b$. Mulitplying this by $\epsilon$ gives $D = \frac{\epsilon_r \gamma}{2 \pi r_2}$, no?
EDIT: $\epsilon_r = 1$ in vacuo so remove the $\epsilon_r$, I see your point. Thanks.

I notice you assumed a unit length (1m) for your Gaussian cylinders.

How so? When I found the E fields via the microscopic form of Gauss, $E(2\pi r l) = \gamma l$ and the l cancelled.

 

[CAF123]

Do you agree with the rest of the OP? In particular, my calculations of V and C?

 

[rude man]

I don't think I see it yet - given that I have already calculated E let's find D. $D = εE,$ where $E = \frac{\gamma}{2 \pi \epsilon_o r_2}\,\,, 2b<r_2 < 3b$. Mulitplying this by $\epsilon$ gives $D = \frac{\epsilon_r \gamma}{2 \pi r_2}$, no?
EDIT: $\epsilon_r = 1$ in vacuo so remove the $\epsilon_r$, I see your point. Thanks.

OK. Don't find E and then D, do it the other way around. Your post #5 says it all. It's applicable to all regions.

How so? When I found the E fields via the microscopic form of Gauss, $E(2\pi r l) = \gamma l$ and the l cancelled.

Yes. I didn't see your canceling the l's. Good.

 

[CAF123]

Hi rude man,
Did you see #10?

 

[rude man]

Hi rude man,
Did you see #10?

So4rry, did not.

How on Earth did you come up with the first term below:

Subbing in, I get $$V_{inner} = \frac{\gamma}{2\pi \epsilon_0} \left(\ln\left(\delta^{1/\epsilon_r -1} \right) + \ln(3) \right)$$.

δ doesn't even belong in that expression. You're integrating from b to 2b. The two integrations give expressions that are entirely similar, just different constants involved.

The second term is also not right.

Check your limits of integration. They are very simple functions of b and do not include δ. The way the problem is stated, δ is a variable radius lying in the dielectric region: b < δ < 2b.

 

[CAF123]

Okay, that makes sense and with these corrections I obtain the potential difference between the two cylinders is $$P.D = \frac{\gamma}{2 \pi \epsilon_o} \left(\ln \left(\frac{3}{2}\right) + \frac{\ln(2)}{\epsilon_r}\right) = \frac{\gamma}{2 \pi \epsilon_o} \left( \ln \left(2^{1/\epsilon_r - 1}\right) + \ln(3)\right)$$

 

[rude man]

Okay, that makes sense and with these corrections I obtain the potential difference between the two cylinders is $$P.D = \frac{\gamma}{2 \pi \epsilon_o} \left(\ln \left(\frac{3}{2}\right) + \frac{\ln(2)}{\epsilon_r}\right) = \frac{\gamma}{2 \pi \epsilon_o} \left( \ln \left(2^{1/\epsilon_r - 1}\right) + \ln(3)\right)$$

Correct.

 

[CAF123]

Is it okay to say the capacitance/length is Q/lΔV = γ/ΔV? I know that the capacitance will be increased because of the dielectric but this will not be by a factor of ε_r.

 

[rude man]

Is it okay to say the capacitance/length is Q/lΔV = γ/ΔV? I know that the capacitance will be increased because of the dielectric but this will not be by a factor of ε_r.

Yes.

 

[CAF123]

Many thanks, the last part asks to compute the Poynting vector giving reasons for the direction. It is clear that the Poynting vector points in the $\underline{e}_z$ direction if the E field is in $\underline{e}_r$ and B in $\underline{e}_{\phi}$. I think this makes sense - the set up is remiscent of a cable and data flow/light/ would travel in this direction, along the wire.
Is this the comment to be made?

Also, I don't know whether it makes sense to define two Poynting vectors here. The E field inside and outside the dielectric is different and so I am not sure what one to use in $\underline{S} = 1/\mu_o \underline{E} \times \underline{B}$

 

[rude man]

Many thanks, the last part asks to compute the Poynting vector giving reasons for the direction. It is clear that the Poynting vector points in the $\underline{e}_z$ direction if the E field is in $\underline{e}_r$ and B in $\underline{e}_{\phi}$. I think this makes sense - the set up is remiscent of a cable and data flow/light/ would travel in this direction, along the wire.
Is this the comment to be made?

Also, I don't know whether it makes sense to define two Poynting vectors here. The E field inside and outside the dielectric is different and so I am not sure what one to use in $\underline{S} = 1/\mu_o \underline{E} \times \underline{B}$

CAF, put this into the right thread. I need to review the prior posts.

rude man

EDIT: never mind, I guess it's the right thread, but please provide the wording for this ensuing question.
EDIT EDIT: never mind that too!

 

[rude man]

Many thanks, the last part asks to compute the Poynting vector giving reasons for the direction. It is clear that the Poynting vector points in the $\underline{e}_z$ direction if the E field is in $\underline{e}_r$ and B in $\underline{e}_{\phi}$. I think this makes sense - the set up is remiscent of a cable and data flow/light/ would travel in this direction, along the wire.
Is this the comment to be made?
Also, I don't know whether it makes sense to define two Poynting vectors here. The E field inside and outside the dielectric is different and so I am not sure what one to use in $\underline{S} = 1/\mu_o \underline{E} \times \underline{B}$

[/quote]

The Poynting vector always points in the direction of energy flow. So if the cable is end-excited by a time-varying voltage then the Poynting vector points (mostly) in the z direction outside the cable.

Inside the cable conductors the situation is different. The Poynting vector actually points radially from the outside towards the axis, representing the diminishing flow of heat energy dissipated within the conductor. This is kind of an advanced concept so I would just say that the Poynting vector points basically along the z axis. If the conductivities are infinite then there is no radial internal component and the Poynting vector points exactly along the z axis.

 

[CAF123]

The question is: The cable forms a circuit in which a current $i$ flows down the inner cylinder and back up the outer cylinder. Find the Poynting vector, giving insight into its direction.

How would I find the magnitude of this vector though? I have two regions [b,2b] and [2b,3b] each with a different E-field.
Thanks.

 

[rude man]

Sorry if I confused you with the last couple of messages.

But you haven't described how current is generated in the cable. No current, no poynting vector. Were you given a time-varying current?

There is a small radial component of the poynting vector (points radially towards the cable axis) due to disspiative forces. This component is identically zero if the conductivities are infinite.

Anyway, yes, there are really two separate components to the poynting vector. Just add them. I'm wondering how you will determine the E and H fields within the two regions.

 

[rude man]

we're playing message tag here I guess!

OK, a steady current i? So there's a length of cable with a resistor R at the far end and a dc voltage V at the near end, producing a current i = V/R.

You know E in both regions since you've placed voltage V between the inner & outer conductors. That gives you the radial E(r) as you say. Then, H(r) is circular about the axis, again as you say. H is unaffected by differences in ε and you can use Ampere's law to get it as a function of r. Then P = E x H will tell you the direction and magnitude. Magnitude will vary with r. You have to integrate P over the entire cross-section of the cable to get the total energy flow rate (power). P has units of watts/sq. m.

 

[CAF123]

Sorry if I confused you with the last couple of messages.

But you haven't described how current is generated in the cable. No current, no poynting vector. Were you given a time-varying current?

There is a small radial component of the poynting vector (points radially towards the cable axis) due to disspiative forces. This component is identically zero if the conductivities are infinite.

Anyway, yes, there are really two separate components to the poynting vector. Just add them. I'm wondering how you will determine the E and H fields within the two regions.

I think our posts have passed each other again. There is no indication of a time varying current. I already know E from the first part and B is given by Amperes' Law in vacuum.

In the dielectric, $\underline{E} = \frac{1}{\epsilon_r} \frac{\gamma}{2 \pi \epsilon_o r_1} \underline{e}_r$ and $\underline{B} = \frac{\mu_o I}{2 \pi r_1} \underline{e}_{\phi}$ so $$\underline{S} = \frac{1}{\mu_o} \frac{1}{\epsilon_r} \frac{\gamma \cdot \mu_o \cdot I}{4 \pi^2\epsilon_o r_1^2} \underline{e}_z$$ Similarly for the other region but without the factor of 1/ε _r in the E field and using r2 in place of of r1.
EDIT: It's happened again - sorry!

 

[rude man]

No sweat. You have the right idea.

You should state that your formulas for P are valid for b< r1 < 2b in the dielectric and 2b < r2 < 3b in the air.

Did they ask you for power flow?

 

[CAF123]

No sweat. You have the right idea.
You should state that your formulas for P are valid for b< r1 < 2b in the dielectric and 2b < r2 < 3b in the air.

So I should specify two Poynting vectors - one for vacuum and one for dielectric? The wording of the question seems to hint a single vector '...Find the Poynting vector...'.

Did they ask you for power flow?

No, but they then ask about the flux through some current loop and consequently the self inductance, but I think I have this. I may have some conceptual questions about this later though.

 

[rude man]

So I should specify two Poynting vectors - one for vacuum and one for dielectric? The wording of the question seems to hint a single vector '...Find the Poynting vector...'.


No, but they then ask about the flux through some current loop and consequently the self inductance, but I think I have this. I may have some conceptual questions about this later though.

There is really only one poynting vector. It's just P = E x H. It's not a constant even within one zone. E and H both roll off with r as you have found. So P rolls off as 1/r². There is really only one r. We called it r₁ in one region and r₂ in the other, but that is not necessary. In other words, the P vector assumes different magnitudes, depending on r.

 

[CAF123]

There is really only one poynting vector. It's just P = E x H. It's not a constant even within one zone. E and H both roll off with r as you have found. So P rolls off as 1/r². There is really only one r. We called it r₁ in one region and r₂ in the other, but that is not necessary. In other words, the P vector assumes different magnitudes, depending on r.

This makes sense - so in fact E, D and P are all piece wise defined then such that as we move across the boundary of air/dielectric, all quantities remain continuous. Since quantities are within the same direction over the boundary, we already know that E,D and P will be continuous.

 

[rude man]

This makes sense - so in fact E, D and P are all piece wise defined then such that as we move across the boundary of air/dielectric, all quantities remain continuous.

H and D are continuous across the dielectric-air boundary but E changes abruptly from D/ε to D/ε₀ as we go across the r=2b boundary. Same for P of course since P = E x H.

 

[CAF123]

H and D are continuous across the dielectric-air boundary but E changes abruptly from D/ε to D/ε₀ as we go across the r=2b boundary.

Would this not only affect the magnitude, rather than the direction of the vector? (Since in both regions, we found that all quantities were in the same direction, I.e E was in radial direction inside the dielectric and in the radial direction outside. Similarly for B in the phi direction.)

 

[rude man]

Yes.

E is always radial, H is always along phi (in cylindrical coordinates) so P is always along z. I meant that the magnitudes change, not the direction.

 

[CAF123]

Yes.

E is always radial, H is always along phi (in cylindrical coordinates) so P is always along z. I meant that the magnitudes change, not the direction.

Many thanks for all your help. Last question: in the first part, if we were to do as you suggest and find D first (which I think makes more sense) rather than E, then how do you know that D is radial? Do we assume that the dielectric material has a linear susceptibility and hence that the D field inside the dielectric is related to the E field in a vacuum by the relation D = εE?

 

[rude man]

Many thanks for all your help. Last question: in the first part, if we were to do as you suggest and find D first (which I think makes more sense) rather than E, then how do you know that D is radial? Do we assume that the dielectric material has a linear susceptibility and hence that the D field inside the dielectric is related to the E field in a vacuum by the relation D = εE?

D is actually the sum of two vectors, one of which is ε₀E. The other is the so-called polarization vector P. P is associated with polarized charges only. In an isotropic medium, where a single ε_r can be assigned (i.e. regardless of direction), E and P point in the same direction. In an introductory course you are not likely to encounter non-isotropic dielectrics for which P can point in a different direction than E, and can depend on location and/or direction.