# Electric Flux Density for Coaxial Cable

Discussion in 'Homework Help' started by tquiva, Oct 30, 2011.

1. ### tquiva Thread Starter Member

Oct 19, 2010
176
1
I am given the following problem:

And here is my attempt:

Basically, my answer to the problem is not the same as the book's answer, and I'm pretty sure I'm using the right equations. What could I possibly be doing wrong here? I've been trying to figure this out for three hours now, and am now frustrated. I apologize for my ignorance here, but any advice or tips will be greatly appreciated.

Thank you

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Last edited: Oct 30, 2011
2. ### tquiva Thread Starter Member

Oct 19, 2010
176
1
can someone please help me if they know how to do this? I'm still trying to figure it out and am now close to giving up...

3. ### juve786 Member

Jul 13, 2010
27
0
This is quite hard to explain without a diagram. But you want to look on the internet for the electric field of line charge and there should be something to help, hopefully with a diagram.

To narrow what you're looking for:

The E-field at a point outside a wire (looking at it sideways) will not just be from the charge distributed from the nearest point, but from each tiny segment of the entire wire at various angles.

Say for a wire of length x with a uniform distribution of ρ, each tiny segment of the wire of length ∂x will have a charge of ρ∂x. Using the formula for E, each little segment will have an E-field of ∂E in the direction of the angle from the segment of wire to the point (say distance y perpendicular from the wire) you are evaluating. Each length/distance you use in that formula should be obtained from Pythagoras' Theorem - √(x^2 + y^2). You want the vertical E-field, so taking the vertical component - ∂ESinθ will give you the E-Field you are looking for from one tiny segment of the wire.
Integrate to get the vertical E-Field you are looking for.

∂E should equal 1/[(4πε)(x^2+y^2)] * sinθ
Using SOH/CAH/TOA - sinθ can be re-written as y/√(x^2+y^2).

Substitute and integrate to get a final value of ρ/(2πεy) --> note the difference to the books answer. This should be the answer where y = ρ(distance) and ρ = ρe(charge density). (SORRY FOR THE SYMBOL SWAPPING! - dont mix them up! )

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Have you learnt about Gauss' Law? This problem will be far far easier to solve using that.

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
789
Hi tquiva,

I haven't thought this through entirely but ....

I would think the electric field strength [V/m] would change with the simple inverse of the radial distance ρ (rather than the inverse square of ρ) from the inner conductor surface - and where the maximum field strength occurs at the inner conductor surface. Not sure what your formula is suggesting as it doesn't contain any reference to the radial distance - just the radial unit vector.

So there's another option.

I'd be surprised if the E field strength increased with radial distance - which is what the book solution (as supplied) suggests.

Last edited: Jan 17, 2012
5. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
789
Some ongoing thoughts ...

The question is a little ambiguous particularly with reference to the charge density distribution parameter $\rho_e$ - what does $\rho_e$ represent and what are its units? It's also a little confusing using the same base Greek symbol for both the charge density and the radial distance.

In any case for a simple single dielectric coaxial capacitor of inner radius a, outer radius b, dielectric constant $\mathbf{\epsilon}$, length L and total charge Q, the radial electrical field strength is given by

$\vec{E_{\rho}}=\frac{Q}{2\pi \epsilon \rho L}\vec{\mathbf{ a_{\rho}}}$

recognizing that

$\epsilon=\epsilon_o \epsilon_r$

If we then defined

$\rho_e=\frac{Q}{L}$

then

$\vec{E_{\rho}}=\large{\frac{\rho_e}{2\pi \epsilon \rho}}\vec{\mathbf{ a_{\rho}}}$

which I would suggest is a more credible solution.

For the inner dielectric region of permittivity $\mathbf{\epsilon_1}$ one would substitute $\mathbf{\epsilon_1}$ for the general value $\mathbf{\epsilon}$ in the aforementioned solution equation.

For completeness we could equate the capacitance C, charge Q and terminal voltage V as Q=CV where

$C=\frac{2\pi\epsilon L}{\ln{(\frac{b}{a})}}$

For the case at hand with two dielectric regions the E field would differ in each region in accordance with the relative permittivity values.

Last edited: Jan 17, 2012