Poynting Vector (Griffiths 8.1)


by Rubiss
Tags: griffiths, poynting, vector
Rubiss
Rubiss is offline
#1
Mar29-12, 02:25 PM
P: 21
1. The problem statement, all variables and given/known data
Problem 8.1 from the third edition of Introduction to Electrodynamics by Griffiths:

Calculate the power (energy per unit time) transported down the cables of Ex. 7.13 and Prob. 7.58, assuming the conductors are held at potential difference V, and carry current I (down one and back up the other)


2. Relevant equations

[tex]
P=\frac{dW}{dt}=\int_{S}\vec{S} \cdot d\vec{a}=\frac{1}{\mu_{0}}\int_{S}(\vec{E}\times \vec{B}) \cdot d\vec{a}
[/tex]

3. The attempt at a solution

I have already solved this problem, but am unsure of some of the technicalities. Here we go:

The only region where there is both an electric and magnetic field is in the region a<s<b. For both s<a and s>b, there is no magnetic field or electric field.

Because the two conductors are held at a potential difference V, the electric field is
[tex]
\vec{E}=\frac{V}{s \cdot \ln{(\frac{b}{a})}} \hat{s} \mbox{ for a<s<b}
[/tex]

The magnetic field between the conductors is
[tex]
\vec{B}=\frac{\mu_{0}I}{2\pi s} \hat{\phi} \mbox{ for a<s<b}
[/tex]

Because we now know the magnetic and electric field, we can easily calculate the Poynting vector and therefore the power:

[tex]
P=\frac{dW}{dt}=\int_{S}\vec{S} \cdot d\vec{a}=\frac{1}{\mu_{0}}\int_{S}(\vec{E}\times \vec{B}) \cdot d\vec{a}

= \frac{1}{\mu_{0}} \int_{a}^{b} \Big[\Big(\frac{V}{s \cdot \ln{(\frac{b}{a})}}\hat{s}\Big)\times \Big(\frac{\mu_{0}I}{2\pi s} \hat{\phi}\Big)\Big] \cdot (2\pi s ds \hat{z})

= \frac{IV}{\ln{\frac{b}{a}}} \int_{a}^{b} \frac{ds}{s} = IV

[/tex]

as expected. Now here's my question. My professor said you could also just use the standard formula for the electric field of a line charge (or cylinder, if you are outside of it):
[tex]
\vec{E}=\frac{\lambda}{2\pi\epsilon_{0}s}\hat{s}
[/tex]
and obtain (if you go through exactly what I did above)

[tex]
P=\frac{\lambda I}{2\pi\epsilon_{0}} \ln{\frac{b}{a}}
[/tex]

But I don't like my professor's way for a couple of reasons. One reason is that it introduces a linear charge density that was not given in the problem. For instance, it could well have been a surface current (K) flowing over the surface. Secondly, the only reason there is an electric field between the conductors is because there is a potential difference, correct? There is no way that you should be able to use the formula for an electrostatic field (the standard formula that my professor used) when we have a current (that is, moving charges). How can my professor say that a current is creating an electrostatic field? Isn't that impossible? Does it have anything to do with
[tex]
\vec{I}=\lambda \vec{v} = (-\lambda)(-\vec{v})
[/tex]
?

Also, how do we know that the direction of the electric field is in the positive s-hat direction, and not the negative s-hat direction?

I have the same argument for the second part to this problem (problem 7.58). I used E=V/d for the electric field (because the potential difference is given), but my professor said you could also use E=(sigma)/(epsilon-naught). I don't like this way either. There is obviously a surface current flowing over the strip which equals
[tex]
\vec{K}=\frac{d\vec{I}}{dl_{\perp}}=\frac{\vec{I}}{w}=\sigma \vec{v}
[/tex]
So there must be a surface charge. However, this surface charge is moving, so how can my professor say that the field is equal to the standard electrostatic (sigma)/(epsilon-naught)? Also, how do we know that the electric field points upward (and not downward)?

These are my questions. Hopefully someone will be able to point out where I am wrong. Also, I just remembered another quick question. Right above the problem in Griffiths is figure 8.1 that goes along with example 8.1. At the end of the example, the conclude that the power is S(2*pi*a*L)=IV. Aren't S and da in the opposite direction? Why is there no minus sign from the dot product?
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sunjin09
sunjin09 is offline
#2
Mar30-12, 09:59 PM
P: 312
electric currents are the relative motion between positive and negative charges in a conductor, but the total charge can be neutral, as long as the continuity equations are satisfied. In the case of (qausi)statics, the NET charge is independent of the currents. (The continuity equation degenerates to 0=0.) In order for a potential difference to exist there must be NET charge distributions, since these net charges don't change with time, they can be treated as static charges that aren't part of the current flow. (Physically they can flow, causing currents, but then the neutral currents must adjust accordingly so that the total current is as stated.)
N_J_R
N_J_R is offline
#3
Jan8-14, 09:59 PM
P: 2
Isn't the electric field:

[itex]\vec{E}[/itex] = [itex]\frac{V}{s\cdot ln(\frac{b}{a})}[/itex][itex]\hat{s}[/itex] ([itex]\ast[/itex])

obtained by treating the nested cylinders as a wire and then taking the line integral from a to b of the electric field?

[itex]\vec{E}[/itex] = [itex]\frac{\lambda}{2 \pi \epsilon_0 s}[/itex][itex]\hat{s}[/itex] (1) This is the E-Field of a long wire. Calculated by using a Gaussian surface.

V = [itex]∫^{b}_{a}[/itex] [itex]\vec{E}[/itex] [itex]\cdot[/itex] d [itex]\vec{l}[/itex] insert equation (1)

V = [itex]∫^{b}_{a}[/itex] [itex]\frac{\lambda}{2 \pi \epsilon_0 s}[/itex]ds

V = [itex]\frac{\lambda ln(\frac{b}{a})}{2 \pi \epsilon_0}[/itex] (2)

Now rewrite equation (1) with equation (2):

[itex]\vec{E}[/itex] = [itex]\frac{V}{s\cdot ln(\frac{b}{a})}[/itex][itex]\hat{s}[/itex]

This is the only way I know how to get ([itex]\ast[/itex]). Please let me know how you calculated it. (Sorry for the lack of LaTeX skills!)


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