Displacement current and conduction current

In summary: The book you are looking for is "Vector Analysis" by Thomas Young. It is available for purchase online at Amazon.com.
  • #1
RajChakrabrty
27
0

Homework Statement



conductivity=1/100 siemens/meter
epsilon=3*epsilon[0]
at which frequency displacement current is equal to conduction current?

Homework Equations


The conduction current is LaTeX Code: I =C * \\frac {dV}{dt}

The displacement current D can be calculated from

LaTeX Code: \\frac{\\partial D}{\\partial t} = - \\frac{\\epsilon}{w} * \\frac{dV}{dt}


The Attempt at a Solution



can not attempt using these formulae.
 
Physics news on Phys.org
  • #2
Let's look at a conducting element of length [tex]\ell[/tex] and cross section A:

[tex]V=IR=\frac{I}{G}[/tex]
[tex]I_{conduction}=V\frac{CA}{\ell}[/tex]

For simple AC current across a conducting element: [tex]V=V_0 \cos {\omega t}[/tex]

The electric field across the conductor is proportional to the voltage: [tex]E\ell = V[/tex]

From here on out, you can find the displacement current and compare it with the previous result.

My result was (Looking only at the absolute values of the currents, disregarding the phase difference between the two):

[tex]\omega _c \approx 3.76\cdot 10^8 [sec ^ {-1}][/tex]

Which agrees well with common sense that says that for most frequencies that we deal with we don't concern ourselves with the displacement current at all.
 
Last edited:
  • #3
RajChakrabrty said:

Homework Statement



conductivity=1/100 siemens/meter
epsilon=3*epsilon[0]
at which frequency displacement current is equal to conduction current?

Homework Equations


The conduction current is LaTeX Code: I =C * \\frac {dV}{dt}

The displacement current D can be calculated from

LaTeX Code: \\frac{\\partial D}{\\partial t} = - \\frac{\\epsilon}{w} * \\frac{dV}{dt}


The Attempt at a Solution



can not attempt using these formulae.

I also find this problem a little vague in the wording, and hence difficult to answer, but I'll take my best guess. I would approach this problem from the point of view of complex conductivity for sinusoidal fields. For sinusoidal excitation of linear media with nonzero conductivity (i.e. the real world and low field strengths), one can use complex conductivity as follows:

[tex] \sigma + j \omega \epsilon[/tex]

where [tex] j = \sqrt{-1}[/tex], [tex] \sigma [/tex] is the conductivity of the medium, [tex] \epsilon [/tex] is the permitivity of the medium and [tex]\omega[/tex] is the field frequency of oscillation of the electric field.

Solving this problem amounts to setting the real and imaginary parts of the complex conductivity equal to each other. It seems to me that this would result in displacement currents (as typical for dielectrics) equal to conduction current (as typical for conductors). The resulting phase angle (field to current) would be 45 degrees in this case. Normally displacement current is 90 degree's out of phase (i.e. time derivative involved), while pure conduction is in phase, ideally.

EDIT: I just noticed RoyalCat's update with a numerical answer. I also get 376 Mrad/s for frequency (or 60 MHz). I guess this is the correct answer.
 
Last edited:
  • #4
you have said that you have got an answer of 60MHz .
can you tell me the formulae you have used,to find out that answer?
 
  • #5
Nice solution, stevenb. I'm not quite as experienced with the P, D and H fields so reading up on complex permittivity was enlightening.

This is a great example of how in-depth analysis of a certain problem, structure and all, can be circumvented by using the specific concepts specific to that kind of analysis. :)
(With much less mathematical hassle to boot!)

As for the formulas we used, stevenb's derivation relies on the concept of complex permittivity, which helps relate the phase difference brought on by displacement current to the conduction current. (Correct me if I'm oversimplifying or flat out wrong here :))

I analyzed the conduction current and displacement current separately. My conduction current analysis is further up in my first post.

As for the displacement current, I started with the relation [tex]V=E\ell[/tex] which simply utilizes the definition of the electrical potential for a constant E field: [tex]V_B - V_A=\int_A^B \vec E \cdot \vec d\ell[/tex]

I then related the current to the electrical flux: [tex]I_{displacement} = \epsilon _ r \epsilon _0 \frac{d\Phi_E}{dt}[/tex]

Where the electrical flux in this case, assuming uniformity of the E field (Which breaks down substantially at the frequencies we're talking about) is simply [tex]E\cdot A[/tex]

And lastly, to convert from angular frequency to regular frequency: [tex]\omega = 2\pi f[/tex]
 
  • #6
RajChakrabrty said:
you have said that you have got an answer of 60MHz .
can you tell me the formulae you have used,to find out that answer?

Very simply, some people prefer to work with cyclic frequency (i.e. cylcles per second). Angular frequency and cyclic frequecy are related through the circle relation.

[tex] \omega = 2\pi f[/tex]
 
  • #7
thanks a lot to both of you,stevenb and RoyalCat.
I got the answer from complex relation,
[tex]\sigma[/tex]+jw[tex]\epsilon[/tex]

and equating imaginary with real part.

but can you tell me how does this relation come?
any reference, if possible.
thanks again to both of you.
 
  • #8
RajChakrabrty said:
...I got the answer from complex relation,
[tex]\sigma[/tex]+jw[tex]\epsilon[/tex]

and equating imaginary with real part.

but can you tell me how does this relation come?
any reference, if possible.
...

I'll look for a good reference on-line. I did a quick search but didn't find anything that I liked. I cobbled together the attached PDF from memory. Hopefully, I didn't make a mistake.

There is one book I think is good, but it is expensive and probably too specialized for what your are doing. However, if you can find it in your library, it is worth reading the introductory sections. (note that if you find the first edition in the library, that's good too because the fundamentals don't change)

"Handbook of Biological Effects of Electromagnetic Fields", 2nd edition, Charles Polk and Elliot Postow, CRC Press, 1996
 

Attachments

  • ComplexCond.pdf
    12.7 KB · Views: 440

1. What is the difference between displacement current and conduction current?

Displacement current is the flow of electric charge that occurs due to a changing electric field, while conduction current is the flow of electric charge through a conductive material. Displacement current does not involve physical movement of charges, but rather a change in the electric field. Conduction current, on the other hand, involves the physical movement of electrons through a conductor.

2. How is displacement current related to Maxwell's equations?

Displacement current is included in Maxwell's equations as a term that describes the relationship between changing electric fields and magnetic fields. In the absence of a conduction current, displacement current is necessary to explain the behavior of electromagnetic waves and their propagation through space.

3. What is an example of a situation where displacement current is present?

One example of a situation where displacement current is present is in a capacitor. When the electric field in the capacitor changes due to the charging or discharging of the capacitor, displacement current is created.

4. How does displacement current affect the behavior of electric and magnetic fields?

Displacement current plays a crucial role in the relationship between electric and magnetic fields. It produces an electric field that is perpendicular to the changing magnetic field, and this interaction is what allows for the propagation of electromagnetic waves.

5. Can displacement current be measured?

Displacement current cannot be measured directly, as it does not involve the physical movement of charges. However, its effects can be observed and measured indirectly through its influence on electric and magnetic fields and on the behavior of electromagnetic waves.

Similar threads

  • Introductory Physics Homework Help
Replies
6
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
751
  • Introductory Physics Homework Help
Replies
7
Views
3K
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
11
Views
3K
  • Introductory Physics Homework Help
Replies
1
Views
683
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
665
  • Introductory Physics Homework Help
Replies
4
Views
507
  • Introductory Physics Homework Help
Replies
6
Views
910
Back
Top