Electric Fields and Electrodes

Discussion in 'General Electronics Chat' started by Mazaag, Sep 10, 2007.

  1. Mazaag

    Thread Starter Senior Member

    Oct 23, 2004
    255
    0
    Hi guys,

    I have a few questions on Electric fields and electrodes....

    Lets say I had an AC source, connecting to electrodes , positive and negative , with a dielectric in between (say AIR).

    When I switch on the source, there will be an electric field across the plates which alternates at the same frequency as the source.

    My quesions are as follows,

    - Will electric field density be , on average , equal at both ends of the elctrodes?

    - What will happen to the density at both electrodes if i made one of the electrodes smaller in surface area compared to the other ?

    - How is the power across from the positive electrode to the negative electrode change? Is there a drop due to resistive losses in the dielectric? and what is the CAUSE of these resistive drops (like how is the energy lost and why ) ?

    Thanks guys
     
  2. Dave

    Retired Moderator

    Nov 17, 2003
    6,960
    144
    Are we talking about electric-field flux density, i.e. the density of the electric-field flux lines? If so, for a homogeneous dielectric and ignoring fringing at the electrode edges then yes.

    Again if we are talking about electric-field flux density, then the flux-lines will be more dense at the surface of the smaller electrode.

    If the dielectric has a discernible conductivity then power will be dissipated by the dielectric based on the relationship: P = \sigmaE, where \sigma is the conductivity. The conductivity in turn is a function of the frequency and the loss factor (imaginary) component of the relative permittivity (which in turn is frequency dependant). This is a little more complicated than first seems, because if the loss factor of the dielectric is 0 and the relative permittivity is purely real, then the conductivity and hence power dissipated will be 0. If the conductivity is not zero, power is dissipated and you can treat the dielectric as a parallel RC circuit arrangement.

    Dave
     
  3. Mazaag

    Thread Starter Senior Member

    Oct 23, 2004
    255
    0
    Just out of curiosity, what else could we be talking about ? (like is there anything else that I am not aware of ? )


    So if the dielectric has no conductivity, then it will not dissipate any power and in a sense " conduct " all the energy efficiently ? isnt that counter-intuitive..? (like thinking in I^2 R terms, i would have imagined the conductance to be in the denominator rather than numerator..)
     
  4. Dave

    Retired Moderator

    Nov 17, 2003
    6,960
    144
    I am just clarifying terminology. One could confuse density with intensity in such questions.

    The energy would be stored and not "conducted", so to speak. In this case the voltage and current phases are at right-angles to each other - as in a perfect capacitor.

    Dave
     
  5. Mazaag

    Thread Starter Senior Member

    Oct 23, 2004
    255
    0
    What is the relationship that links Electric field Intensity to E.F Density?
     
  6. recca02

    Senior Member

    Apr 2, 2007
    1,211
    0
    i think they are pretty much the same.
    and that is the reason why electric field density sounds odd and hence a confusion can arrive.
    electric field is force per unit charge.
    electric field flux is electric field passing thru the area.
    here is a link u might want to have a look at.
    http://physics.bu.edu/~duffy/PY106/Electricfield.html
     
  7. Mazaag

    Thread Starter Senior Member

    Oct 23, 2004
    255
    0
    for a charge (lets say positive) , is it neccessary for a negative charge to be in the vicinty in order for an Electric Field (field lines) to exist ?
     
  8. mOOse

    Member

    Aug 22, 2007
    20
    0
    If all other charges in the universe were evenly arranged on a spherical shell
    around the proton, its field lines would radiate in straight lines perpendicular
    to the proton's "surface", no matter how large the radius of the containing
    shell. Pushing it back to infinity, there seems no reason to think the field
    would change that general character. The field flux remains constant, I
    believe, but the field intensity diminishes asymptotically to zero.

    Obviously for a small charge and/or for great distances, the electric field
    could all but disappear, and surely be indetectable.
     
  9. Dave

    Retired Moderator

    Nov 17, 2003
    6,960
    144
    Intensity implies magnitude (non-vector), whereas density implies regionalised quantity (possibly vector, at least better considered this way).

    Dave
     
Loading...