1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The point in the D and H fields

  1. May 16, 2012 #1
    In lectures we have been given:

    ## \vec{\nabla} \cdot \vec{D} = \rho_{\text{free}} ##
    ## \vec{\nabla} \times \vec{H} = \vec{J}_{\text{free}} + \frac{\partial \vec{D}}{\partial t} ##
    ## \vec{\nabla} \times \vec{E} = - \frac{\partial \vec{B}}{\partial t} ##
    ## \vec{\nabla} \cdot \vec{B} = \vec{0} ##

    I need examples to work out why ##\vec{D}## and ##\vec{H}## are such a convenience for me. This is something I can't see at all!

    ## \vec{D} = \epsilon_r \epsilon_0 \vec{E} ## ? So it should be useful in dielectrics, and ## \vec{D} = \epsilon_0 \vec{E} + \vec{P} ## so polarisation is defined as the E field caused by a relative dielectric constant in some dielectric material?

    The there's ## \vec{H} = \frac{\vec{B}}{\mu_0} - \vec{M} ## and also I guess ## \frac{\vec{B}}{\mu_0 \mu_r} = \vec{H}## ? So the magnetisation is the magnetic field due to some medium with relative permeability?

    But why are these vector fields useful for calculations?
  2. jcsd
  3. May 16, 2012 #2
    The basic physical difference between the E and D fields is that the E field represents a force along a line element while the D field represents a flux density through a surface. The spatial relationship between them is simple in a vacuum but not necessarily so in media. The same applies to the H and B fields.

    In media that relationship is given by the constitutive relations and it can be challenging to determine given only charge distribution and charge motion, but there are techniques. Without being able to calculate the D field, you would probably not be able to eliminate the degrees of freedom to arrive at a resolution of the constitutive relations.
  4. May 16, 2012 #3


    User Avatar
    Science Advisor

    The alternative would consist in working with the internal current densities J_int. But in contrast to polarization it is not a good approximation to assume that the internal currents are proportional to the internal field (after all J_int=dP/dt). Same for magnetization.
    Obviously this does not hold always true. E.g. in superconductors, the internal current is proportional to the local value of the magnetic vector potential
    so one won't usually describe a superconductor in terms of D or H fields.
  5. May 16, 2012 #4


    User Avatar
    Science Advisor
    Homework Helper

    Hi Gregg! :smile:
    Imagine an ordinary parallel-plate capacitor with more than one dielectric between the plates.

    Calculating the E field directly could be horrendous.

    Calculating the D field is really easy (you ignore the dielectrics!), and then just divide by the relevant ε to get E. :wink:
  6. May 16, 2012 #5
    Interesting. Thanks for the reference.
  7. May 16, 2012 #6
    Why does the D field and H field not completely replace the B and E fields then in these equations? Is it not useful to say ## \vec{{\nabla}} \times \vec{D} ## ?
  8. May 16, 2012 #7
    Very good point Tim. I was going to say something along that line but you already said it. So I'll just add a few more words.

    If the differing dielectrics are in series, then the D field value is the same for both media. The E values differ, and can be obtained by multiplying the D value, the same for both media, by the appropriate dielectric constant. At the boundary the normal values of D are equal.

    If said differing dielectrics are in parallel, then the E field values are the same, the D values differ in proportion to epsilon. The E field is just V/s in the direction perpendicular to the plates. V is the voltage, and s is the spacing between the plates. At the boundary the tangential values of E are equal.

    Likewise, a similar relation holds for magnetic fields, H & B, in the presence of differing media. For series media, Bnormal1 = Bnormal2. The H values differ & are obtained from H1 = B1/mu1, & H2 = B2/mu2. At a boundary for parallel media, Htangential1 = Htangential2. Then B values are obtained by multiplying H by the appropriate mu value.

    I agree that B, D, E, & H, are all important. Which quantities should be employed is determined by the nature of the problem under scrutiny. For 2 different media in parallel, I determine E & H, then compute D & B by considering the constants epsilon & mu. For 2 media in series, I determine B & D, then compute E & H by considering mu & epsilon.

    Did I help? Best regards.

    Last edited: May 16, 2012
  9. May 16, 2012 #8
    We can express any and all of Maxwell's equations using any combination of quantities we wish. E.g.: curl E = -dB/dt, can also be expressed as curl D = -epsilon*dB/dt, or if you prefer curl D = -epsilon*mu*dH/dt, then again curl E = -mu*dH/dt.

    I don't believe that any of the above is more or less correct, it just happens that I learned ME (Maxwell's equations) in a form I describe as "medium-independent". For Faraday's law using E & B results in no epsilon or mu factor. With Amperes law, using H, D, & J (current density) also results in no factors accounting for relative permittivity/permeability of the medium.

    In the case above with the capacitor having more than 1 dielectric material, series or parallel, the use of D as opposed to E is determined by boundary conditions. In series 2 media have equal D values along the normal direction. In parallel 2 media have equal E values along the tangential direction.

    I prefer the "lazy approach". I express equations in their simplest "medium-independent" form whenever possible. I would rather have no mu or epsilon appearing in the equation. That being said, we need to examine all 4 quantities, not just 2, because they all convey useful info. I will elaborate further if needed.

    Last edited: May 16, 2012
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook