# A question about the fields D and H for fields in matter

Let me give you an example of a reasoning I made in a simple case. Afterwards comes the question:

Start of reasoning.

Consider two standard ideal capacitor plates with a dielectric material in between them. Let's call the external field caused by the plates $\vec{E_{ext}}$ and the average macroscopic internal field caused by polarisation $\vec{E_{int}}$ The total field at any point is then on average $\vec{E_{tot}}=\vec{E_{ext}}+\vec{E_{int}}$

One can prove that for homogenous polarisation $\vec{E_{int}}=-\frac{\vec{P}}{\epsilon_{0}}$ and thus $\vec{E_{tot}}=\vec{E_{ext}}-\frac{\vec{P}}{\epsilon_{0}}$

Let's now look at the $\vec{D}$ field at any point in the material. The definition is $\vec{D}=\epsilon_{0}\vec{E_{tot}}+\vec{P}$. Plugging the previous expression into D, results in:

$\vec{D}=\epsilon_{0} \vec{E_{ext}}$

End of reasoning.

QUESTION:

1) So in this case $\vec{D}$ seems to be not only independent of polarization but really equal to the external field up to a constant. Is this the correct way to interpret the $\vec{D}$-field, as something that is equal to the E field that is caused by sources of free charges and neglect any E fields caused by polarization?

2) If I replace D with H and E with B , is the same interpretation correct?

## Answers and Replies

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mfb
Mentor
1) So in this case $\vec{D}$ seems to be not only independent of polarization but really equal to the external field up to a constant. Is this the correct way to interpret the $\vec{D}$-field, as something that is equal to the E field that is caused by sources of free charges and neglect any E fields caused by polarization?
The polarization changes E, and D is defined specifically to be invariant of that (as long as the setup is as nice as yours).
2) If I replace D with H and E with B , is the same interpretation correct?
D with B and E with H. Yes, that is not intuitive.

Edit: removed weird sentence fragment

Last edited:
For fields orth. The polarization changes E, and D is defined specifically to be invariant of that (as long as the setup is as nice as yours).
D with B and E with H. Yes, that is not intuitive.
May I ask what you mean by your last line? Is H no longer invariant in nice setups?

mfb
Mentor
B takes the place of D, and H the one of E, so yes.

TSny
Homework Helper
Gold Member
In general, the "source" of D is not just the free charges.

The polarization P will also have an effect on the what the value of D will be. For certain simple cases with lots of symmetry, D does end up being the same as if you treated only the free charges as "producing" D. But, in general, this is not true. Examples where D depends on the properties of the dielectric as well as the free charges can be found in standard textbooks.

It is true that the divergence of D is given by $\vec{\nabla} \cdot \vec{D} = \rho_{free}$, but that equation is not sufficient to find D except for simple configurations with sufficient symmetry. In general, a vector field is determined by both its divergence and curl. You can see that the curl of D depends on the curl of P. So, P can act as a "source" for the curl of D.

In general, the "source" of D is not just the free charges.

The polarization P will also have an effect on the what the value of D will be. For certain simple cases with lots of symmetry, D does end up being the same as if you treated only the free charges as "producing" D. But, in general, this is not true. Examples where D depends on the properties of the dielectric as well as the free charges can be found in standard textbooks.

It is true that the divergence of D is given by $\vec{\nabla} \cdot \vec{D} = \rho_{free}$, but that equation is not sufficient to find D except for simple configurations with sufficient symmetry. In general, a vector field is determined by both its divergence and curl. You can see that the curl of D depends on the curl of P. So, P can act as a "source" for the curl of D.
Thanks. Same reasoning applies to B and H right?

TSny
Homework Helper
Gold Member
Thanks. Same reasoning applies to B and H right?
Yes, that's right.

Yes, that's right.
May I then ask about the formula (assuming no E field is changing) ''closed loop integral of H'' = sum of macroscopic currents through the loop.

This formula seems to only care about what the macro current is through the loop. For example, I take a toroid-shaped solenoid and put material inside this solenoid and then take the loop integral of H. I will find that the field H does not depend on the material inside the solenoid and only on the properties of the solenoid itself. The previous poster seemed to imply this was the opposite case for H fields. Am I misunderstanding something?

mfb
Mentor
You have a different setup now, a solenoid is not a capacitor.
H does not change in that case, but B does.