Suppose a cylinder of radius $R$ and length $L$, with longitudinal polarization $\vec{P}=P(r)\hat{z}$, where $(r,\theta,z)$ are usual cylindrical coordinates.

I want to compute the electric field $\vec{E}$ on the cylinder axis.

My reasoning is this: volumetric bound charge is zero, surface bound charge is $P(r)$ on one cap, $-P(r)$ on the other, so basically we have the field due to two charged disks, end of story.

I hesitate because I suspect the displacement field $\vec{D}$ should be useful, but I don't know how to compute it.

How do we compute the displacement field in this case?

  • $\begingroup$ If you already know how to solve the problem without $\mathbf{D}$, what's the problem? $\mathbf{D}$ is useful when you have a definite relationship between polarization and electric field, as in the case of a linear medium, because it leads to simpler differential equations. But in this case, the polarization is fixed and you're told it from the start, so $\mathbf{D}$ has no use. $\endgroup$
    – knzhou
    Jun 23, 2022 at 16:11
  • $\begingroup$ @knzhou Since in this problem there is no free charge, can I compute $D$ without knowing $E$ beforehand? $\endgroup$
    – thedude
    Jun 23, 2022 at 16:30
  • 1
    $\begingroup$ I don't think so. Or, to be more precise, it is certainly possible but it would be much less convenient than calculating $\mathbf{E}$ first and using that to get $\mathbf{D}$. $\endgroup$
    – knzhou
    Jun 23, 2022 at 16:49

1 Answer 1


Your first argument makes sense to me, except one point: the surface bound charge density is $P(r)$ on the top and $-P(r)$ at the bottom.

Regarding whether the displacement field $\vec{D}$ should be useful, let us recall its definition $$\vec{D}=\varepsilon_0\vec{E}+\vec{P}$$ which implies $\nabla \cdot \vec{D} =\rho-\rho_b=\rho_f$ where $\rho_f$ is the free charge density. Therefore, $\vec{D}$ is useful when we try to solve problem where the quantity and distribution of free charge, and the electric field caused by it are wanted. Since here we are interested in the bound charge, it is reasonable that the displacement field does not become useful.

  • $\begingroup$ Since in this problem there is no free charge, can I compute $D$ without knowing $E$ beforehand? $\endgroup$
    – thedude
    Jun 23, 2022 at 16:29
  • $\begingroup$ Well, the physics here is that the resulting electric field is caused only by the bound charge. There is no free charge at all! If you think about this, $\vec{D}$ should just be $0$. $\endgroup$
    – Andy Chen
    Jun 23, 2022 at 16:46
  • 2
    $\begingroup$ Not quite. The absence of bound charge means $\nabla \cdot \mathbf{D} = 0$, but we have $\nabla \times \mathbf{D} \neq 0$, so it's some complicated annoying thing, not zero. $\endgroup$
    – knzhou
    Jun 23, 2022 at 16:50
  • $\begingroup$ Thank @knzhou for the correction. It is just $\nabla \cdot D = 0$, and since there is no magnetic field here, we have $\nabla \times E = 0$ and $\nabla \times D = \nabla \times P$. $\endgroup$
    – Andy Chen
    Jun 23, 2022 at 16:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.