I'm confused with the answer to the following exercise:

enter image description here

I don't see how the placement of the sphere would change the $\mathbf D$-field given this arrangement. I think of the $\mathbf D$-field as effectively the electric field exclusively for free charges. Since the sphere is neutral, it ought to not have a net free charge density, and thus when considering the $\mathbf D$-field at point P, we ought only concern ourselves with the infinite plane. Where is my thinking going wrong?

  • $\begingroup$ Based on cyllindrical symmetry of the whole problem, I would say that electric field would be pointing up. Could it be different in magnitude? I don't have the time to solve this, but this plot proxy.duckduckgo.com/iu/… suggests that dielectric sphere coud distort electric field magnitude-wise. $\endgroup$ – Cryo May 1 at 21:07

I think of the D-field as effectively the electric field exclusively for free charges.

This is a common misconception. It is true that displacement field $\mathbf D$ obeys the equation $$ \nabla \cdot \mathbf D = \rho_{free}. $$ But this does not mean that we can use the Coulomb formula and plug in $\rho_{free}$ to obtain the field $\mathbf D(\mathbf x)$! Displacement field has in general non-zero curl, so its divergence is not sufficient to reconstruct it. In other words, displacement field is not determined merely by distribution of free charge.

The easy way to answer the question is to notice that at the point P, the field $\mathbf D$ is proportional to electric field with the same constant of proportionality whether the sphere is present or not. But the electric field is different when the sphere is present, the polarized sphere will change the uniform field of the charged plane into a non-uniform ( complicated spatially varying) field. So the displacement field $\mathbf D$ has to change as well, and due to the proportionality, outside the sphere it will have the same pattern of lines of force as $\mathbf E$ has.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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