# Calculating fields in the presence of insulators

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## Main Question or Discussion Point

Hello! If I have some conductors in space, each at a certain potential (and assume everything is inside a conducting sphere, in order to have some well defined boundary conditions), we can calculate the potential everywhere (inside the sphere) by solving Laplace's equation. Hence a particle placed inside, will have a well defined trajectory. Of course one will need a numerical approach for an arbitrary distribution, but you can get a trajectory as accurately as you want. I am a bit confused about what happens if you place a neutral insulator material inside (assuming that everything else is the same). Will the Laplace's equation be exactly the same outside the insulator, hence the trajectory of the particle will be the same (assuming it doesn't hit the insulator), or do I have to take the insulator material into account somehow? Thank you!

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Perhaps, you could give a concrete example? In your first sentence, I could read it to mean that you are placing conductors inside conductors. Also, LaPlace's equation is used for charge-free regions. You probably want to refer to Gauss's law or Poisson's equation.

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Perhaps, you could give a concrete example? In your first sentence, I could read it to mean that you are placing conductors inside conductors. Also, LaPlace's equation is used for charge free regions. You probably want to refer to Gauss's law or Poisson's equation.
For the first sentence I just wanted to assume we have some boundary condition. You can ignore the sphere and assume the potential at infinity is zero and we have some conductors in space, each at a given potential. Also, yes, we have no charge. As I said, the dielectrics have zero net charge. I am not thinking of a specific example actually. Any configuration would work, my question is mostly conceptual, not practical.

Well, it's sometimes easier for people to give a good answer if you have a concrete example.

Electric potentials come from charges, which means that you have a problem when you are asking for charge neutrality and a potential on a metal.

If you have an insulator in an electric field, it will get polarized but retain charge neutrality. Integrating Gauss's law over a surface around the insulator will give still give you a net zero charge inside.

Well, it's sometimes easier for people to give a good answer if you have a concrete example.

Electric potentials come from charges, which means that you have a problem when you are asking for charge neutrality and a potential on a metal.
I meant that the insulator is not charged, not the conductors. Maybe to put the problem in an easier way: say we have a 3D region in space where we know the electric field (potential) at each point. We then place a neutral dielectric material somewhere inside that 3D region. Will the electric field/potential change outside the dielectric material? Of curse the dielectric will be polarized, so the field at a point inside the dielectric will change, compared to the case when it was no dielectric there. My question is only for the region outside the dielectric. Thank you!

Yes, the field will change outside the dielectric.

Yes, the field will change outside the dielectric.
But I am not sure I understand why. If I put a Gaussian surface around the dielectric, the number of charges inside is zero and it was zero before the dielectric was placed there, too. So the net flux out of ANY surface around that dielectric (basically around the region of space where the dielectric is now) is the same before and after adding the dielectric. So why does the field changes?

Well, the number of charges is not zero. More specifically the net charge is zero; the postive and negative charges are balanced in the neutral dielectric. And, yes, the net flux is determined by Gauss law, so if you have zero net charge inside the surface before and after, then the net flux is the zero before and after.

Why does it change? The charges were overlapped and their fields canceled, but now they have moved slightly. One end of the Gaussian surface now has slightly more outward pointing electric field and the other end has slightly more inward pointing electric field.

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