# Electric field inside a non-conducting shell with a charge inside the cavity

I have read about conducting charged shells and spheres and know somewhat about the electric fields associated with them. But I have never found anything on non-conducting shells. I have searched online and gone through several famous physics and electrodynamics/electrostatic books. Hence I have devised a thought experiment to resolve my doubts on the topic.

Consider a charged symmetrical non-conducting shell having a charge $$Q$$ on its surface and $$q$$ at a point inside the cavity. If I try to find the electric field a point inside the cavity, I will find that they point outwards (due to $$q$$). But since $$Q$$ is already uniformly distributed on the surface, how can the field lines due to $$q$$ pass though the outer surface?

Another question that I have is does the uniformly distributed charge $$Q$$ on the surface create an electric field inside the cavity? If it does then where do the field lines due to $$Q$$ inside the cavity go(end)?

What would be the situation inside the cavity if instead of a non-conductor, we simply have a shell of charges and somehow managed to keep them in the situation described above?

There are many things to look at. Firstly, a conducting material and non-conducting material are different. For conducting materials, typically all the charges are found on the surfaces. For non-conducting shells (e.g. plastic), the materials contain molecules which act like dipoles which product a field that counteract the applied field. (See http://hyperphysics.phy-astr.gsu.edu/hbase/electric/dielec.html)

You may or may not be aware of Gauss's Law for Electric Fields (it's one of maxwells equations). But essentially, the electric field inside the shell is only due to the charge q. (See https://en.wikipedia.org/wiki/Gauss%27s_law) The reason for this is simple, you split up all the charges on the surface of the shell and draw the electric field at point inside the shell contributed by each little charge. The net electric field which is a vector sum would be zero.

The field outside the shell is a superposition of the electric fields due to the charge q and the shell charge Q. This is a consequence of the linearity of maxwells equations. This just means, that we can add electric field vectors.

• Ok, so no field line inside the shell due to $Q$. What about the field lines due to $q$ ? How do the lines pass though the shell of uniformly distributed charge? Souldn't it be impossible since field lines never overlap? Nov 12 '20 at 3:09
• Think of it like this. Every single charge is contributing an electric field. That's nature. Inside the sphere, the electric field due to the shell they cancel out, so you just get an electric field due to q in the cavity.
– Ali
Nov 12 '20 at 3:15
• I am not sure what you mean by "cancel out". Opposite charges form a dipole of sorts and E-lines star at +q and end at -q. But if like charges are present then the lines curve away from each other. i.stack.imgur.com/UmU7M.jpg Nov 12 '20 at 3:23
• That picture represents the "net" electric field. For a positive charge, the electric field points outwards radially, for a negative charge they point in. When you add the fields you get the picture above. For example, consider two positive charges next to each other. Look at a point half way between the charges. The charge on the left will produce a field to the right say E. The charge on the right will produce a field to the left however pointing in the opposite direction -E. The total is zero. Going back to your picture, there is gap between the two positive charges.
– Ali
Nov 12 '20 at 3:30
• Every single charge present in your system will contribute an electric field vector to a point. You have to sum all the vectors. This is easier said than done. In a non-conducting shell, you have many charges and those charges are not on the surface. This is why we prescribe something called a dielectric constant which tells us how much a field is "reduced" after it passed through the non-conducting material. By "reduced", I mean that you have your original field E, your non-conducting material produce an opposing field E' as a result of E being applied and then your field coming out is E-E'
– Ali
Nov 12 '20 at 3:36

A "nonconducting", or rather dielectric medium, will have its own susceptibility, because of which its response to external fields will be nontrivial. For a general position of the charge $$q$$ inside the shell, the medium will develop some non-homogeneous polarization, which will have to be solved for using Maxwell's equations.

As for getting a feel for the field lines, you could start by assuming isotropic, linear electric susceptibility ($$\neq -1$$, which is the case for metals/conductors), and think in terms of polarization of the medium, i.e. see where a net accumulation of bound charges occurs.