In one of the problems involving Gauss’s law, we have to calculate the electric field outside and inside a symmetric slab of some thickness and certain net charge. Of course, the electric field we get outside the slab is independent of the distance between the point and the slab which is understandable but the solution continues by calculating the electric field inside the slab, how is this possible?

In electrostatics we are dealing with non moving charges and if we have a bunch of charges next to each other the only possible way for them to still stationary is that the force affecting each one of them is zero (or to say the total sum of forces is zero) but if we have an electric field inside the slab that means the charges are being acted on by a force and hence disrupting the equilibrium of electrostatics.

So how can we have an electric field inside a slab and still have static charges that are standing still and in equilibrium?

To give you a good idea about the kind of problem i’m talking about look at this problem: https://youtu.be/1eAauKHR72A

Notice that we have already assumed the electric field is only in the upward and downward direction (meaning that the left and right direction have been canceled by each other to preserve the equilibrium of motion)


1 Answer 1


It depends on what material you are talking about.

In a conductor, you are right: charges are free to move so at equilibrium the electric field inside the material must be 0.

In a dielectric material, we assume instead that charges are "fixed", held together by other forces, so that even if an electric field is present they do not move as inter-molecular forces are still keeping them together. It is an assumption, of course, the microscopic picture is much more complicated. But if you make this assumption, then you can allow the existence of a field inside your slab.

  • $\begingroup$ Well now it makes more sense but if that was the case why are we assuming the electric field is in the upward and downward direction only? What about the left and right direction? Are we dismissing them because they are relatively small compared with the other dimensions so we assume that they are practically zero? I’ve added a link to a problem to give you a better idea $\endgroup$ Aug 28, 2021 at 10:23
  • $\begingroup$ That will in general depend on the problem, but in the case of an infinite slab, the field must be in the vertical direction for symmetry reason (you can't define left/right on an infinite slab) - in other words, the horizontal components cancel each other at each point. $\endgroup$
    – JalfredP
    Aug 28, 2021 at 10:32
  • $\begingroup$ Thank you so much that was really helpful, the major problem i’m facing with this subject is the amount of assumptions that the instructors take for granted when solving a problem without further elaboration on how or why that was the case, your answers gave me a good insight about some of these assumptions. Do you know any kind of resources whether it be a textbook or a website or whatever that explains these assumptions with details? $\endgroup$ Aug 28, 2021 at 10:48

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