I have some problems with the next exercise. It states:

Two infinite conducting parallel plates I and II, with thickness $t_1$ and $t_2$ respectively are separated by a distance $L$ from its nearer faces. The surface charge density of the plate I is equal to $q_1$ ($q_1$ = the sum of the interior surface density plus its exterior surface density) and the surface charge density of the plate II is equal to $q_2$.

a) Show that the surface charge density of the interior faces are equal in magnitude and with opposite signs.

b) Show that the surface charge density of the exterior faces are equal.

I will set some notation first. I will denote by $a$ to the exterior surface charge density of the plate I, by $b$ the interior surface charge density of the plate I, by $c$ the interior surface charge density of the plate II, and by $d$ to the exterior surface charge density of the plate II. This is illustrated in the following image:

Notation for this problem

Then the conditions are $a+b=q_1$ and $c+d=q_2$. And we have to show that $a=d$ and $b=-c$.

By the symmetry of the problem we can deduce that the electric field must be perpendicular to the surface in every point and that the surface charge density must be uniform in all the conductors. Since the surface charge density is uniform we also know that the electric field must be constant in the regions above the plates, between the plates and down the plates. Since the plates are conductors we also know that the field inside them must be zero.

With all of that in mind, I can show that $b=-c$, by taking as my gaussian surface a cylinder with its faces in the middle of the conductors (as shown by the green rectangle in the last image). Using Gauss' Law we have that the flux must be equal to zero, but since the flux is proportional to the charge inside we conclude that $b= -c$. So far so good.

But when I try to show that $a=d$ I get stuck. All the gaussian surfaces that I take give me the same result $a+d= q_1 + q_2$ or some variation of that. I was hoping that you could help me solve that problem. Thank you ! (:

  • $\begingroup$ @AaronStevens I am also skeptical about the result. And I am not convince that it is true, I tried to find a case where it was false, but if you assume the result to be true you get a system of linear equations that always has a solution. So apparently for every $q_1$ and $q_2$ you always get a solution and no contradiction arises. $\endgroup$
    – Ponciopo
    Commented Sep 20, 2018 at 2:32

1 Answer 1


The key is to consider a point inside one of the conductors

Now, we know the field must be $0$ in the conductor. We also know that the field due to an infinite sheet of charge is given by $$E=\frac{\sigma}{2\epsilon_0}$$

Each surface forms an infinite sheet of charge. So if you add up the field from each (taking direction into account) in the conductor, and use the information presented above, you should be able to show $a=d$. I will leave the details to you.

  • $\begingroup$ I think I get your point, if you have two charge-parallel-planes and want the field inside to be zero you must conclude that $a$=$d$. But I don't see it that clearly with conductors. Take for example the case of a conductor filling half of space with surface charge, in that case you only have one surface and by considering the surface charge as a charge plane you deduce that the field inside the conductor must be different from zero... So, is it right to talk about the surface charge of a conductor in the same may as a charge surface? $\endgroup$
    – Ponciopo
    Commented Sep 21, 2018 at 1:48
  • $\begingroup$ @Ponciopo If you have a conductor that goes to infinity then there will be a no surface charge. Let's assume we have your conductor that fills half of space, with the other half of space as empty space, and we put excess charge onto the conductor. Then the excess charge will all fly off to infinity in order to maximize the separation between charges. You will not get any charge residing on the surface between the conductor and empty space. $\endgroup$ Commented Sep 21, 2018 at 2:01
  • $\begingroup$ That is a very clever argument! But now I am wondering, wouldn't that also apply to our case with the parallel conductors. Since they are infinite maybe the charges will never find an equilibrium position and will fly away to infinity to maximize separation? $\endgroup$
    – Ponciopo
    Commented Sep 21, 2018 at 2:11
  • $\begingroup$ @Ponciopo The half space conductor we just discussed is not physically possible. In reality all conductors are finite. When we look at infinite sheets of charge, we really mean that we have a finite sheet of charge, but on the spatial scales we are interested in the sheets can be treated as infinite. For example, for the two sheets in your question we could say the sheet separation and thicknesses are much smaller than length and width of the sheets. The sheets are physically finite, which gives them surface charge densities. We can mathematically treat them as infinite sheets of charge. $\endgroup$ Commented Sep 21, 2018 at 3:30

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