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:
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 ! (: