How do I prove the charge on the outer part of the last plate is half the total charge in a system of parallel metal plates?

or in other words, if I have $n$ large parallel metal plates with charges $q_1$, $q_2$, ... $q_n$, how do I prove the charge on the outer part of the last plate $\displaystyle \frac{q_1+q_2+..q_n}{2}$ ?

This is what I tried:

From this I get, $\displaystyle \frac{s \cdot \sum_{i=1}^n Q_i }{A\epsilon_0} = (E_1 + E_2)s$ where A is the area of the plates..how do I proceed from here?

You use Gauss's law assuming the plates have very large area $A$ so that the fringe fields at the borders of the plates can be neglected: $$\epsilon_0 E_1·A +\epsilon_0 E_2·A=q_1+q_2+...+q_n$$ From symmetry $E_1=E_2$ Thus $$E_1A\epsilon_0=E_2A\epsilon_0=\frac {q_1+q_2+...+q_n}{2}$$ By applying Gauss's law to the last metal plate surface, you see that $E_2 A\epsilon_0$ corresponds the total outer surface charge of the last plate.
• Thank you for answering, how is $E_1 = E_2$? It look obvious, but why exactly is it? – Rick Apr 5 '18 at 3:54