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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:

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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?

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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.

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  • $\begingroup$ Thank you for answering, how is $ E_1 = E_2 $? It look obvious, but why exactly is it? $\endgroup$ – Rick Apr 5 '18 at 3:54
  • $\begingroup$ @Rick - You can see this by mirroring the system of plates and charges at the middle symmetry plane. The charge inside the Gauss surface stays the same and the electric fields should stay the same. Alternatively, you can stack the plates and charges in reverse order. $\endgroup$ – freecharly Apr 5 '18 at 5:08
  • $\begingroup$ @Rick- Maybe the most compelling argument is using the Gauss's surface for each charged plate. Here the symmetry of equal fields an both sides should be immediately evident. Then you combine the plates with their charges and superimpose their electric fields. $\endgroup$ – freecharly Apr 5 '18 at 5:20

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