For a system of $n$ parallel metal plates bearing charges $q_1, q_2,\dots, q_n$, why it is so that the facing surfaces of the plates bear equal and opposite charges?

How do we prove that?


This is the direct consequence of gauss law.

Consider a cuboidal gaussian surface as in the following figure.NOTE THAT GAUSSIAN SURFACE LIES JUST INSIDE THE PLATES THICKNESSenter image description hereSince electric field inside a conductor is zero, therefore no field lines pass through the surfaces enclosing charges, thus no flux through them. As for remaining surfaces, if we ignore edge effects, since area vector makes an angle of 90 degrees with field lines , thus their flux is also zero.

Thus $$\phi_{net}=0$$ but from gauss law$$\phi_{net}=q_{enclosed}/\epsilon$$ thus its only possible when q enclosed is zero. Which is further only possible for pair of equal and opposite charges.

  • $\begingroup$ It would be extremely helpful if you could draw a diagram with at least three to four plates. I'm curious as to the nature of charges on the other plates mentioned in the question. $\endgroup$ – Saprativ Ray Apr 21 '16 at 20:04

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