# Finding charges on the plates of a capacitor system connected to a battery

Five identical capacitor plates, each of area A, are arranged such that the adjacent plates are at a distance d apart. The plates are connected to a battery of e.m.f. E volt as shown. The charges on plates 1 and 4 respectively are :

So, I started off this exercise problem by finding equivalent capacitance. Due to charge induction, the charge on plate-2 must be equal and opposite in sign to that of capacitor of plate-1 , plate-3 equal and opposite to that of plate -2 and like this till finally plate-5 has opposite and equal charge of plate -4.

For capacitors, the voltage between plates are completely decided by the charge on plate and capacitance, so the total voltage of the capacitors formed by the plates i.e (1,2) , (2,3),(3,4),(4,5) must be the same. (note : two plates make one capacitor)

Now, this system is similar to four capacitors in series, hence the total capacitance is $$4C$$ where C is capacitance of any two consecutive plates. Using this I found the charge on the equivalent capacitor's plate, which is,

$$4C \cdot E = q$$

But, my problem now is relating charge of the equivalent capacitors plate to the original 'split' configuration of capacitors. As in, how would I find the charge on original plates?

On, second thought maybe I could have skipped the whole equivalent capacitor procedure and maybe directly found the capacitance taking the capacitance of a capacitor with the plate of interest as in parallel with the battery. Now this is easy to do for plate -1 as it only has plate-2 adjacent to it, however for plate-4, it makes a capacitor with plate -3 and plate -5. So how would I deal with this?