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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 :

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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?

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1 Answer 1

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Note that plates 2,3,and 4 have a charge of q/4 on each side of each plate, but plates 1 and 5 have q/4 only on the inner surfaces.

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  • $\begingroup$ So, if I understood correctly both sides of plates contain charges? then in the regular case of capacitor connected to battery, is there charge on the face where we connect the lead from battery to a capacitor? $\endgroup$
    – Babu
    Aug 3, 2020 at 16:37
  • $\begingroup$ In a normal capacitor, the opposite charges attract each other and reside primarily on the two facing sides. $\endgroup$
    – R.W. Bird
    Aug 3, 2020 at 17:55

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