Five identical plates, each having area A are arranged as shown in Fig. Find the equivalent capacity of the structure between A and B?
.

Explanation

Method 1: The charge distribution on difference surface can be done by following conservation of charge and noting that facing surfaces have equal and opposite charges. Thus, the equivalent capacity can be given by .
`C_(eq)=Q/(V_(A)-V_(B))` (i)
Betweem plates `(1)` and `(2)`,
`(Q-x) =C(V''-V')`
Between plates (2) and (3),
`(Q-x)=C(V-V'')` (ii)
Between plates (3) and (4),
`x=C(V-V')` (iii)
Between plates `(4)` and `(5),`
`Q=C(V'-0)` (v)
Adding Eqs. (ii) and (iii), we get
`2(Q-x)=C(V-V')`
From (iv) and (vi),
`x=2(Q-x)` or `x=(2Q)/3`
Adding (v) and (vi), we get
`(V-0)=1/C(Q+x)=(5Q)/(3C)`
or `Q/((V_(A)-V_(B)) =Q/((V-0))=3/(5)C=(3epsilon_(0)A)/(5d)`

Method 2: Let the potentials of `A` and `B` be `V_(A)` and `V_(B)`,
respectively. plates (1) and (2) are connected together and have a common potential, say `V'`. Plate (2) is isolated and has a potential, say `V''`. Now we draw an equivalent circuit diagram by connecting the different plates across the assumed potential diference.
.
As the equivalint circuit is a simple circuit of a combination of capacitors, the equivalent capacitance can be easily calculated as follows:
`1/(C_(eq))=1/C+2/(3C) =5/(3C)` or `C_(eq) =(3C)/5 =(3Aepsilon)/(5d)`.