Four identical metal plates are located in air at equal distances `d` from one another. The area of each plate is equal to `S`. Find teh capacitance of the system between points `a` and `B` if the plates are interconnected as shown
(a) Fig `A`, (b) FIg `B`

Explanation

(a) In the given arrangment, we three capacitors of equal capacitance `C = (epsilon_(0) S)/(d)` and the first and third plates are at the same potential.
Hence, we can resolve the network into a simple from using series and parallel grouping of capacitors, as shown in fig.
Thus the equivalent capacitance
`C_(0) = ((C + C) C)/((C + C) + C) = (2)/(3) C`
(b) Let us mentally impart the chagres `+q` and `-q` to the plates 1 and 2 and then distribute them to other plates using charge conservation and electric induction. (Fig).
As the potential difference between the plates 1 and 2 is zero,
`(q_(1))/(C ) + (q_(2))/(C) - (q_(1))/(C) = 0`, (where `C = (epsilon_(0) S)/(d)`)
or, `q_(2) = 2 q_(3)`.
The potential difference between `A` and `B`,
`varphi = varphi_(A) = varphi_(B) = q_(2)//C`
Hence the sought capacitance,
`C_(0) = (q)/(varphi) = (q_(1) +q_(2))/(q_(2)//C) = (3 q_(1))/(2q_(1)//C) = (3)/(2) C = (3 epsilon_(0) S)/(2d)`