Identical metal plates are located in air at equal distance d from one another. The area of each plate is equl to `A` Evaluate the capacitance of the system between P and `Q` if the plates are interconnected as shown in.
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Explanation

As the battery is connected to `P` and `Q`, there is a charge `+Q` on plate `A` and `-Q` and on plate `B`. The charges on `A` nd `B` will induce a charge`-Q` on `C` and a change `+Q' on `D`, on `C` and `D` remains zero.
Consequently, charge `Q` on `A` is divided inti two paris: `+Q` on the upper side ofplate `A` and `-(Q-Q')` on the lower side. similarly, charge `-Q` on `B` is also on plate B.
Let us assmue that the potential of plater A be V, then the potential of plate B is zero. As plates `C` and `D` are connected, they will be at a common potential, say `V'. So the capacity of the system can be written as follows
`C=Q/(V_(+)-V_(-))= (Q)/(V)` (i)
We can write the equation for facing surfaces (1) and (2) as
`Q'=C_(0)(V-V')`
For facines (3) and (4)
`Q'=(C_(0)(V-V')`
For facing surfaces (5) and (6)
`Q=C_(0)(V'-0)`
From Eqs. (ii),(iii) and (iv)
`Q/V=3/1 C_(0)=3/2(epsilonA)/d `
and `Q'=(C_(0)V)/2=(epsilon_(0)AV)/(2d)`
Hence, the equivalent capacitent capacitance is
`C_(eq)=3/2(3epsilon_(0)A)/d.V`
charges on dirrernt surfaces are geven in the following table:
`|("Surface","(1),(5)","(2),(6)",(3),(4)),("Charge",-(epsi_(0)AV)/(2d),+(epsi_(0)AV)/(2d),(epsi_(0)AV)/(d),-(epsi_(0)AV)/(d))|`
Method 2: An equivalent circuit can drawn as shown in.
.
Hence, the equivalent capacitance is
`C_(eq)=3/2C_(0)=3/2(epsilon_(0)A)/d`
The charge supplied by the battery is
`Q=3/2(epsilon_(0)A)/(d).V`
`:. Q_(1)=Q(C_(0)/(C_(0)+C_(0)//2))=2/3Q` and `Q_(2)Q/3`
From these results, the charge on different surfaces can be calculated.