Five identical conducting plates, `1, 2,3,4` and `5` are fixed parallel pltes equidistant from each other (see figure). A conductor connects plates 2 and 5 while another conductor joins 1 and 3. The junction of 1 and 3 and the plate 4 are connected to a source of constant emf `V_0`. Find

(a) the effective capacity of the system between the terminals of source.
(b) the charges on the plates 3 and 5. Given, `d =` distance between any two successive plates and `A = ` area of either face of each plate.

Explanation

i. The equivalent circuit is shown in. The system consists of four capacitors, i.e. `C_(12), C_(32,) C_(34)`, and `C_(54)`. The capacity of each capacitor is `(Kepsilon_(0)A//d)=C_(0)`

The capacitors `C_(12)` and `C_(32)` are in parallel, and their capacity is `C_(0)+C_(0)=2C_(0)`. The capacitor `C_(54)`. is series with the parallel combination of `C_(12)` and`C_(32)`. Hence, the resultant capacity will be.
`C_(i)=(C_(0)xx1C_(0))/(C_(0)+2C_(0))=(2C_(0))/3`
Further, `C_(34)` is again in parallel with the combination `C_(12),C_(32)`, and `C_(54)`. Hence, the effective capacity is
`C_(eff.)=C_(0)+(C_(0)xx2C_(0))/(C_(0)+2C_(0))=5/3C_(0)5/3Kepsilon_(0)A/d`
ii. Charge on plate 5 is equal to the charge on the upper half of the parallel combination. so
`Q_(5)=V_(0)(2/3C_(0))=2/3(Kepsilon_(0)AV_(0))/d`
Charge on palte 3 on the surface facing 2 is equal to the potential differece across (3-2), i.e.,
`C_(0)=V_(0)C_(0)/(C_(0)+2C_(0))C_(0)=Kepsilon_(0)(AV_(0))/(3d)`
Net charge on plate 3 is
`Q_(3)=(Kepsilon_(0)AV_(0))/d+Kepsilon_(0)(AV_(0))/(3d)=Kepsilon_(0)AV_(0)/d[1+1/3]`
`=4/3K epsilon_(0) A/dV_(0)`.