Question

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 source of emf $V$ as shown in figure. Match the following?

Answer 1

These five plates constitute $(5-1)=4$ identical capacitors in parallel, each of capacity $C=\frac{A{ϵ}_0}{d}$.
As plate 1 is connected to the positive terminal of the battery and is a part of one capacitor only, so charge on it is $q_1=CV=\frac{AV{ϵ}_0}{d}$
thus, $A-2$
The plate 4 is connected to the negative terminal of the battery and is common to the two identical capacitors in parallel. so charge on it is $q_4=-2CV=-\frac{2AV{ϵ}_0}{d}$
So $B-1$
From the circuit, plate 2 and 3 will make a capacitor and it is connected across battery. So the potential across plate 2 and 3 will be V.
So $C-4$
Plates 1 and 5 get connected through connecting wire, so the potential difference between 1 and 5 is zero.
Thus, $D-3$