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.

 

 

 

 

 

	Column I		Column I
i)	Charge on plate 1	p)	$-2{\epsilon }_{0}AV/d$
ii)	Charge on plate 4	q)	$+{\epsilon }_{0}AV/d$
iii)	Potential difference between plates 2 and 3	r)	Zero
iv)	Potential difference  between plates 1 and 5	s)	V

These five plates constitute four identical capacitors in parallel, each of capacity ${\epsilon }_{0}A/d.$ Now, 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 $q_1=+\left(\frac{{\epsilon }_{0}AV}{d}\right)$

So, i-b.
However, plate 4 is connected to the negative terminal of the battery and is common to the two identical capacitors in parallel. So,  $q_4=-\left(\frac{2{\epsilon }_{0}AV}{d}\right)$
Hence, ii-a.
From the circuit, it is obvious that between the plates 2 and 3, the battery is connected, so the potential difference will be V. Plates 1 and 5 get connected through connecting wire, so the potential difference between I and 5 is zero.