Five identical capacitor plates each of area A, are arranged such that adjacent plates are at d distance apart. Plates are connected to a source of e.m.f $ε$ as shown in fig. Match the following:
COLUMN - I COLUMN - II
A) Charge on plate 1 P) -2 $ε_{0}$ AV / d
B) Charge on plate 4 Q) + $ε_{0}$ AV / d
C) Potential difference between plates 2 and 3 R) zero
D) Potential difference between plates 1 and 5 S) V

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$A \to P; B \to Q; C \to S; D \to R$

$A \to S; B \to P; C \to R; D \to Q$

$A \to Q; B \to P; C \to S; D \to R$

$A \to P; B \to S; C \to R; D \to Q$

answer is A.

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Detailed Solution

Potential difference across each capacitor is V hence charge on each capacitor = CV $= \frac{{Aε}_{0} V}{d}$

Therefore charge on plate 1 is $= \frac{{Aε}_{0} V}{d}$

And charge on plate 4 is $= \frac{- 2 {Aε}_{0} V}{d}$

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COLUMN - I		COLUMN - II
A)	Charge on plate 1	P)	-2 $ε_{0}$ AV / d
B)	Charge on plate 4	Q)	+ $ε_{0}$ AV / d
C)	Potential difference between plates 2 and 3	R)	zero
D)	Potential difference between plates 1 and 5	S)	V