The parallel plates of a capacitor have an area 0.2 m² and 10^–2m apart. The original potential difference between them is 3000 V and it decreases to 1000 V when a dielectric sheet is inserted between the plates filling the full space between plates.
(take, $ε_{0}$ = 9 × 10^–12C²N^–1m^–2).
Match the column-I with Column-II and select the correct option among the codes given below.
COLUMN-I COLUMN-II
A) Original capacity in pF P) $3 \times 10^{5}$
B) Capacity after insertion of dielectric (pF) Q) $54 \times 10^{8}$
C) Permittivity of dielectric (C²N^–1m^–2) R) $27 \times 10^{- 12}$
D) Charge on each plate (C) S) $540$
E) Initial value of field between plates (V/m) T) $180$

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

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

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

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

answer is A.

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Detailed Solution

$\begin{array}{l} C_{0} = \frac{ε_{0} A}{d} = 180 pF; Q_{0} = C_{0} V_{0} = 54 \times 10^{- 8} C \\ C^{1} = \frac{Q_{0}}{V^{1}} = \frac{54 \times 10^{- 8}}{1000} = 540 pF \\ ∴ \frac{{kε}_{0} A}{d} = 540 \Rightarrow K = 3 \\ and ε = {kε}_{0} = 27 \times 10^{- 12} C^{2} N^{- 1} m^{- 2} \end{array}$

$E = \frac{V}{d} = \frac{3000}{10^{- 2}} = 3 \times 10^{- 5}$

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COLUMN-I	COLUMN-II
A) Original capacity in pF	P) $3 \times 10^{5}$
B) Capacity after insertion of dielectric (pF)	Q) $54 \times 10^{8}$
C) Permittivity of dielectric (C²N^–1m^–2)	R) $27 \times 10^{- 12}$
D) Charge on each plate (C)	S) $540$
E) Initial value of field between plates (V/m)	T) $180$