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MHT CET 2024 (16 May, Shift 1) · Dielectrics

Dielectrics question from MHT CET 2024 (16 May, Shift 1)

Q45MHT CET 2024 · 16 MAY · SHIFT 1Medium

Dielectrics

Two identical capacitors A and B are connected in series to a battery of E.M.F., 'E'. Capacitor B contains a slab of dielectric constant $K,\ Q_{A}$ and $Q_{B}$ are the charges stored in A and B. When the dielectric slab is removed, the corresponding charges are $Q'_{A}$ and $Q'_{B}$ . Then

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Answer: (d)

In series $Q_A = Q_B$ always.
With slab: $C_{eq} = \frac{KC \cdot C}{(K+1)C} = \frac{KC}{K+1}$ , so $Q = \frac{KCE}{K+1}$ .
Without slab: $C_{eq} = \frac{C}{2}$ , so $Q' = \frac{CE}{2}$ .
$\frac{Q'_B}{Q_B} = \frac{Q'_A}{Q_A} = \frac{K+1}{2K}$
So the answer is (d).

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