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MHT CET 2026 (16 Apr, Shift 1) · Dielectrics

Dielectrics question from MHT CET 2026 (16 Apr, Shift 1)

Q22MHT CET 2026 · 16 APR · SHIFT 1Medium

Dielectrics

Two condensers of capacities, '2C' and 'C' are joined in parallel and charged upto potential 'V'. The battery is then disconnected and condenser of capacity 'C' is filled completely with a medium of dielectric constant 'K'. The potential difference across the capacitors in the second case is

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Answer: (a)

  1. Charge stored: Q=(2C+C)V=3CVQ = (2C + C)V = 3CV; battery removed, so QQ stays constant.
  2. New capacitance: C′=2C+KC=(K+2)CC' = 2C + KC = (K+2)C.
  3. V′=QC′=3CV(K+2)C=3VK+2V' = \frac{Q}{C'} = \frac{3CV}{(K+2)C} = \frac{3V}{K+2}
  4. So the answer is (a) 3VK+2\frac{3V}{K+2}.
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