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MHT CET 2026 (19 Apr, Shift 1) · Dielectrics

Dielectrics question from MHT CET 2026 (19 Apr, Shift 1)

Q7MHT CET 2026 · 19 APR · SHIFT 1Easy

Dielectrics

A slab of material of dielectric constant KK has the same area as the plates of a parallel plate capacitor but has a thickness (4/5)d(4/5)d, where dd is the separation of the plates. The capacitance in the presence and absence of dielectric are CC and C0C_0 respectively. The ratio (C/C0)(C/C_0) is

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Answer: (c)

  1. C=ϵ0Ad−t+tKC = \frac{\epsilon_0 A}{d - t + \frac{t}{K}} with t=4d5t = \frac{4d}{5}.
  2. d−t+tK=d5+4d5K=d(K+4)5Kd - t + \frac{t}{K} = \frac{d}{5} + \frac{4d}{5K} = \frac{d(K+4)}{5K}.
  3. CC0=5KK+4\frac{C}{C_0} = \frac{5K}{K+4}.
  4. So the answer is (c).
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