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MHT CET 2026 (20 Apr, Shift 1) · Dielectrics

Dielectrics question from MHT CET 2026 (20 Apr, Shift 1)

Q15MHT CET 2026 · 20 APR · SHIFT 1Medium

Dielectrics

In a parallel plate air capacitor of plate separation 'dd', a dielectric slab of thickness 'tt' is introduced between the plates. The capacitance becomes one-third of the original value. The dielectric constant of the slab will be

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Answer: (b)

  1. With a slab of thickness tt: C′=ϵ0Ad−t+tKC'=\frac{\epsilon_0A}{d-t+\frac{t}{K}}.
  2. C′=C3⇒d−t+tK=3dC'=\frac{C}{3} \Rightarrow d-t+\frac{t}{K}=3d
  3. tK=2d+t⇒K=t2d+t\frac{t}{K}=2d+t \Rightarrow K=\frac{t}{2d+t}
  4. So the answer is (b) t2d+t\frac{t}{2d+t}.
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