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MHT CET 2026 (15 May, Shift 2) · Dielectrics

Dielectrics question from MHT CET 2026 (15 May, Shift 2)

Q4MHT CET 2026 · 15 MAY · SHIFT 2Easy

Dielectrics

A parallel plate capacitor of plate separation 22 mm and plate area 4040 cm22 is filled with dielectric whose area is same as plates of capacitor. Thickness is 11 mm and dielectric constant is 55. New capacitance is (ϵ0\epsilon_0 = permittivity of free space)

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Answer: (b)

  1. Partial slab: C=ϵ0Ad−t+tKC = \frac{\epsilon_0 A}{d - t + \frac{t}{K}}.
  2. d−t+tK=1 mm+0.2 mm=1.2×10−3d - t + \frac{t}{K} = 1\text{ mm} + 0.2\text{ mm} = 1.2 \times 10^{-3} m, A=4×10−3 m2A = 4 \times 10^{-3}\ \text{m}^2.
  3. C=4×10−31.2×10−3ϵ0=10ϵ03C = \frac{4 \times 10^{-3}}{1.2 \times 10^{-3}}\epsilon_0 = \frac{10\epsilon_0}{3} F.
  4. So the answer is (b) 10ϵ03\frac{10\epsilon_0}{3} F.
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