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JEE Main 2024 (8 Apr, Shift 2) · Dielectrics

Dielectrics question from JEE Main 2024 (8 Apr, Shift 2)

Q47JEE Main 2024 · 8 APR · SHIFT 2Difficult

Dielectrics

A capacitor has air as dielectric medium and two conducting plates of area 12 cm212\text{ cm}^2 and they are 0.6 cm0.6\text{ cm} apart. When a slab of dielectric having area 12 cm212\text{ cm}^2 and 0.6 cm0.6\text{ cm} thickness is inserted between the plates, one of the conducting plates has to be moved by 0.2 cm0.2\text{ cm} to keep the capacitance same as in previous case. The dielectric constant of the slab is: (Given ϵ0=8.834×10−12 F/m\epsilon_0 = 8.834 \times 10^{-12}\text{ F/m})

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Answer: (a)

  1. Slab of thickness t=0.6t = 0.6 cm in gap d′d': C=ϵ0Ad′−t+t/KC = \frac{\epsilon_0 A}{d' - t + t/K}.
  2. New gap d′=0.6+0.2=0.8d' = 0.6 + 0.2 = 0.8 cm; same CC needs 0.2+0.6K=0.60.2 + \frac{0.6}{K} = 0.6.
  3. 0.6K=0.4⇒K=1.5\frac{0.6}{K} = 0.4 \Rightarrow K = 1.5
  4. So the answer is (a) 1.50.
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