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JEE Main 2021 (26 Aug, Shift 2) · Dielectrics

Dielectrics question from JEE Main 2021 (26 Aug, Shift 2)

Q82JEE Main 2021 · 26 AUG · SHIFT 2Medium

Dielectrics

A parallel plate capacitor with plate area A has separation d between the plates. Two dielectric slabs of dielectric constant K1 and K2 of same area A/2 and thickness d/2 are inserted in the space between the plates. The capacitance of the capacitor will be given by:

Figure

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Answer: (a)

  1. Left half (area A2\frac{A}{2}, air, gap dd): ϵ0A2d\frac{\epsilon_0A}{2d}.
  2. Right half: K1,K2K_1,K_2 layers of d2\frac{d}{2} in series: ϵ0A/2d2K1+d2K2=ϵ0Ad⋅K1K2K1+K2\frac{\epsilon_0A/2}{\frac{d}{2K_1}+\frac{d}{2K_2}}=\frac{\epsilon_0A}{d}\cdot\frac{K_1K_2}{K_1+K_2}
  3. The two halves are in parallel: C=ϵ0Ad(12+K1K2K1+K2)C=\frac{\epsilon_0A}{d}\left(\frac{1}{2}+\frac{K_1K_2}{K_1+K_2}\right)
  4. So the answer is (a).
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