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JEE Main 2013 (25 Apr, Shift 1) · Dielectrics

Dielectrics question from JEE Main 2013 (25 Apr, Shift 1)

Q109JEE Main 2013 · 25 APR · SHIFT 1Medium

Dielectrics

A parallel plate capacitor having a separation between the plates d, plate area A and material with dielectric constant K has capacitance C0C_0. Now one-third of the material is replaced by another material with dielectric constant 2K, so that effectively there are two capacitors one with area 13\frac{1}{3} A, dielectric constant 2K and another with area 23\frac{2}{3} A and dielectric constant K. If the capacitance of this new capacitor is C then CC0\frac{C}{C_0} is

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Answer: (b)

  1. Side-by-side parts are in parallel: C=ϵ0d(2K⋅A3+K⋅2A3)=43⋅Kϵ0AdC=\frac{\epsilon_0}{d}\left(2K\cdot\frac{A}{3}+K\cdot\frac{2A}{3}\right)=\frac{4}{3}\cdot\frac{K\epsilon_0A}{d}
  2. C0=Kϵ0AdC_0=\frac{K\epsilon_0A}{d}
  3. So the answer is (b) 43\frac{4}{3}.
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