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NEET 2016 (Phase 2) · Dielectrics

Dielectrics question from NEET 2016 (Phase 2)

Q102NEET 2016 · PHASE 2Difficult

Dielectrics

A parallel-plate capacitor of area A, plate separation d and capacitance C is filled with four dielectric materials having dielectric constants k1, k2, k3 and k4 as shown in the figure. If a single dielectric material is to be used to have the same capacitance C in this capacitor, then its dielectric constant k is given by

Figure

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Answer: (c)

  1. Top half (thickness d/2d/2): k1,k2,k3k_1, k_2, k_3 side by side in parallel, each taken as area A/3A/3: Ctop=2ϵ0Ad⋅k1+k2+k33C_{top} = \frac{2\epsilon_0 A}{d} \cdot \frac{k_1 + k_2 + k_3}{3}.
  2. Bottom half: C4=2ϵ0Ak4dC_4 = \frac{2\epsilon_0 A k_4}{d}, in series with the top.
  3. 1C=d2ϵ0A[3k1+k2+k3+1k4]\frac{1}{C} = \frac{d}{2\epsilon_0 A}\left[\frac{3}{k_1 + k_2 + k_3} + \frac{1}{k_4}\right] with C=kϵ0AdC = \frac{k\epsilon_0 A}{d}
  4. 2k=3k1+k2+k3+1k4\frac{2}{k} = \frac{3}{k_1 + k_2 + k_3} + \frac{1}{k_4}, so the answer is (c).
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