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JEE Main 2019 (12 Apr, Shift 1) · Dielectrics

Dielectrics question from JEE Main 2019 (12 Apr, Shift 1)

Q94JEE Main 2019 · 12 APR · SHIFT 1Medium

Dielectrics

Two identical parallel plate capacitors, of capacitance C each, have plates of area A, separated by a distance d. The space between the plates of the two capacitors, is filled with three dielectrics, of equal thickness and dielectric constants K1, K2 and K3. The first capacitor is filled as shown in fig.I, and the second one is filled as shown in fig II. If these two modified capacitors are charged by the same potential V, the ratio of the energy stored in the two, would be (E1 refers to capacitor (I) and E2 to capacitor (II)):

Figure

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Answer: (d)

  1. Fig I (layers stacked between plates) is series: C1=3CK1K2K3K1K2+K2K3+K3K1C_1=\frac{3CK_1K_2K_3}{K_1K_2+K_2K_3+K_3K_1}
  2. Fig II (side by side) is parallel: C2=C(K1+K2+K3)3C_2=\frac{C(K_1+K_2+K_3)}{3}
  3. Same VV, so E1E2=C1C2=9K1K2K3(K1+K2+K3)(K2K3+K3K1+K1K2)\frac{E_1}{E_2}=\frac{C_1}{C_2}=\frac{9K_1K_2K_3}{(K_1+K_2+K_3)(K_2K_3+K_3K_1+K_1K_2)}
  4. So the answer is (d).
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