Two parallel plates with dielectric placed between the plates are as shown in figure. The resultant capacity of capacitor will be [A = area of plate. t₁, t₂ and t₃ are thickness of dielectric slabs, k₁, k₂ and k₃ are dielectric constants.] (a) (Aε₀)/([(t₁+t₂+t₃)/(k₁+k₂+k₃)]) (b) (Aε₀(k₁k₂k₃))/(t₁t₂t₃) (c) Aε₀[(k₁)/(t₁) + (k₂)/(t₂) + (k₃)/(t₃)] (d) (Aε₀)/([(t₁)/(k₁) + (t₂)/(k₂) + (t₃)/(k₃)])

Answer: (d). Slabs placed one after another between X and Y are in series. 1/C=(t₁)/(k₁Aε₀)+(t₂)/(k₂Aε₀)+(t₃)/(k₃Aε₀) C=(Aε₀)/((t₁)/(k₁)+(t₂)/(k₂)+(t₃)/(k₃)). So the answer is (d).

Two parallel plates with dielectric placed between the plates are as… | MHT CET 2021 (23 Sep, Shift 1)

Q81MHT CET 2021 · 23 SEP · SHIFT 1Medium

Dielectrics

Two parallel plates with dielectric placed between the plates are as shown in figure. The resultant capacity of capacitor will be [A = area of plate. $t_1, t_2$ and $t_3$ are thickness of dielectric slabs, $k_1, k_2$ and $k_3$ are dielectric constants.]

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Answer: (d)

Slabs placed one after another between X and Y are in series.
$\frac{1}{C}=\frac{t_1}{k_1A\epsilon_0}+\frac{t_2}{k_2A\epsilon_0}+\frac{t_3}{k_3A\epsilon_0}$
$C=\frac{A\epsilon_0}{\frac{t_1}{k_1}+\frac{t_2}{k_2}+\frac{t_3}{k_3}}$ . So the answer is (d).

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Dielectrics question from MHT CET 2021 (23 Sep, Shift 1)

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