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JEE Main 2022 (24 Jun, Shift 1) · Dielectrics

Dielectrics question from JEE Main 2022 (24 Jun, Shift 1)

Q67JEE Main 2022 · 24 JUN · SHIFT 1Medium

Dielectrics

A parallel plate capacitor is formed by two plates each of area 30π\pi cm2 separated by 1 mm. A material of dielectric strength 3.6 ×\times 107 Vm−1{}^{-1} is filled between the plates. If the maximum charge that can be stored on the capacitor without causing any dielectric breakdown is 7 ×\times 10−-6C, the value of dielectric constant of the material is:

[Use 14πε0=9×109\frac{1}{4\pi\varepsilon_0} = 9 \times 10^{9} Nm2 C−2{}^{-2}]

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Answer: (d)

  1. Field in dielectric E=QKAϵ0⇒K=QAϵ0EmaxE=\frac{Q}{KA\epsilon_0} \Rightarrow K=\frac{Q}{A\epsilon_0E_{max}}
  2. Aϵ0=30π×10−44π×9×109=56×10−13A\epsilon_0=\frac{30\pi\times10^{-4}}{4\pi\times9\times10^9}=\frac{5}{6}\times10^{-13}
  3. Aϵ0Emax=56×10−13×3.6×107=3×10−6A\epsilon_0E_{max}=\frac{5}{6}\times10^{-13}\times3.6\times10^7=3\times10^{-6}
  4. K=7×10−63×10−6≈2.33K=\frac{7\times10^{-6}}{3\times10^{-6}}\approx2.33. So the answer is (d).
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