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JEE Main 2019 (8 Apr, Shift 1) · Dielectrics

Dielectrics question from JEE Main 2019 (8 Apr, Shift 1)

Q96JEE Main 2019 · 8 APR · SHIFT 1Medium

Dielectrics

Voltage rating of a parallel plate capacitor is 500V. Its dielectric can withstand a maximum electric field of 106 V/m. The plate area is 10–4 m2. What is the dielectric constant if the capacitance is 15 pF? (given ε\varepsilon0 = 8.86 × 10–12 C2/Nm2)

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Answer: (a)

  1. Minimum separation d=VEmax=500106=5×10−4 md=\frac{V}{E_{max}}=\frac{500}{10^6}=5\times10^{-4}\text{ m}
  2. C=Kϵ0Ad⇒K=Cdϵ0A=15×10−12×5×10−48.86×10−12×10−4C=\frac{K\epsilon_0A}{d} \Rightarrow K=\frac{Cd}{\epsilon_0A}=\frac{15\times10^{-12}\times5\times10^{-4}}{8.86\times10^{-12}\times10^{-4}}
  3. K≈8.47K\approx8.47. So the answer is (a) 8.5.
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