Capacitors

JEE Main level, single correct, medium

107 of 180

Question 107 of 180

Voltage rating of a parallel plate capacitor is 500V.500V. Its dielectric can withstand a maximum electric field of 106V/m.{10}^{6}V/m. The plate area is 10−4m2{10}^{-4}{m}^{2} . What is the dielectric constant if the capacitance is 15pF15pF ?
(givenϵ0=8.86×10−12C2/Nm2)\left(given{ϵ}_{0}=8.86\times {10}^{-12}{C}^{2}/N{m}^{2}\right)

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Answer: (B) 8.58.5

Electric field inside capacitor =Vd=\dfrac{V}{d}
=VdAkϵ0Akϵ0=VCAkϵ0(∵kAϵ0d=C)=\dfrac{V}{d}\dfrac{Ak{\epsilon }_{0}}{Ak{\epsilon }_{0}}=\dfrac{VC}{Ak{\epsilon }_{0}}\left(∵\dfrac{kA{\epsilon }_{0}}{d}=C\right)
106=500×15×10−1210−4×k×8.85×10−12{10}^{6}=\dfrac{500\times 15\times {10}^{-12}}{{10}^{-4}\times k\times 8.85\times {10}^{-12}}
k=15×5008.85×10−4×106=758.85k=\dfrac{15\times 500}{8.85\times {10}^{-4}\times {10}^{6}}=\dfrac{75}{8.85}
k=8.47≈8.5k=8.47\approx 8.5

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