Voltage rating of a parallel plate capacitor is 500 V. Its dielectric can withstand a maximum electric field of $10^{6} \frac{\mathrm{~V}}{\mathrm{~m}}$. The plate area is $10^{-4} \mathrm{~m}^{2}$. What is the dielectric constant if the capacitance is 15 pF? (given $\epsilon_{0}=8.86 \times 10^{-12} \mathrm{C}^{2} / \mathrm{Nm}^{2}$)

JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Capacitance

Voltage rating of a parallel plate capacitor is 500 V. Its dielectric can withstand a maximum electric field of

1 0^{6} \frac{V}{m}

. The plate area is

1 0^{- 4} m^{2}

. What is the dielectric constant if the capacitance is 15 pF? (given

ϵ_{0} = 8.86 \times 1 0^{- 12} C^{2} / Nm^{2}

)

Answer: (D)

The plate separation is fixed by the breakdown limit, where the dielectric just holds off the rated voltage:

d = \frac{V _{m a x}}{E _{m a x}} = \frac{500}{1 0 ^{6}} = 5 \times 1 0^{- 4}

m. From

C = \frac{k ϵ _{0} A}{d}

, solve for the dielectric constant,

k = \frac{C d}{ϵ _{0} A} = \frac{15 \times 1 0 ^{- 12} \times 5 \times 1 0 ^{- 4}}{8.86 \times 1 0 ^{- 12} \times 1 0 ^{- 4}}

. The powers of ten cancel to leave

k = \frac{75}{8.86} \approx 8.5

JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Single correct option