A parallel plate capacitor is having separation between plates 0.885 mm . It has a capacitance of $1 \mu \mathrm{~F}$ when the space between the plates is filled with an insulating material of resistivity $1 \times 10^{13} \Omega \mathrm{~m}$ and resistance $17.7 \times 10^{14} \Omega$. Relative permittivity of the insulating material is $\alpha \times 10^{7}$. The value of $\alpha$ is ______. (Take permittivity of free space $=8.85 \times 10^{-12} \mathrm{~F} / \mathrm{m}$)

JEE Main · 2026 · 08 Apr 2026 · Shift 2 · Capacitance

A parallel plate capacitor is having separation between plates 0.885 mm . It has a capacitance of

1 μ F

when the space between the plates is filled with an insulating material of resistivity

1 \times 1 0^{13} Ω m

and resistance

17.7 \times 1 0^{14} Ω

. Relative permittivity of the insulating material is

α \times 1 0^{7}

. The value of

α

is ______.
(Take permittivity of free space

= 8.85 \times 1 0^{- 12} F / m

)

Answer: 2

For the dielectric-filled capacitor,

C = \frac{K ϵ _{0} A}{d}

and its leakage resistance is

R = \frac{ρ d}{A}

, so the product

R C = ρ K ϵ_{0}

is independent of the geometry. Hence

K = \frac{R C}{ρ ϵ _{0}} = \frac{17.7 \times 1 0 ^{14} \times 1 0 ^{- 6}}{1 0 ^{13} \times 8.85 \times 1 0 ^{- 12}} = 2 \times 1 0^{7}

. Writing

K = α \times 1 0^{7}

, the value of

α

2

JEE Main · 2026 · 08 Apr 2026 · Shift 2 · Integer answer