A parallel plate capacitor is having separation between plates 0.885 mm. It has a capacitance of 1 when the space between the plates is filled with an insulating material of resistivity 1× 10¹³ Ω m and resistance 17.7× 10¹⁴ Ω. Relative permittivity of the insulating material is α× 10⁷. The value of α is ______. (Take permittivity of free space =8.85× 10⁻¹² F/m) Numerical answer type. Work it out, then open the answer.

Answer: 2. R = (ρ d)/(A) ⇒ A = (ρ d)/(R) = 10¹³×0.885×10⁻³17.7×10¹⁴ = 5×10⁻⁶ m² C = (Kε₀A)/(d) ⇒ K = (Cd)/(ε₀A) K = 10⁻⁶×0.885×10⁻³8.85×10⁻¹²×5×10⁻⁶ = 2×10⁷ So α = 2.

A parallel plate capacitor is having separation between plates 0.885… | JEE Main 2026 (8 Apr, Shift 2)

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Capacitors in DC Circuits & RC

A parallel plate capacitor is having separation between plates 0.885 mm. It has a capacitance of $1\ \mu\text{F}$ when the space between the plates is filled with an insulating material of resistivity $1\times 10^{13}\ \Omega\text{ m}$ 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 $=8.85\times 10^{-12}\text{ F/m}$ )

Numerical answer type. Work it out, then open the answer.

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Answer: 2

$R = \dfrac{\rho d}{A} \Rightarrow A = \dfrac{\rho d}{R} = \dfrac{10^{13}\times0.885\times10^{-3}}{17.7\times10^{14}} = 5\times10^{-6}\text{ m}^2$
$C = \dfrac{K\varepsilon_0A}{d} \Rightarrow K = \dfrac{Cd}{\varepsilon_0A}$
$K = \dfrac{10^{-6}\times0.885\times10^{-3}}{8.85\times10^{-12}\times5\times10^{-6}} = 2\times10^{7}$
So $\alpha = 2$ .

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