A parallel plate capacitor is having separation between plates 0.885 mm . It has a capacitance of 1μ F when the space between the plates is filled with an insulating material of resistivity 1times10 13 Ω m and resistance 17.7times10 14 Ω . Relative permittivity of the insulating material is α × 10 7 . The value of α is (Take permittivity of free space = 8.85times10 -12 F/m ).

Physics · Capacitors and R-C Circuits

JEE Main 2026 — 8 April, Evening Shift — Question 17

A parallel plate capacitor is having separation between plates $0.885\mathrm{mm}$ . It has a capacitance of $1\mu\mathrm{F}$ when the space between the plates is filled with an insulating material of resistivity $1\times10^{13}\Omega\mathrm{m}$ and resistance $17.7\times10^{14}\Omega$ . Relative permittivity of the insulating material is $\alpha \times 10^{7}$ . The value of $\alpha$ is (Take permittivity of free space $= 8.85\times10^{-12}\mathrm{F/m}$ ).

Answer: 2

Numerical answer — enter this value.

Step-by-step solution

$RC = \rho \kappa \epsilon_0$ ⇒ $\kappa = \frac{RC}{\rho\epsilon_0} = \frac{17.7\times10^{14}\times10^{-6}}{10^{13}\times8.85\times10^{-12}} = 2\times10^7$ , so $\alpha = 2$ .

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Exam: JEE Main 2026
Paper: 8 April, Evening Shift
Subject: Physics
Chapter: Capacitors and R-C Circuits
Topic: Effect of Dielectrics