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JEE MAIN 2026 · Electrostatics

JEE Main 2026 Session 2 Question Paper: 8 Apr, Shift 2 (B.E./B.Tech), question 46

Updated
Numerical answer+4 / -1Answer matches NTA’s final key

A parallel plate capacitor is having separation between plates 0.885 mm. It has a capacitance of 1 μF when the space between the plates is filled with an insulating material of resistivity 1×10131 \times 10^{13} Ωm and resistance 17.7×101417.7 \times 10^{14} Ω. Relative permittivity of the insulating material is α×107\alpha \times 10^7. The value of α\alpha is ______.

(Take permittivity of free space =8.85×10−12= 8.85 \times 10^{-12} F/m)

Your answer

Solution

Solution

For the same slab, C=ϵ0ϵrAdC = \dfrac{\epsilon_0\epsilon_r A}{d} and R=ρdAR = \dfrac{\rho d}{A}. Multiplying removes the geometry:

RC=ρ ϵ0ϵr  ⇒  ϵr=RCρ ϵ0=17.7×1014×10−61013×8.85×10−12=17.7×10888.5=2×107.RC = \rho\,\epsilon_0\epsilon_r \;\Rightarrow\; \epsilon_r = \frac{RC}{\rho\,\epsilon_0} = \frac{17.7 \times 10^{14} \times 10^{-6}}{10^{13} \times 8.85 \times 10^{-12}} = \frac{17.7 \times 10^{8}}{88.5} = 2 \times 10^7.

(The plate separation is not needed.)

Answer: α=2\alpha = 2.

Answer checked against the official answer key. The solution is our own working.

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