Skip to content
dronacharya.ai
← JEE Main 2026 · 8 Apr · Shift 2
PhysicsInteger+4 / -1Question 46 of 75

JEE Main 2026 · 8 Apr · Shift 2, Physics question 46

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\times10^{13} Ωm and resistance 17.7×101417.7\times10^{14} Ω. Relative permittivity of the insulating material is α×107\alpha\times10^7. The value of α\alpha is ________.

(Take permittivity of free space =8.85×10−12=8.85\times10^{-12} F/m)

Answer and worked solution

Official answer: 2

Area. A=ρdR=1013×8.85×10−417.7×1014=5×10−6A=\dfrac{\rho d}R=\dfrac{10^{13}\times8.85\times10^{-4}}{17.7\times10^{14}}=5\times10^{-6} m².

Relative permittivity.
εr=Cdε0A=10−6×8.85×10−48.85×10−12×5×10−6=2×107.\varepsilon_r=\frac{Cd}{\varepsilon_0A}=\frac{10^{-6}\times8.85\times10^{-4}}{8.85\times10^{-12}\times5\times10^{-6}}=2\times10^7.

Answer: 2

Topics:DielectricsOhm's law and resistivity
← Previous questionNext question →