JEE Main 2026 Physics question
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 1×10¹³ Ωm 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)
Answer: 2
Solution
RC=(ρd/A)(ε₀ε_rA/d)=ρε₀ε_r, independent of geometry. So ε_r=RC/(ρε₀)=(17.7×10¹⁴×10⁻⁶)/(10¹³×8.85×10⁻¹²)=2×10⁷, giving α=2.