A parallel-plate capacitor has plate separation 0.885 mm. When an insulating material fills the gap, the capacitance is 1 μF. The material has resistivity 1×10¹³ Ω·m, and the measured resistance between the plates is 17.7×10¹⁴ Ω. If the relative permittivity is α×10⁷, find α. Take ε₀ = 8.85×10⁻¹² F·m⁻¹.
Physics · Capacitor · capacitor with dielectric
JEE Mains 2026 8 April Evening Shift

Question

A parallel-plate capacitor has plate separation 0.885 mm. When an insulating material fills the gap, the capacitance is 1 μF. The material has resistivity 1×10¹³ Ω·m, and the measured resistance between the plates is 17.7×10¹⁴ Ω. If the relative permittivity is α×10⁷, find α. Take ε₀ = 8.85×10⁻¹² F·m⁻¹.

Physics · Capacitor · capacitor with dielectric

JEE Mains 2026 8 April Evening Shift

G1 Physics · Accepted Answer

For a parallel-plate sample of thickness d and area A, R = ρd ÷ A, and C = Kε₀A ÷ d, where K is the relative permittivity. Multiplying: RC = [ρd ÷ A][Kε₀A ÷ d] = ρKε₀. Therefore, K = RC ÷ (ρε₀). Substitute: R = 17.7×10¹⁴ Ω, C = 1 μF = 1×10⁻⁶ F, ρ = 1×10¹³ Ω·m, ε₀ = 8.85×10⁻¹² F·m⁻¹. Thus, K = [17.7×10¹⁴×10⁻⁶] ÷ [10¹³×8.85×10⁻¹²]. The numerator is 17.7×10⁸. The denominator is 8.85×10¹ = 88.5. Hence, K = 17.7×10⁸ ÷ 88.5 = 2×10⁷. Given, K = α×10⁷, we obtain α = 2.