Question

A parallel-plate capacitor consists of two square metal plates 500 mm on a side separated by 10 mm. A slab of Teflon (ε_r = 2.0) 6 mm thick is placed on the lower plate leaving an air gap 4 mm thick between it and the upper plate. If 100 V is applied across the capacitor, find the electric field (E₀) in the air, electric field E_t in Teflon, flux density Da in air, flux density D_t in Teflon and potential difference V_t across Teflon slab.

Answer 1

C = ε₀A/(d₁/ε_e1 + d₂/ε_r2) = (8.854 x 10^-12 x (0.5)²)/((6 x 10^-3/2) + (4 x 10^-3/1)) = 3.16 x 10^-10F

 = CV = 3.16 × 10⁻¹⁰ × 100 = 31.6 × 10⁻⁹ C 

D = Q/A = 31.6 × 10⁻⁹/(0.5)² = 1.265 × 10⁻⁷ C/m² 

The charge or flux density will be the same in both media i.e. D_a = D_t = D 

In air, E₀ = D/ε₀ = 1.265 × 10⁻⁷/8.854 × 10⁻¹² = 14,280 V/m 

In Teflon, E_t = D/ε₀ε_r = 14,280/2 = 7,140V/m 

V_t = E_t × d_t = 7,140 × 6 × 10⁻³ = 42.8 V