A parallel plated capacitor has area 2 ~m ² separated by 3 dielectric slabs. Their relative permittivity is 2,3,6 and thickness is 0.4 ~mm , 0.6 ~mm , 1.2 ~mm , respectively. The capacitance is

Step-by-step solution

The given situation is shown below, d₁=0.4 ~mm =4 10⁻⁴ ~m d₂=0.6 ~mm =6 10⁻⁴ ~m d₃=1.2 ~mm =1.2 10⁻³ ~m The given arrangement will be equal to three capacitors connected in series. i.e. C₁= ₀ K₁ A d₁ , C₂= ₀ K₂ A d₂ and C₃= ₀ K₃ A d₃ Equivalent capacitance is given as, aligned & 1 C_ eq = 1 C₁ + 1 C₂ + 1 C₃ = 1 ₀ A ( d₁ K₁ + d₂ K₂ + d₃ K₃ ) & 1 C_ eq = 1 ₀ A [ 4 10⁻⁴ 2 + 6 10⁻⁴ 3 + 1.2 10⁻³ 6 ] & 1 C_ eq = 1 ₀ A [6 10⁻⁴ ] aligned C_ eq = ₀ A 6 10⁻⁴ = 8.85 10⁻¹² 2 6 10⁻⁴ =2.95 10⁻⁸ ~F