The space between the plates of a parallel plate capacitor is completely filled with two dielectric slabs of equal thickness d 2 . The first slab has a dielectric constant K₁ = 3 and the second slab has a dielectric constant K₂ = 6 . If the original capacitance with air between the plates was C₀ , what is the ratio of the new equivalent capacitance to the original capacitance?

Step-by-step solution

The original capacitance with air is C₀ = ₀ A d . The two dielectric slabs can be considered as two capacitors connected in series, each with thickness d 2 . Capacitance of the first part: C₁ = K₁ ₀ A d/2 = 2 3 ₀ A d = 6 C₀ Capacitance of the second part: C₂ = K₂ ₀ A d/2 = 2 6 ₀ A d = 12 C₀ The equivalent capacitance C_ eq for the series combination is given by: 1 C_ eq = 1 C₁ + 1 C₂ C_ eq = C₁ C₂ C₁ + C₂ = 6 C₀ 12 C₀ 6 C₀ + 12 C₀ = 72 C₀ 18 = 4 C₀ The ratio of the new capacitance to the original capacitance is 4:1