The space between the plates of a parallel-plate capacitor of capacitance C (without any dielectric) is now filled with three dielectric slabs of dielectric constants K₁ = 2, K₂ = 3, and K₃ = 5 (as shown in the figure). If new capacitance is nC/3 then the value of n i $s _______.$

Question

G1 Physics · Accepted Answer

Let the original air-capacitor capacitance be C = ε₀A/d. The arrangement can be treated as two parallel branches, each branch corresponding to area A/2. Right branch: The upper half has dielectric constant K₁ = 2, area A/2 and thickness d/2. Its capacitance is C₁ = 2ε₀(A/2)/(d/2) = 2C. The lower half has dielectric constant K₃ = 5 with the same area and thickness. Its capacitance is C₃ = 5ε₀(A/2)/(d/2) = 5C. These two are in series: C(right) = (2C×5C)/(2C+5C) = 10C/7. Left branch: The upper hal…

The space between the plates of a parallel-plate capacitor of capacitance C (without any dielectric) is now filled with three dielectric slabs of dielectric constants K₁ = 2, K₂ = 3, and K₃ = 5 (as shown in the figure). If new capacitance is nC/3 then the value of n i s _______.