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 figure). If new capacitance is n 3 C then the value of n is _ _ _ _ .

Step-by-step solution

The initial capacitance is C = ₀ A d . From the figure, the capacitor is divided into three parts. The bottom half is divided into two capacitors C₂ and C₃ in parallel, each with area A/2 and thickness d/2 . C₂ = K₂ ₀ (A/2) d/2 = K₂ ₀ A d = 3C C₃ = K₃ ₀ (A/2) d/2 = K₃ ₀ A d = 5C The equivalent capacitance of the bottom part is C_ eq, bottom = C₂ + C₃ = 3C + 5C = 8C . The top half is a single capacitor C₁ with area A and thickness d/2 . C₁ = K₁ ₀ A d/2 = 2K₁ ₀ A d = 2(2)C = 4C . The total capacitance C' is the serie