The plates of a parallel plate capacitor are separated by d. Two slabs of different dielectric constant K₁ and K₂ with thickness 3/8d and d/2, respectively are inserted in the capacitor. Due to this, the capacitance becomes two times larger than when there is nothing between the plates. If K₁=1.25K…

Question

The plates of a parallel plate capacitor are separated by d. Two slabs of different dielectric constant K₁ and K₂ with thickness 3/8d and d/2, respectively are inserted in the capacitor. Due to this, the capacitance becomes two times larger than when there is nothing between the plates. If K₁=1.25K₂, the value of K₁ is:

Accepted Answer

(1) 2.66. The slabs fill 3/8d+1/2d=7/8d, so an air gap of d/8 remains. Layers across the gap act in series: C=(ε₀A)/((3d)/(8K₁)+d/(2K₂)+d/8), and C=2C₀=(2ε₀A)/d. So 3/(8K₁)+1/(2K₂)+1/8=1/2. K₂=(K₁)/(1.25), so 1/(2K₂)=(1.25)/(2K₁)=5/(8K₁). 3/(8K₁)+5/(8K₁)=1/(K₁)=1/2-1/8=3/8. K₁=8/3=2.67, printed as 2.66.

Step-by-step solution

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