The plates of a parallel plate capacitor are separated by d. Two slabs of different dielectric constant K1 and K2 with thickness…

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=1.25\,K2, the value of K1 is:

1. A. 2.66 (Correct answer)
2. B. 2.33
3. C. 1.60
4. D. 1.33

Correct Answer

Option A — 2.66

Detailed Solution & Explanation

The correct answer is 2.66.

Key Points

• The dielectrics fill $\dfrac{3d}{8}$ and $\dfrac{d}{2}$, leaving an air gap of $\dfrac{d}{8}$ ($t_3=\dfrac{d}{8}$, $K_3=1$).
• Series combination: $C_{eq}=\dfrac{\varepsilon_0 A}{\frac{t_1}{K_1}+\frac{t_2}{K_2}+\frac{t_3}{K_3}}$ with $K_2=\dfrac{K_1}{1.25}$.
• Setting $C_{eq}=2C_0=\dfrac{2\varepsilon_0 A}{d}$ gives $2=\dfrac{1}{\frac{3}{8K_1}+\frac{5}{8K_1}+\frac{1}{8}}$.
• Solving: $K_1=\dfrac{8}{3}=2.66$.