Q. The plates of a parallel plate capacitor are separated by $d$. Two slabs of different dielectric constant $K_1$ and $K_2$ with thickness $\frac{3}{8}d$ and $\frac{d}{2}$, respectively are inserted in the capacitor. Due to this, the capacitance becomes two times larger than when there is nothing between the plates. If $K_1 = 1.25\,K_2$, the value of $K_1$ is:
A) 2.66
B) 2.33
C) 1.60
D) 1.33
Explanation
[Solution by Claude - not from source] Air gap $=d-\frac38d-\frac12d=\frac d8$. $\frac1C\propto\frac18+\frac{3}{8K_1}+\frac{1}{2K_2}=\frac12\Rightarrow\frac{3}{K_1}+\frac{4}{K_2}=3$. With $K_2=0.8K_1$: $\frac{3}{K_1}+\frac{5}{K_1}=3\Rightarrow K_1=\frac83\approx2.66$. Answer (1).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.