Question
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 $K_{1} = 1.25K_{2}$, the value of $K_{1}$ is:
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Solution
Using $C_{eq} = \frac{\varepsilon_{0}A}{\frac{t_{1}}{K_{1}} + \frac{t_{2}}{K_{2}} + \frac{t_{3}}{K_{3}}}$
here $C_{0} = \frac{\varepsilon_{0}A}{d},t_{1} = \frac{3d}{8},t_{2} = \frac{d}{2},t_{3} = \frac{d}{8}$
$K_{1} = K_{1},\ K_{2} = \frac{K_{1}}{1.25}$ and $K_{3} = 1$
Given $C_{\text{eq}\text{~}} = 2C_{0}$
$${\Rightarrow 2C_{0} = \frac{\varepsilon_{0}A}{\frac{3d}{8K_{1}} + \frac{d \times 1.25}{2K_{1}} + \frac{d}{8}} }{\Rightarrow \frac{2\varepsilon_{0}A}{d} = \frac{\varepsilon_{0}A}{\frac{3d}{8K_{1}} + \frac{d}{2K_{1}} \times \frac{5}{4} + \frac{d}{8}} }{\Rightarrow 2 = \frac{1}{\frac{3}{8K_{1}} + \frac{5}{8K_{1}} + \frac{1}{8}} \Rightarrow K_{1} = \frac{8}{3} = 2.66}$$
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