Calculate the ratio of the capacitance of two capacitors filled...

Here, ${C}_{1} = \dfrac{{ε}_{0} {K}_{1} \text{A}}{2 \text{d}} + \dfrac{{ε}_{0} {K}_{2} \text{A}}{2 \text{d}}$

= \dfrac{{ε}_{0} \text{KA}}{2 \text{d}} + \dfrac{{ε}_{0} \text{KA}}{8 \text{d}}\left(\begin{cases}\text{Given :} & {K}_{1} = \text{K} \\ \text{and} & {K}_{2} = \dfrac{\text{K}}{4}\end{cases}\right)

$= \dfrac{{ε}_{0} \text{KA}}{\text{d}} \left(\dfrac{1}{2} + \dfrac{1}{8}\right) = \dfrac{{ε}_{0} \text{KA}}{\text{d}} \left(\dfrac{4 + 1}{8}\right)<br/><br/> =\dfrac{{5 ε}_{0} \text{KA}}{8 \text{d}} ...... (i)<br/><br/>and${C}_{2} = $\dfrac{{ε}_{0} \text{A}}{\dfrac{\text{d}}{2 {K}_{1}} + \dfrac{\text{d}}{2 {K}_{2}}} = \dfrac{{ε}_{0} \text{A}}{\dfrac{\text{d}}{2 \text{K}} + \dfrac{\text{d}}{\dfrac{2 \text{K}}{4}}}<br/><br/>$= \dfrac{\left(2 \text{K}\right) {ε}_{0} \text{A}}{5 \text{d}} .... ($ii)<br/><br/>Dividing eqs. (i) by (ii)<br/><br/>$\dfrac{{C}_{1}}{{C}_{2}} = \dfrac{\dfrac{5 {ε}_{0} \text{KA}}{8 \text{d}}}{\dfrac{2 \text{K} {ε}_{0} \text{A}}{5 \text{d}}} = \dfrac{5 \times 5}{2 \times 8} = \dfrac{2 5}{1 6}