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:

• Correct answer
  A

  2.66
• B

  2.33
• C

  1.60
• D

  1.33

Solution

$C_1=\varepsilon_0 A / d$
$\begin{aligned} & \mathrm{C}_2=\frac{\varepsilon_0 \mathrm{~A}}{\frac{3 \mathrm{~d}}{8 \mathrm{~K}_1}+\frac{\mathrm{d}}{2 \mathrm{~K}_2}+\left(\mathrm{d}-\frac{3 \mathrm{~d}}{8}-\frac{\mathrm{d}}{2}\right)} \\ & \mathrm{C}_2=\frac{\varepsilon_0 \mathrm{~A} / \mathrm{d}}{\frac{3}{8 \mathrm{~K}_1}+\frac{1}{2 \mathrm{~K}_2}+\frac{1}{8}} \\ & =\frac{\varepsilon_0 \mathrm{~A} / \mathrm{d}}{\frac{3}{8 \mathrm{~K}_1}+\frac{1}{2\left(\frac{4}{5} \mathrm{~K}_1\right)}+\frac{1}{8}} \\ & \mathrm{C}_2=2 \mathrm{C}_1 \\ & \frac{\varepsilon_0 \mathrm{~A} / \mathrm{d}}{\frac{3}{8 \mathrm{~K}_1}+\frac{5}{8 \mathrm{~K}_1}+\frac{1}{8}}=2 \varepsilon_0 \mathrm{~A} / \mathrm{d} \\ & \Rightarrow \frac{1}{\frac{1}{\mathrm{~K}_1}+\frac{1}{8}}=2 \Rightarrow \frac{1}{\mathrm{~K}_1}+\frac{1}{8}=\frac{1}{2} \Rightarrow \mathrm{~K}_1=\frac{8}{3}=2.66\end{aligned}$

Hence, the correct answer is option: 'A'.