(3)

Capacitance without dielectric

$C_0={\epsilon }_0\frac{A}{d}$

Capacitance with first dielectric slab,

$C_1=\frac{K_1{\epsilon }_{0}A}{\frac{3}{8}d}=\frac{8K_1{\epsilon }_{0}A}{3d}$

Capacitance with second dielectric slab,

$C_2=\frac{K_2{\epsilon }_{0}A}{\frac{d}{2}}=\frac{2K_2{\epsilon }_{0}A}{d}$

Capacitance with air,

$C_3=\frac{{\epsilon }_{0}A}{(d-\frac{3d}{8}-\frac{d}{2})}=\frac{{\epsilon }_{0}A\times 8}{d}$

Equivalent Capacitance, $\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}$

or    $\frac{1}{C_{eq}}=\frac{3d}{8K_1{\epsilon }_{0}A}+\frac{d}{2K_2{\epsilon }_{0}A}+\frac{d}{{\epsilon }_{0}A\times 8}$

or    $\frac{1}{C_{eq}}=\frac{d}{{\epsilon }_{0}A}[\frac{{\displaystyle 3}}{{\displaystyle 8K_1}}+\frac{{\displaystyle 1}}{{\displaystyle 2K_2}}+\frac{{\displaystyle 1}}{{\displaystyle 8}}]$                             ...(ii)

Given $C_{eq}=2C_0$

or   $\frac{C_0}{C_{eq}}=\frac{1}{2}$

$\therefore$    $\frac{{\epsilon }_{0}A}{d}\times \frac{d}{{\epsilon }_{0}A}[\frac{3}{8K_1}+\frac{1}{2K_2}+\frac{1}{8}]=\frac{1}{2}$

$\Rightarrow \frac{3}{8\times 1.25K_2}+\frac{1}{2K_2}=\frac{1}{2}-\frac{1}{8}$

$\Rightarrow \frac{3}{10K_2}+\frac{1}{2K_2}=\frac{3}{8}$

$\Rightarrow \frac{3+5}{10K_2}=\frac{3}{8}$

$\Rightarrow 64=30K_2$

$\Rightarrow K_2=\frac{64}{30}=2.13\Rightarrow \frac{K_1}{1.25}\Rightarrow K_1=2.66$

(1)

As, Wheatstone bridge is balanced, $\frac{C_1}{C_2}=\frac{C_3}{C_4}$ and $C_5$ becomes ineffective.

Capacitor $C_1$ and $C_3$ are in series.

$\therefore$   $C_{13}=\frac{C_1{C}_3}{C_1+C_3}=1$ $\mu \text{F}$

Similarly, capacitor, $C_2$ and $C_4$ are in series,

$\therefore$   $C_{24}=\frac{C_2{C}_4}{C_2+C_4}=1$ $\mu \text{F}$

Hence, equivalent capacitance between A and B is

$C_{13}+C_{24}=2$$\mu \text{F}$

(4)

(3)

$C_p=3+3=6$$\mu \text{F;}$ $C_{\text{eq}}=\frac{C_p\times 3}{3+C_p}=\frac{3\times 6}{6+3}=2$$\mu \text{F}$

(3)

Here, AB arm is short, so the two capacitors C and C in parallel

$C_{\text{eq}}=C+C=2C$

(3)

Here, $C_1=\frac{2{\epsilon }_0{k}_{1}A}{3d}, C_2=\frac{2{\epsilon }_0{k}_{2}A}{3d}$

$C_3=\frac{2{\epsilon }_0{k}_{3}A}{3d}, C_4=\frac{2{\epsilon }_0{k}_{4}A}{d}$

Given system of $C_1,C_2,C_3$ and $C_4$ can be simplified as

$\therefore$   $\frac{1}{C_{AB}}=\frac{1}{C_1+C_2+C_3}+\frac{1}{C_4}$

Suppose, $C_{AB}=\frac{k{\epsilon }_{0}A}{d}$

$\Rightarrow \frac{1}{\left(\frac{k{\epsilon }_{0}A}{d}\right)}=\frac{1}{\frac{2{\epsilon }_{0}A}{3d}(k_1+k_2+k_3)}+\frac{1}{\frac{2{\epsilon }_{0}A}{d}{\displaystyle {{\displaystyle k}}_4}}$

$\Rightarrow \frac{1}{k}=\frac{3}{2(k_1+k_2+k_3)}+\frac{1}{2k_4}$     $\therefore$   $\frac{2}{k}=\frac{3}{k_1+k_2+k_3}+\frac{1}{k_4}$