A parallel plate capacitor having area $A$, the separation between the plates $d$ and capacitance  $C$  is filled with three different dielectric materials which are having dielectric constants  ${{K}_{1}},{{K}_{2}},{{K}_{3}}$​ as shown in the figure. If a single dielectric material is to be used to have the similar capacitance $C$  in this capacitor then it's dielectric constant $K$ is given by

$\begin{align}
  & A.\dfrac{1}{K}=\dfrac{1}{{{K}_{1}}}+\dfrac{1}{{{K}_{2}}}+\dfrac{1}{2{{K}_{3}}} \\
 & B.\dfrac{1}{K}=\dfrac{1}{{{K}_{1}}+{{K}_{2}}}+\dfrac{1}{2{{K}_{3}}} \\
 & C.\dfrac{1}{K}=\dfrac{{{K}_{1}}{{K}_{2}}}{{{K}_{1}}+{{K}_{2}}}+2{{K}_{3}} \\
\end{align}$

Hint: The capacitance of a capacitor is found by taking the product of the permittivity of material, area of the plates and the dielectric constant which is then divided by the separation between plates. The effective capacitance in a parallel circuit is the sum of individual capacitance. The effective capacitance in a series circuit is the sum of the reciprocal of the individual capacitance. These all may help you to solve this question.

Complete step-by-step answer:
Let us take the capacitance of the first area of the cross section as ${{C}_{1}}$ which is having a dielectric constant of ${{K}_{1}}$.
The capacitance of the second area of cross section be ${{C}_{2}}$ which is having a dielectric constant of ${{K}_{2}}$.${{K}_{1}}$
The capacitance of the third area of cross section be ${{C}_{3}}$ which is having a dielectric constant of ${{K}_{3}}$.${{K}_{1}}$
Therefore the capacitance of the first dielectric material can be written as,
As the area of the material is $\dfrac{A}{2}$, plate separation is given as $\dfrac{d}{2}$
Hence we can write that,
${{C}_{1}}={{K}_{1}}\left( \dfrac{A}{2} \right)\dfrac{{{\varepsilon }_{0}}\times 2}{d}=\dfrac{A{{\varepsilon }_{0}}{{K}_{1}}}{d}$
Then the capacitance of the second dielectric material can be written as,
As the area of the material is $\dfrac{A}{2}$, plate separation is given as $\dfrac{d}{2}$
Hence we can write that,
${{C}_{2}}={{K}_{2}}\left( \dfrac{A}{2} \right)\dfrac{{{\varepsilon }_{0}}\times 2}{d}=\dfrac{A{{\varepsilon }_{0}}{{K}_{2}}}{d}$
Finally the capacitance of the third dielectric material can be written as,
As the area of the material is $A$, plate separation is given as $d$,
Hence we can write that,
${{C}_{2}}={{K}_{3}}\times A\times \dfrac{{{\varepsilon }_{0}}}{d}=\dfrac{2A{{\varepsilon }_{0}}{{K}_{3}}}{d}$
As we can see, ${{C}_{2}}$ and ${{C}_{1}}$ are in parallel combination,
Thus we can write that,
${{C}_{eq}}={{C}_{1}}+{{C}_{2}}=\dfrac{{{\varepsilon }_{0}}A}{d}\left( {{K}_{1}}+{{K}_{2}} \right)$
This equivalent capacitance and ${{C}_{3}}$are in series, therefore we can write that,
$\dfrac{1}{C}=\dfrac{1}{{{C}_{eq}}}+\dfrac{1}{{{C}_{3}}}$
This can be written as,
$\dfrac{d}{KA{{\varepsilon }_{0}}}=\dfrac{d}{A{{\varepsilon }_{0}}\left( {{K}_{1}}+{{K}_{2}} \right)}+\dfrac{d}{2A{{\varepsilon }_{0}}{{K}_{3}}}$
Cancelling the common terms will give the answer,
$\dfrac{1}{K}=\dfrac{1}{\left( {{K}_{1}}+{{K}_{2}} \right)}+\dfrac{1}{2{{K}_{3}}}$

So, the correct answer is “Option B”.

Note: Dielectric constant is also known as relative permittivity or the specific inductive capacity. It is the quantity which will measure the ability to store the electric field in a material. This is referred to as a dimensionless quantity.