#### A parallel plate capacitor is made of two dielectric blocks in series. One of the blocks has thickness $d_{1}$ and dielectric constant $k_{1}$ and the other has thickness $d_{2}$ and dielectric constant $k_{2}$ as shown in the figure. This arrangement can be thought of as a dielectric slab of thickness $d=d_{1}+d_{2}$ and effective dielectric constant k. The k is$a) \frac{2K_{1}K_{2}}{\left ( K_{1}+K_{2} \right )}$$b) \frac{K_{1}d_{1}+K_{2}d_{2}}{\left ( d_{1}+d_{2} \right )}$$c) \frac{\left ( K_{1}d_{1}+K_{2}d_{2} \right )}{\left ( K_{1}+K_{2} \right )}$$d)\frac{K_{1}K_{2}\left ( d_{1}+d_{2} \right )}{\left ( K_{1}d_{1}+K_{2}d_{2} \right )}$

The answer is the option (d)

${{C}_{eq}}=\dfrac{{{k}_{1}}{{k}_{2}}{{\varepsilon }_{0}}A}{{{k}_{1}}{{d}_{2}}+{{k}_{2}}{{d}_{1}}}.............(1)$

$C=\dfrac{k{{\varepsilon }_{0}}A}{{{d}_{1}}+{{d}_{2}}}.........(2)$

From (1) and (2)

$\dfrac{{{k}_{1}}{{k}_{2}}{{\varepsilon }_{0}}A}{{{k}_{1}}{{d}_{2}}+{{k}_{2}}{{d}_{1}}}=\dfrac{k{{\varepsilon }_{0}}A}{{{d}_{1}}+{{d}_{2}}}$

$K=\frac{K_{1}K_{2}\left ( d_{1}+d_{2} \right )}{\left ( K_{1}d_{1}+K_{2}d_{2} \right )}$