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(D) The electric field $E$ between the plates of a parallel plate capacitor is given by $E = \frac{\sigma}{\varepsilon_0 K}$,where $\sigma$ is the sur

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(D) The electric field $E$ between the plates of a parallel plate capacitor is given by $E = \frac{\sigma}{\varepsilon_0 K}$,where $\sigma$ is the surface charge density and $K$ is the dielectric constant of the medium.In the regions where the dielectric is absent (air/vacuum),$K = 1$,so the electric field is $E_0 = \frac{\sigma}{\varepsilon_0}$.In the regions where the dielectric slabs are present,the electric field is $E_1 = \frac{\sigma}{\varepsilon_0 K_1}$ and $E_2 = \frac{\sigma}{\varepsilon_0 K_2}$.Since $K_1  \frac{1}{K_2}$,which implies $E_1 > E_2$.Also,both $E_1$ and $E_2$ are less than the electric field $E_0$ in the air gap ($E_1 Therefore,the electric field is highest in the air gaps,and among the dielectric regions,the field is higher in the slab with the smaller dielectric constant $(K_1)$.Comparing this with the given options,the graph that shows the highest field in the air gaps and a higher field in the $K_1$ region compared to the $K_2$ region is represented by option $(D)$.

Two thin dielectric slabs of dielectric constants $K_1$ and $K_2$ $(K_1 < K_2)$ are inserted between the plates of a parallel plate capacitor,as shown in the figure. The variation of the electric field $E$ between the plates with distance $d$ as measured from plate $P$ is correctly shown by:

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