Option A: 
Option B: 
Option C: 
Option D: 
Solution:
The correct option is C
Electric field is given as $\mathrm{E}^{\prime} \infty \frac{\mathrm{E}_{0}}{\mathrm{~K}}$
As $\mathrm{K}_{1}<\mathrm{K}_{2}$ so $\mathrm{E}_{1}>\mathrm{E}_{2}$
Two thin dielectric slabs of dielectric constants $\mathrm{K}_{2}$ and $\mathrm{K}_{2}\left(\mathrm{~K}_{1}<\mathrm{K}_{2}\right)$ are interested between plates of a parallel plate capacitor, as shown in the figure. The variation of electric field 'E' between the plates with distance'd' as measured from plate P is correctly shown by:
Two thin dielectric slabs of dielectric constants $\mathrm{K}_{2}$ and $\mathrm{K}_{2}\left(\mathrm{~K}_{1}<\mathrm{K}_{2}\right)$ are interested between plates of a parallel plate capacitor, as shown in the figure. The variation of electric field 'E' between the plates with distance'd' as measured from plate P is correctly shown by: