Two thin dielectric slabs of dielectric constants ${\mathrm{K}}_1 \text{and }{\mathrm{K}}_2({\mathrm{K}}_1<{\mathrm{K}}_2)$ are inserted between plates of a parallel plate capacitor, as shown in figure. The variation of electric field E between the plates with distance d as measured from plate P is correctly shown by

Let the electric field due to applied external source (battery) between the parallel plates of capacitor, be E₀.After insertion of dielectric the electric field in the material of dielectric slab) present inside capacitor,

$E=\frac{E_0}{K}$ …… (1)

From, the given arrangement we infer that there are three regions in which dielectric material is not present.
Hence in these regions, E= E₀
Given: 
K₂ > K₁ Therefore in the two region which contains dielectric slabs,
 $E_1=\frac{E_0}{K_1}$ = constant
$E_2=\frac{E_0}{K_2}$= constant

$\therefore E_2<E_1<E_0$

Thus following graph is best representation of E vs d.