Two thin dielectric slabs of dielectric constants K1 and K2 (K1

Question

Two thin dielectric slabs of dielectric constants K1 and K2 (K1 < K2) are inserted 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 (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

A uniform electric field is present between the plates of capacitors in case of absence of dielectric material.
When a dielectric material is inserted in between the plates of capacitor, the dielectric becomes electrically polarized. The polarization charges induced on the two faces of the slab produce their own electric field

E_{0}

, which opposes the external field

E_{0}

. Hence, the resultant field

E

within the dieclectric is smaller than

E_{0}

but is in the same direction as

E_{0}

and is given by

E = \frac{E _{0}}{K}

.
As in the given question the electric field in the region of dielectric will be less as compared to other regions but will not be zero. As

K_{2} > K_{1}

, the drop in the electric field for

K_{2}

dielectric is more than

K_{1}

.

Q. Two thin dielectric slabs of dielectric constants K1​ and K2​ (K1​ < K2​) are inserted 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

Q. Two thin dielectric slabs of dielectric constants $K_{1}$ and $K_{2}$ ( $K_{1}$ < $K_{2}$ ) are inserted 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