Question

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 below. 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.

Answer 1

The correct option is C

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 ${\overrightarrow{E}}_o$​​ , which opposes the external field ${\overrightarrow{E}}_o$​​. Hence, the resultant field $\overrightarrow{E}$ within the dielectric is smaller than ${\overrightarrow{E}}_o$​ but is in the same direction as ${\overrightarrow{E}}_o$​​ and is given by $E=\frac{{\overrightarrow{E}}_o}{K}$​.
The electric field inside the dielectric will be less than the electric field outside the dielectrics. The electric field inside the dielectrics cannot not be zero.
As $K_2>K_1$, the drop in electric field for $K_2$ dielectric must be more than $K_1$
Hence, graph (c) is correct.