Two thin dielectric slabs of dielectric constants $K_1$ and $K_2$ ($K_1

Question

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 the plate P is correctly shown by:

Accepted Answer

The electric field between the plates of a capacitor in vacuum (or air) is $E_0$. When a dielectric slab of dielectric constant $K$ is introduced, the electric field inside the dielectric decreases to $E = \frac{E_0}{K}$ [NCERT Class 12, Sec 2.10].

1.  **Inverse Relationship:** The electric field is inversely proportional to the dielectric constant ($E \propto 1/K$).
2.  **Comparison:** We are given $K_1 < K_2$. Therefore, $\frac{1}{K_1} > \frac{1}{K_2}$.
3.  **Field Magnitude:** This implies that the electric field inside the first slab ($E_1 = E_0/K_1$) is greater than the electric field inside the second slab ($E_2 = E_0/K_2$).
4.  **Graph Analysis:** The correct graph should show regions of constant electric field. The field will be highest in the air regions ($E_0$), intermediate in the slab with $K_1$, and lowest in the slab with $K_2$. Thus, the graph steps down from $E_0$ to $E_1$, goes back to $E_0$ (if there is an air gap), and then steps down to a lower value $E_2$ for the second slab.

Step-by-Step Solution

Practice All PHYSICS Questions