Solution:

Given Information:

1. Parallel plate capacitor: Contains two dielectric layers.
   • First layer ($k=2$) extends from $0$ to $d$.
   • Second layer ($k=4$) extends from $d$ to $3d$.
2. Capacitor is connected to a battery: This means the potential difference $V$ across the plates is fixed.

Key Concepts:

1. Electric Field in a Dielectric:
   • The electric field $E$ in a dielectric is inversely proportional to the dielectric constant $k$: $$E = \frac{\sigma}{\varepsilon_0 k}$$ where $\sigma$ is the surface charge density.
2. Continuity of Potential:
   • Since the potential $V$ is constant across the capacitor, the sum of the potential drops across the two dielectric layers must equal $V$. For a uniform electric field in each region: $$V = E_1 \cdot d + E_2 \cdot 2d$$ where $E_1$ and $E_2$ are the electric fields in the regions with $k=2$ and $k=4$, respectively.
3. Relation Between Fields:
   • The electric displacement $\mathbf{D} = \varepsilon_0 k E$ must be continuous across the boundary of the dielectrics: $$k_1 E_1 = k_2 E_2$$ Substituting $k_1 = 2$ and $k_2 = 4$, we find: $$2E_1 = 4E_2 \quad \Rightarrow \quad E_2 = \frac{E_1}{2}$$

Explanation of the Graph:

1. Region 1 ($0 \leq x < d$):
   • In this region, the dielectric constant $k=2$, so the electric field $E_1$ is relatively stronger compared to the next region.
2. Region 2 ($d \leq x \leq 3d$):
   • Here, $k=4$, and since $E_2 = \frac{E_1}{2}$, the electric field is halved.

Thus, the electric field decreases discontinuously at $x = d$ due to the change in the dielectric constant, leading to the stepwise graph shown in the second figure.