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.