Let’s solve the problem step by step:

Given:

• Initial capacitance:
  $C$ 
• The separation between the plates is doubled, and a dielectric medium is inserted.
• The new capacitance becomes
  $2C$ 

  .

Step 1: Capacitance formula

The capacitance of a parallel plate capacitor is given by:

$$C = \frac{\varepsilon_0 A}{d}$$

where:

• 
  $C$ 

  is the capacitance,

• 
  $\varepsilon_0$ 

  is the permittivity of free space,

• 
  $A$ 

  is the area of the plates, and

• 
  $d$ 

  is the separation between the plates.

Step 2: Effect of doubling the separation and adding a dielectric

When the separation

$d$

is doubled, the capacitance would normally decrease by a factor of 2 (since capacitance is inversely proportional to

$d$

).

Now, when a dielectric of dielectric constant

$K$

is inserted, the capacitance increases by a factor of

$K$

. So, the new capacitance

$C'$

is:

$$C' = K \cdot \frac{\varepsilon_0 A}{2d} = K \cdot \frac{C}{2}$$

We are told that the new capacitance is

$2C$

, so:

$$K \cdot \frac{C}{2} = 2C$$

Step 3: Solve for $K$

$$K = 4$$

Final Answer:

The dielectric constant of the medium is 4.

Both the parts can be taken as separate capacitors connected in parallel.

So, C=C1+ C2=\(\frac{\varepsilon_{0A}}{d}\frac{\left( K_{1} +K_{2}\right)}{2}\)

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.

Given:

• Initial charge on the capacitor:
  $Q_1 = 60 \, \mu C$ 
• After inserting the dielectric, the total charge from the battery:
  $Q_2 = 120 \, \mu C$ 

  (additional charge drawn is $120 \, \mu C$ 

  , so the total charge on the capacitor is $120 \, \mu C + 60 \, \mu C = 180 \, \mu C$ 

  ).

Key concept:

• The charge on a capacitor is given by: 

  $$Q = C \cdot V$$where

  $C$is the capacitance and

  $V$is the potential difference across the plates.

• The dielectric increases the capacitance of the capacitor. If the dielectric constant is
  $K$ 

  , the capacitance becomes $K \times C$ 

  . Since the battery is still connected, the potential difference $V$ 

  remains constant, and the charge increases proportionally with the increase in capacitance.

Step 1: Relationship between charge and capacitance

Before the dielectric, the charge was

$Q_1 = 60 \, \mu C$

, and after inserting the dielectric, the charge is

$Q_2 = 180 \, \mu C$

.

The ratio of the final charge to the initial charge is proportional to the dielectric constant

$K$

:

$$\frac{Q_2}{Q_1} = K$$

Step 2: Solve for $K$

Substitute the given values:

$$K = \frac{180 \, \mu C}{60 \, \mu C} = 3$$

Final Answer:

The dielectric constant of the material inserted is 3.

Here's the short solution for the circuit:

1. Just after the switch is closed:
   • The capacitors act as open circuits because they are initially uncharged (capacitor voltage cannot change instantaneously).
   • This means no current flows through the branches containing the capacitors.
2. Current Distribution:
   • The total resistance in the circuit is only the sum of the resistors in the main loop (2Ω + 8Ω), as the branches with capacitors are effectively open.
   • Total resistance =
     $2\Omega + 8\Omega = 10\Omega$ 

     .

   • Current
     $I_1 = \frac{\text{Voltage}}{\text{Resistance}} = \frac{6V}{2\Omega + 8\Omega} = 0.6A$ 

     .

3. Branch currents:
   • 
     $I_3 = 0$ 

     since the capacitor branch is open.

   • 
     $I_2 = 0$ 

     for the same reason as above.

Final Answer:

• \( I_1 = 0.6 A, \ I_2 = 0\)

At t =0 capacitor behaves as closed circuit to the 1000 ohm resistor connected in parallel with capacitor will get short circuited.

current  through the other resistor = 2/1000 = 2mA

At t = infinite capacitor behaves as open circuit so equivalent resistance becomes R=1000+1000 = 2000 Ohm

current  through the  resistor = 2/2000 = 1 mA

In steady state no current flows through the capacitor so, current through 4 ohm resistor will be zero.

In absence of 4 ohm resistor, total resistance of circuit is (1.2+2.8)= 4 ohm.

Total current given by battery = 6/4=1.5 Ampere.

In parallel combination current divides in reverse ratio of resistors: (3/5)*1.5= 0.9 Ampere

To solve this, we will use the concept of common potential and energy conservation. Let's derive it step-by-step:

1. Initial Energy Stored in Capacitor 1
   Let the initial capacitance of the first capacitor be $C$ 

   , and the battery's voltage be $V$ 

   .
   The energy stored in the capacitor is: 

   $$U = \frac{1}{2} C V^2$$ 

2. When the second capacitor is connected
   After disconnecting the battery, an identical capacitor (with capacitance $C$ 

   ) is connected across the first one. The total charge remains conserved because the battery is removed. Let the final voltage be $V_f$ 

   .Total charge initially:

    

   $$Q_{\text{initial}} = CV$$After connecting the second capacitor, the total capacitance becomes:

    

   $$C_{\text{total}} = C + C = 2C$$Common potential

   $V_f$:

    

   $$V_f = \frac{\text{Total charge}}{\text{Total capacitance}} = \frac{CV}{2C} = \frac{V}{2}$$ 

3. Final Energy Stored in Both Capacitors
   The final energy stored in the system is the sum of the energy in both capacitors: 

   $$U_{\text{final}} = \frac{1}{2} C V_f^2 + \frac{1}{2} C V_f^2 = C V_f^2$$Substituting

   $V_f = \frac{V}{2}$:

    

   $$U_{\text{final}} = C \left( \frac{V}{2} \right)^2 = C \frac{V^2}{4}$$Simplifying:

    

   $$U_{\text{final}} = \frac{1}{2} \cdot C V^2 \cdot \frac{1}{2} = \frac{U}{2}$$ 

Thus, the final energy stored in the system is

$\frac{U}{2}$

.