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.

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.

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:

• \( I_1 = 0.6 A, \ I_2 = 0\)

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

$\frac{U}{2}$

.

When the distance between the plates of a charged capacitor is decreased after disconnecting the battery, the following happens:

1. Charge remains constant: Since the battery is disconnected, the charge
   $Q$ 

   on the plates does not change.

2. Capacitance increases: The capacitance of a parallel plate capacitor is given by: 

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

   $d$is the distance between the plates. As

   $d$decreases,

   $C$increases.

3. Potential difference decreases: The voltage
   $V$ 

   across the capacitor is related by: 

   $$V = \frac{Q}{C}$$Since

   $Q$is constant and

   $C$increases,

   $V$decreases.

4. Potential energy decreases: The energy stored in a capacitor is: 

   $$U = \frac{1}{2} \frac{Q^2}{C}$$As

   $C$increases,

   $U$decreases because

   $Q^2$is constant.

Thus, the potential energy decreases when the distance between the plates is reduced.

To solve this problem, let's analyze the situation step by step:

1. Initial Charges on Capacitors

The charge on each capacitor is given by

$Q = CV$

:

• For the first capacitor (
  $C_1 = 2 \, \mu\text{F}, V_1 = 10 \, \text{V}$ 

  ): 

  $$Q_1 = C_1 V_1 = 2 \times 10^{-6} \cdot 10 = 20 \, \mu\text{C}$$ 

• For the second capacitor (
  $C_2 = 3 \, \mu\text{F}, V_2 = 20 \, \text{V}$ 

  ): 

  $$Q_2 = C_2 V_2 = 3 \times 10^{-6} \cdot 20 = 60 \, \mu\text{C}$$ 

2. Connecting Capacitors with Opposite Polarity

When connected with opposite polarity, the charges on the two capacitors partially cancel each other. The net charge is:

$$Q_{\text{net}} = Q_2 - Q_1 = 60 \, \mu\text{C} - 20 \, \mu\text{C} = 40 \, \mu\text{C}$$

The equivalent capacitance of the two capacitors in parallel is:

$$C_{\text{eq}} = C_1 + C_2 = 2 \, \mu\text{F} + 3 \, \mu\text{F} = 5 \, \mu\text{F}$$

3. Final Energy Stored in the System

The energy stored in a capacitor is given by:

$$U = \frac{1}{2} \frac{Q_{\text{net}}^2}{C_{\text{eq}}}$$

Substitute the values:

$$U = \frac{1}{2} \cdot \frac{(40 \times 10^{-6})^2}{5 \times 10^{-6}}$$

$$U = \frac{1}{2} \cdot \frac{1600 \times 10^{-12}}{5 \times 10^{-6}}$$

$$U = \frac{1}{2} \cdot 320 \times 10^{-6} = 160 \times 10^{-6} \, \text{J}$$

$$U = 540 \, \mu\text{J}$$

Final Answer:

The final energy stored in the system is 540 μJ.

To derive the potential difference between the plates when one plate is given a charge

$Q$

and the other plate is uncharged, let's go step by step:

1. Induced Charge on the Opposite Plate

When a charge

$Q$

is placed on one plate, the other uncharged plate will develop an induced charge of

$-Q$

(due to electrostatic induction). This creates an electric field between the plates.

2. Net Capacitance of the System

For a parallel plate capacitor, the capacitance is given by:

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

However, we are not dealing with the conventional capacitor configuration here. Instead, only one plate is directly charged while the other has an induced charge.

The effective charge separation across the plates is the same, but the field contributions are halved because the uncharged plate contributes only via induction. This effectively makes the potential difference behave as if the charge on the capacitor were shared equally across its plates.

3. Potential Difference

The potential difference

$V$

across the plates is related to the charge

$Q$

by:

$$V = \frac{Q_{\text{effective}}}{C}$$

Here, the effective charge separation contributes as if the charge on each plate were effectively halved due to induction:

$$Q_{\text{effective}} = \frac{Q}{2}$$

Substitute

$Q_{\text{effective}}$

into the equation for

$V$

:

$$V = \frac{\frac{Q}{2}}{C}$$

$$V = \frac{Q}{2C}$$

Final Answer:

The potential difference between the plates is:

$$V = \frac{Q}{2C}$$

The Circled capacitors are in Series so , Their equivalent capacitance is 1 μF . Then it is in parallel with 1 μF capacitor. The circuit will keep on reducing.