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.

Let's break down the situation step by step:

Given:

• A capacitor is charged using a battery and then disconnected (so no current can flow after disconnection).
• A dielectric slab is inserted between the plates of the capacitor after disconnecting the battery.

Key concepts:

• Capacitance with Dielectric: When a dielectric slab is inserted, the capacitance of the capacitor increases. The new capacitance
  $C'$ 

  is related to the original capacitance $C$ 

  by the dielectric constant $K$ 

  : 

  $$C' = K \cdot C$$where

  $K$is the dielectric constant of the material.

• Charge on the Plates: Since the capacitor is disconnected from the battery, no additional charge can flow onto the plates. Thus, the charge
  $Q$ 

  remains the same, given by: 

  $$Q = C \cdot V$$where

  $V$is the potential difference across the plates. Since the charge remains constant, the equation becomes:

   

  $$Q = C' \cdot V'$$where

  $V'$is the new potential difference across the plates.

• Potential Difference: Since the capacitance increases and the charge stays constant, the potential difference
  $V'$ 

  must decrease (because $Q = C' \cdot V'$ 

  and $C' > C$ 

  ).

• Stored Energy: The energy stored in a capacitor is given by: 

  $$U = \frac{Q^2}{2C}$$Since the capacitance increases and the charge is constant, the stored energy

  $U$decreases, as it is inversely proportional to the capacitance.

Conclusion:

• Decrease in potential difference: The potential difference across the plates decreases because the capacitance increases while the charge remains constant.
• Reduction in stored energy: The energy stored in the capacitor decreases because the capacitance increases.
• No change in charge: The charge on the plates remains the same since the capacitor is disconnected from the battery.

Thus, the correct answer is: "Decrease in the potential difference across the plates, reduction in stored energy, but no change in the charge on the plates."

To solve this, we use the formula for the capacitance of a parallel plate capacitor:

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

Where:

• 
  $C$ 

  is the capacitance,

• 
  $k$ 

  is the dielectric constant,

• 
  $\varepsilon_0$ 

  is the permittivity of free space,

• 
  $A$ 

  is the area of the plates, and

• 
  $d$ 

  is the separation between the plates.

Given:

• Initial capacitance without dielectric:
  $C_1 = 10^{-12} \, \text{F}$ 
• Final capacitance with wax dielectric:
  $C_2 = 2 \times 10^{-12} \, \text{F}$ 
• The separation between plates is doubled, so
  $d_2 = 2d_1$ 

  .

Step 1: Relate the initial and final capacitances

The capacitance of a capacitor is directly proportional to the dielectric constant and inversely proportional to the distance between the plates. So when the separation doubles and a dielectric with dielectric constant

$k$

is inserted, the capacitance will change as follows:

$$C_2 = k \times \frac{C_1}{2}$$

Step 2: Solve for $k$

$$2 \times 10^{-12} = k \times \frac{10^{-12}}{2}$$

$$2 \times 10^{-12} = \frac{k \times 10^{-12}}{2}$$

$$k = 4$$

Final Answer:

The dielectric constant of wax is 4.

Here’s a shorter solution:

Given:

• 
  $V = 100 \, \text{V}$ 

  (total battery voltage)

• Dielectric constant
  $K = 3$ 

  for capacitor $A$ 

• Identical capacitors
  $A$ 

  and $B$ 

Step 1: Capacitance

• 
  $C_A = 3C_0$ 

  (because of the dielectric in $A$ 

  )

• 
  $C_B = C_0$ 

  (no dielectric in $B$ 

  )

Step 2: Total capacitance in series:

$$\frac{1}{C_{\text{eq}}} = \frac{1}{C_A} + \frac{1}{C_B} = \frac{1}{3C_0} + \frac{1}{C_0} = \frac{4}{3C_0}$$

$$C_{\text{eq}} = \frac{3C_0}{4}$$

Step 3: Voltage division:

The voltage is divided in proportion to the inverse of capacitances:

$$V_A = \frac{C_B}{C_A + C_B} \times 100 = \frac{C_0}{3C_0 + C_0} \times 100 = \frac{1}{4} \times 100 = 25 \, \text{V}$$

$$V_B = 100 - 25 = 75 \, \text{V}$$

Final Answer:

• 
  $V_A = 25 \, \text{V}$ 
• 
  $V_B = 75 \, \text{V}$ 

Given:

• Initial capacitance with air as the dielectric:
  $C = 9 \, \text{pF}$ 
• Dielectric constants:
  $k_1 = 3$ 

  , $k_2 = 6$ 

• Thicknesses of the dielectrics:
  $d_1 = \frac{d}{3}$ 

  and $d_2 = \frac{2d}{3}$ 

Step-by-step solution:

1. Capacitance formula:

The capacitance of a parallel plate capacitor is:

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

Where:

• 
  $k$ 

  is the dielectric constant,

• 
  $\varepsilon_0$ 

  is the permittivity of free space,

• 
  $A$ 

  is the area of the plates,

• 
  $d$ 

  is the separation between the plates.

When we insert dielectrics in series, we treat the system as two capacitors in series with different dielectric constants.

2. Capacitance of each section:

• For the dielectric with
  $k_1 = 3$ 

  and thickness $d_1 = \frac{d}{3}$ 

  , the capacitance $C_1$ 

  is:

$$C_1 = \frac{k_1 \varepsilon_0 A}{d_1} = \frac{3 \varepsilon_0 A}{d/3} = \frac{9 \varepsilon_0 A}{d}$$

• For the dielectric with
  $k_2 = 6$ 

  and thickness $d_2 = \frac{2d}{3}$ 

  , the capacitance $C_2$ 

  is:

$$C_2 = \frac{k_2 \varepsilon_0 A}{d_2} = \frac{6 \varepsilon_0 A}{2d/3} = \frac{9 \varepsilon_0 A}{d}$$

3. Total capacitance:

The total capacitance of the system is found by treating the two capacitances in series. For capacitors in series, the total capacitance

$C_{\text{total}}$

is given by:

$$\frac{1}{C_{\text{total}}} = \frac{1}{C_1} + \frac{1}{C_2}$$

Substitute

$C_1 = C_2 = \frac{9 \varepsilon_0 A}{d}$

:

$$\frac{1}{C_{\text{total}}} = \frac{1}{\frac{9 \varepsilon_0 A}{d}} + \frac{1}{\frac{9 \varepsilon_0 A}{d}} = \frac{2}{\frac{9 \varepsilon_0 A}{d}}$$

Thus,

$$C_{\text{total}} = \frac{9 \varepsilon_0 A}{2d}$$

4. Relating to the original capacitance:

The original capacitance with air as the dielectric is:

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

Therefore, the total capacitance becomes:

$$C_{\text{total}} = \frac{9}{2} C = 4.5 C = 4.5 \times 9 \, \text{pF} = 40.5 \, \text{pF}$$

Final Answer:

The total capacitance with the two dielectrics is 40.5 pF.

Thus, 40.5 pF is the correct answer.

The question involves calculating the capacitance of a parallel-plate capacitor when a dielectric slab of thickness

$t$

and dielectric constant

$K = 2$

is partially inserted between the plates.

The short solution is based on treating the system as a combination of two capacitors in series:

1. Capacitor 1 (region with dielectric): The thickness of this region is
   $t$ 

   , and the capacitance is given by: 

   $$C_1 = \frac{\varepsilon_0 A K}{t} = \frac{2\varepsilon_0 A}{t}$$ 

2. Capacitor 2 (region without dielectric): The thickness of this region is
   $d - t$ 

   , and the capacitance is: 

   $$C_2 = \frac{\varepsilon_0 A}{d - t}$$ 

Since the two regions are in series, the equivalent capacitance is given by:

$$\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2}$$

Substituting

$C_1$

and

$C_2$

:

$$\frac{1}{C} = \frac{t}{2\varepsilon_0 A} + \frac{d - t}{\varepsilon_0 A}$$

Simplifying:

$$\frac{1}{C} = \frac{t + 2(d - t)}{2\varepsilon_0 A} = \frac{2d - t}{2\varepsilon_0 A}$$

Therefore:

$$C = \frac{2\varepsilon_0 A}{2d - t}$$

When the dielectric slab thickness is

$t = \frac{d}{2}$

, the capacitance simplifies to:

$$C = \frac{\varepsilon_0 A}{d - \frac{t}{2}}$$

This matches the given answer. Let me know if further clarification is needed!

The incorrect statement is "The charge on the capacitor is not conserved".

Here’s the explanation:

1. Initially, when the capacitor is connected to the battery with emf
   $V$ 

   , it stores a charge

   $Q_{\text{initial}} = C \times V$ 

   , where

   $C$ 

   is the capacitance of the air capacitor.

2. After disconnecting the capacitor from the battery, the charge on the capacitor is conserved because the capacitor is isolated. No charge can flow in or out.
3. When the dielectric slab of dielectric constant
   $K$ 

   is inserted, the capacitance of the capacitor increases to

   $K \times C$ 

   , but the charge remains the same because the capacitor is disconnected from the battery. The charge is now distributed on the new capacitance, and the voltage across the plates decreases.

4. The statement "charge is not conserved" is incorrect because charge is conserved in this isolated system. The only change is in the voltage across the capacitor due to the increased capacitance.

Thus, charge on the capacitor is conserved and the incorrect statement is the one claiming otherwise.