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.

The solution can be derived as follows:

1. Initial Capacitance (with air as the dielectric):
   For a parallel plate capacitor with air as the dielectric, the capacitance $C_{\text{initial}}$ 

   is given by: 

   $$C_{\text{initial}} = C$$The battery applies a potential difference

   $E$, so the initial charge on the capacitor is:

    

   $$Q_{\text{initial}} = C_{\text{initial}} \times E = C \times E$$ 

2. Capacitance with Dielectric Slab:
   When a dielectric slab of dielectric constant $K$ 

   is inserted between the plates, the capacitance increases by a factor of $K$ 

   , so the new capacitance $C_{\text{new}}$ 

   becomes: 

   $$C_{\text{new}} = K \times C$$ 

3. Final Charge on Capacitor:
   The battery remains connected and maintains a constant voltage $E$ 

   . Therefore, the final charge on the capacitor is: 

   $$Q_{\text{final}} = C_{\text{new}} \times E = K \times C \times E$$ 

4. Additional Charge Drawn from the Battery:
   The additional charge drawn from the battery is the difference between the final charge and the initial charge: 

   $$\Delta Q = Q_{\text{final}} - Q_{\text{initial}} = (K \times C \times E) - (C \times E)$$Simplifying:

    

   $$\Delta Q = (K - 1) \times C \times E$$ 

Thus, the additional charge drawn from the battery is:

$$(K - 1) \times C \times E$$