Solution:

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.