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.