A capacitor stores `60muC` charge when connected across a battery. When the gap between the plates is filled with a dielectric, a charge of `120muC` flows through the battery. The dielectric constant of the material inserted is:

Options

A. 1
B. 2
C. 3
D. 4

Answer

To find the dielectric constant of the material inserted into the capacitor, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Initial Charge**:
The initial charge stored in the capacitor when connected to the battery is given as:
\[ Q_i = 60 \, \mu C \]

2. **Identify Final Charge**:
When the dielectric is inserted, the charge that flows through the battery is given as:
\[ Q_{flow} = 120 \, \mu C \]
Therefore, the final charge on the capacitor after inserting the dielectric is:
\[ Q_f = Q_i + Q_{flow} = 60 \, \mu C + 120 \, \mu C = 180 \, \mu C \]

3. **Relate Charges with Dielectric Constant**:
When a dielectric is inserted, the new charge \( Q_f \) can be expressed in terms of the initial charge \( Q_i \) and the dielectric constant \( k \):
\[ Q_f = k \cdot Q_i \]

4. **Substitute Known Values**:
We know \( Q_f = 180 \, \mu C \) and \( Q_i = 60 \, \mu C \). Substitute these values into the equation:
\[ 180 \, \mu C = k \cdot 60 \, \mu C \]

5. **Solve for Dielectric Constant \( k \)**:
Rearranging the equation to solve for \( k \):
\[ k = \frac{180 \, \mu C}{60 \, \mu C} = 3 \]

6. **Conclusion**:
The dielectric constant of the material inserted is:
\[ k = 3 \]

### Final Answer:
The dielectric constant of the material is \( k = 3 \).

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