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. none

Explanation

To solve the problem, we need to find the dielectric constant (K) of the material inserted between the plates of the capacitor. We start with the information given:

1. The initial charge stored in the capacitor (Q1) is 60 µC.
2. After inserting the dielectric, the new charge stored (Q2) is 180 µC (60 µC + 120 µC).

### Step-by-Step Solution:

**Step 1: Understand the relationship between charge, capacitance, and voltage.**
The charge stored in a capacitor is given by the formula:
\[ Q = C \cdot V \]
Where:
- \( Q \) is the charge,
- \( C \) is the capacitance,
- \( V \) is the voltage.

**Step 2: Establish the initial and final conditions.**
Initially, we have:
\[ Q_1 = C_1 \cdot V \]
After inserting the dielectric:
\[ Q_2 = C_2 \cdot V \]
Where \( C_2 \) is the capacitance with the dielectric.

**Step 3: Relate the capacitance with and without the dielectric.**
The capacitance of a capacitor with a dielectric is given by:
\[ C_2 = K \cdot C_1 \]
Where \( K \) is the dielectric constant.

**Step 4: Substitute the capacitance relationship into the charge equations.**
From the equations for charge:
\[ Q_1 = C_1 \cdot V \]
\[ Q_2 = C_2 \cdot V = K \cdot C_1 \cdot V \]

**Step 5: Express the charges in terms of capacitance and voltage.**
We can express \( Q_2 \) as:
\[ Q_2 = K \cdot Q_1 \]

**Step 6: Substitute the known values.**
We know:
- \( Q_1 = 60 \, \mu C \)
- \( Q_2 = 180 \, \mu C \)

Substituting these values into the equation:
\[ 180 \, \mu C = K \cdot 60 \, \mu C \]

**Step 7: Solve for the dielectric constant (K).**
To find \( K \), rearrange the equation:
\[ K = \frac{Q_2}{Q_1} = \frac{180 \, \mu C}{60 \, \mu C} = 3 \]

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

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