A parallel plate air capacitor has capacitance 18`mu`F. If the distance between the plates is tripled and dielectric medium is introduced the capacitance becomes 72`mu`F. The dielectric constant of the medium is :-

Options

A. 4
B. 9
C. 2
D. 12

Explanation

To find the dielectric constant of the medium introduced in a parallel plate capacitor, we can follow these steps:

### Step 1: Understand the initial conditions
The initial capacitance \( C_1 \) of the air capacitor is given as:
\[ C_1 = 18 \, \mu F \]

The formula for the capacitance of a parallel plate capacitor in air is:
\[ C = \frac{A \epsilon_0}{D} \]
where:
- \( A \) is the area of the plates,
- \( \epsilon_0 \) is the permittivity of free space,
- \( D \) is the distance between the plates.

### Step 2: Analyze the changes made to the capacitor
When the distance \( D \) is tripled, the new distance becomes:
\[ D' = 3D \]

Now, a dielectric medium with dielectric constant \( K \) is introduced, and the new capacitance \( C_2 \) is given as:
\[ C_2 = 72 \, \mu F \]

The formula for the capacitance with the dielectric is:
\[ C_2 = \frac{K A \epsilon_0}{D'} \]

### Step 3: Substitute the new distance into the capacitance formula
Substituting \( D' \) into the capacitance formula gives:
\[ C_2 = \frac{K A \epsilon_0}{3D} \]

### Step 4: Relate the two capacitances
From the first case, we know:
\[ C_1 = \frac{A \epsilon_0}{D} = 18 \, \mu F \]

We can express \( A \epsilon_0 \) in terms of \( C_1 \):
\[ A \epsilon_0 = C_1 D = 18 \, \mu F \cdot D \]

Now substituting this into the equation for \( C_2 \):
\[ C_2 = \frac{K (18 \, \mu F \cdot D)}{3D} \]
\[ C_2 = \frac{K \cdot 18 \, \mu F}{3} \]
\[ C_2 = 6K \, \mu F \]

### Step 5: Set up the equation with the given capacitance
We know \( C_2 = 72 \, \mu F \), so we can set up the equation:
\[ 6K = 72 \]

### Step 6: Solve for \( K \)
Dividing both sides by 6 gives:
\[ K = \frac{72}{6} = 12 \]

### Conclusion
The dielectric constant \( K \) of the medium is:
\[ K = 12 \]