The two plates of a parallel plate capacitor are `4 mm` apart. A slab of dielectric constant 3 and thickness `3 mm` is introduced between the plates is so adujected that the capacitance of the capacitor becomes `(2)/(3) rd` of its original value. What is the new distance between the plates ?

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the original capacitance
The capacitance \( C \) of a parallel plate capacitor is given by the formula:
\[ C = \frac{\varepsilon_0 A}{D} \]
where:
- \( \varepsilon_0 \) is the permittivity of free space,
- \( A \) is the area of the plates,
- \( D \) is the distance between the plates.

Given that the original distance \( D = 4 \, \text{mm} \).

### Step 2: Introduce the dielectric slab
When a dielectric slab of thickness \( T = 3 \, \text{mm} \) and dielectric constant \( K = 3 \) is introduced, the new capacitance \( C' \) can be calculated using the formula:
\[ C' = \frac{\varepsilon_0 A}{D' - T} + \frac{\varepsilon_0 A}{T/K} \]
where \( D' \) is the new distance between the plates after adjustment.

### Step 3: Set up the equation for the new capacitance
According to the problem, the new capacitance \( C' \) is \( \frac{2}{3} \) of the original capacitance \( C \):
\[ C' = \frac{2}{3} C \]

Substituting the expression for capacitance, we have:
\[ \frac{\varepsilon_0 A}{D' - T} + \frac{\varepsilon_0 A}{T/K} = \frac{2}{3} \cdot \frac{\varepsilon_0 A}{D} \]

### Step 4: Substitute known values
Substituting \( D = 4 \, \text{mm} \), \( T = 3 \, \text{mm} \), and \( K = 3 \):
\[ \frac{\varepsilon_0 A}{D' - 3} + \frac{\varepsilon_0 A}{3/3} = \frac{2}{3} \cdot \frac{\varepsilon_0 A}{4} \]

This simplifies to:
\[ \frac{1}{D' - 3} + 1 = \frac{2}{12} \]
which simplifies to:
\[ \frac{1}{D' - 3} + 1 = \frac{1}{6} \]

### Step 5: Solve for \( D' \)
Rearranging the equation:
\[ \frac{1}{D' - 3} = \frac{1}{6} - 1 \]
\[ \frac{1}{D' - 3} = \frac{1 - 6}{6} = -\frac{5}{6} \]
Taking the reciprocal gives:
\[ D' - 3 = -\frac{6}{5} \]
Thus:
\[ D' = 3 - \frac{6}{5} = \frac{15}{5} - \frac{6}{5} = \frac{9}{5} = 1.8 \, \text{mm} \]

### Step 6: Convert to mm
Since we need the final answer in mm, we convert:
\[ D' = 1.8 \, \text{mm} \]

### Final Answer
The new distance between the plates is:
\[ D' = 1.8 \, \text{mm} \]