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 adjusted that the capacitance of the capacitor becomes `(2)/(3) rd` of its original value. What is the new distance between the plates ?

Options

A. 9 mm
B. 21 mm
C. 5 mm
D. 8 mm

Answer

To solve the problem step by step, we will use the formula for the capacitance of a parallel plate capacitor and the effects of introducing a dielectric slab.

### Given:
- Initial distance between plates, \( d = 4 \, \text{mm} \)
- Thickness of the dielectric slab, \( t = 3 \, \text{mm} \)
- Dielectric constant, \( K = 3 \)
- New capacitance is \( \frac{2}{3} \) of the original capacitance.

### Step 1: Write the formula for the capacitance of a parallel plate capacitor.
The capacitance \( C \) of a parallel plate capacitor is given by:
\[ C = \frac{\varepsilon A}{d} \]
where \( \varepsilon \) is the permittivity, \( A \) is the area of the plates, and \( d \) is the distance between the plates.

### Step 2: Set up the equation for the new capacitance with the dielectric slab.
When a dielectric slab is introduced, the new capacitance \( C' \) can be expressed as:
\[ C' = \frac{\varepsilon A}{D - t + \frac{t}{K}} \]
where \( D \) is the new distance between the plates.

### Step 3: Relate the new capacitance to the original capacitance.
According to the problem, the new capacitance is \( \frac{2}{3} \) of the original capacitance:
\[ C' = \frac{2}{3} C \]
Substituting the expressions for \( C \) and \( C' \):
\[ \frac{\varepsilon A}{D - t + \frac{t}{K}} = \frac{2}{3} \cdot \frac{\varepsilon A}{d} \]

### Step 4: Cancel out common terms.
Since \( \varepsilon A \) is common in both sides, we can cancel it out:
\[ \frac{1}{D - t + \frac{t}{K}} = \frac{2}{3d} \]

### Step 5: Cross-multiply to solve for \( D \).
Cross-multiplying gives:
\[ 3d = 2(D - t + \frac{t}{K}) \]
Expanding the right side:
\[ 3d = 2D - 2t + \frac{2t}{K} \]

### Step 6: Rearrange the equation to isolate \( D \).
Rearranging gives:
\[ 2D = 3d + 2t - \frac{2t}{K} \]
\[ D = \frac{3d + 2t - \frac{2t}{K}}{2} \]

### Step 7: Substitute the known values.
Substituting \( d = 4 \, \text{mm} \), \( t = 3 \, \text{mm} \), and \( K = 3 \):
\[ D = \frac{3(4) + 2(3) - \frac{2(3)}{3}}{2} \]
Calculating each term:
\[ D = \frac{12 + 6 - 2}{2} = \frac{16}{2} = 8 \, \text{mm} \]

### Final Answer:
The new distance between the plates is \( D = 8 \, \text{mm} \).

---