In a capacitor of capacitance `20mu F` the distance the plates is 5mm. If a dielectric slab of width 2 dielectric constant 3 is inserted between the plates, new capacitor will be

Options

A. `22.3mu F`
B. `25mu F`
C. `19mu F`
D. `27.2mu F`

Explanation

To find the new capacitance of a capacitor when a dielectric slab is inserted, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Original capacitance, \( C = 20 \, \mu F = 20 \times 10^{-6} \, F \)
- Distance between the plates, \( d = 5 \, mm = 5 \times 10^{-3} \, m \)
- Width of dielectric slab, \( t = 2 \, mm = 2 \times 10^{-3} \, m \)
- Dielectric constant, \( k = 3 \)

2. **Calculate Remaining Distance**:
- The remaining distance between the plates after inserting the dielectric slab is:
\[ d' = d - t = 5 \times 10^{-3} \, m - 2 \times 10^{-3} \, m = 3 \times 10^{-3} \, m \]

3. **Capacitance Formula**:
- The capacitance of a capacitor with a dielectric can be expressed as:
\[ C' = \frac{\epsilon_0 A}{d'} + \frac{\epsilon_0 A}{d - t} \cdot \frac{1}{k} \]
- Here, \( C' \) is the new capacitance, \( \epsilon_0 \) is the permittivity of free space, and \( A \) is the area of the plates.

4. **Expressing New Capacitance**:
- The new capacitance can be expressed as:
\[ C' = \frac{\epsilon_0 A}{d'} + \frac{\epsilon_0 A}{d - t} \cdot \frac{1}{k} \]
- Substitute \( d' \) and \( d - t \):
\[ C' = \frac{\epsilon_0 A}{3 \times 10^{-3}} + \frac{\epsilon_0 A}{3 \times 10^{-3}} \cdot \frac{1}{3} \]

5. **Combine Terms**:
- Factor out \( \epsilon_0 A \):
\[ C' = \epsilon_0 A \left( \frac{1}{3 \times 10^{-3}} + \frac{1}{3} \cdot \frac{1}{3 \times 10^{-3}} \right) \]
- This simplifies to:
\[ C' = \epsilon_0 A \left( \frac{1 + \frac{1}{3}}{3 \times 10^{-3}} \right) = \epsilon_0 A \left( \frac{4/3}{3 \times 10^{-3}} \right) \]

6. **Relate to Original Capacitance**:
- The original capacitance can be expressed as:
\[ C = \frac{\epsilon_0 A}{d} = \frac{\epsilon_0 A}{5 \times 10^{-3}} \]
- Therefore, we can express \( C' \) in terms of \( C \):
\[ C' = C \cdot \frac{4/3}{5/3} = C \cdot \frac{4}{5} \]

7. **Calculate New Capacitance**:
- Substitute \( C = 20 \, \mu F \):
\[ C' = 20 \, \mu F \cdot \frac{4}{5} = 16 \, \mu F \]

### Final Answer:
The new capacitance \( C' \) after inserting the dielectric slab is **16 µF**.