A parallel state air capacitor has capacitance of 100 `muF`. The plates are at a distance d apart. If a slab of thickness t `(t le d)` and dielectric constant 5 is introduced between the parallel plates, then the capacitance can be

Options

A. `50muF`
B. `100muF`
C. `200muF`
D. `500muF`

Explanation

To solve the problem of finding the new capacitance of a parallel plate capacitor when a dielectric slab is introduced, we can follow these steps:

### Step 1: Understand the initial conditions
We have a parallel plate capacitor with:
- Capacitance \( C_0 = 100 \, \mu F \)
- Distance between the plates \( d \)

The capacitance of a parallel plate capacitor in air (or vacuum) is given by the formula:
\[ C_0 = \frac{\epsilon_0 A}{d} \]
where \( \epsilon_0 \) is the permittivity of free space and \( A \) is the area of the plates.

### Step 2: Introduce the dielectric slab
A dielectric slab of thickness \( t \) (where \( t \leq d \)) and dielectric constant \( k = 5 \) is introduced between the plates.

### Step 3: Determine the effective capacitance
When a dielectric slab is introduced, the capacitance can be calculated using the formula:
\[ C' = \frac{k \cdot \epsilon_0 A}{d'} \]
where \( d' \) is the effective distance between the plates after inserting the dielectric slab.

### Step 4: Calculate the effective distance
The effective distance \( d' \) can be calculated as:
\[ d' = d - t + \frac{t}{k} \]
This accounts for the portion of the distance that is filled with the dielectric and the portion that remains air.

### Step 5: Substitute values
Since we know that \( C_0 = \frac{\epsilon_0 A}{d} \), we can express the new capacitance as:
\[ C' = \frac{5 \cdot \epsilon_0 A}{d - t + \frac{t}{5}} \]

### Step 6: Analyze the limits
If \( t \) is much less than \( d \), we can approximate \( d' \) as:
\[ d' \approx d - t \]
Thus, the new capacitance can be approximated as:
\[ C' \approx \frac{5 \cdot \epsilon_0 A}{d - t} \]

### Step 7: Compare with original capacitance
Since \( C_0 = \frac{\epsilon_0 A}{d} \), we can express \( C' \) in terms of \( C_0 \):
\[ C' \approx 5 \cdot C_0 \cdot \frac{d}{d - t} \]

### Step 8: Final approximation
Given that \( t \) is much smaller than \( d \), \( \frac{d}{d - t} \) approaches 1. Therefore:
\[ C' \approx 5 \cdot C_0 \approx 5 \cdot 100 \, \mu F = 500 \, \mu F \]

However, since \( t \) is not equal to \( d \) and is less than \( d \), the actual capacitance \( C' \) will be slightly less than \( 500 \, \mu F \).

### Conclusion
The new capacitance \( C' \) will be greater than \( 100 \, \mu F \) but less than \( 500 \, \mu F \). The closest reasonable value for \( C' \) in this case is approximately \( 200 \, \mu F \).