A dielectric of thickness 5cm and dielectric constant 10 is introduced between the plates of a parallel plate capacitor having plate area 500 sq. cm and separation between the plates 10cm. The capacitance of the capacitor with dielectric slab is
`(epsi_(0) = 8.8 xx 10^(-12)C^(2)//N-m^(2))`

Options

A. 4.4pF
B. 6.2pF
C. 8pF
D. 10pF

Explanation

To find the capacitance of a parallel plate capacitor with a dielectric slab inserted, we can follow these steps:

### Step 1: Identify the Parameters
- Dielectric thickness \( t_d = 5 \, \text{cm} = 0.05 \, \text{m} \)
- Dielectric constant \( k = 10 \)
- Plate area \( A = 500 \, \text{cm}^2 = 500 \times 10^{-4} \, \text{m}^2 = 5 \times 10^{-2} \, \text{m}^2 \)
- Total separation between plates \( d = 10 \, \text{cm} = 0.1 \, \text{m} \)

### Step 2: Calculate the Capacitance of the Dielectric Section (C1)
The capacitance of the section with the dielectric is given by:
\[ C_1 = \frac{k \epsilon_0 A}{t_d} \]
Where \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{F/m} \).

Substituting the values:
\[ C_1 = \frac{10 \times 8.85 \times 10^{-12} \times 5 \times 10^{-2}}{0.05} \]
Calculating \( C_1 \):
\[ C_1 = \frac{10 \times 8.85 \times 10^{-12} \times 5 \times 10^{-2}}{0.05} = 8.85 \times 10^{-12} \times 10^{-1} = 8.85 \times 10^{-12} \times 10 = 88.5 \times 10^{-12} = 88.5 \, \text{pF} \]

### Step 3: Calculate the Capacitance of the Air Section (C2)
The remaining distance between the plates that is filled with air is:
\[ t_a = d - t_d = 0.1 - 0.05 = 0.05 \, \text{m} \]
The capacitance of the air section is given by:
\[ C_2 = \frac{\epsilon_0 A}{t_a} \]
Substituting the values:
\[ C_2 = \frac{8.85 \times 10^{-12} \times 5 \times 10^{-2}}{0.05} \]
Calculating \( C_2 \):
\[ C_2 = \frac{8.85 \times 10^{-12} \times 5 \times 10^{-2}}{0.05} = 8.85 \times 10^{-12} \times 10^{-1} = 8.85 \times 10^{-12} \times 10 = 8.85 \times 10^{-12} \times 10 = 8.85 \, \text{pF} \]

### Step 4: Calculate the Equivalent Capacitance (Ceq)
Since \( C_1 \) and \( C_2 \) are in series, the equivalent capacitance \( C_{eq} \) can be calculated using:
\[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} \]
Substituting the values:
\[ \frac{1}{C_{eq}} = \frac{1}{88.5 \times 10^{-12}} + \frac{1}{8.85 \times 10^{-12}} \]
Calculating:
\[ \frac{1}{C_{eq}} = \frac{1}{88.5} + \frac{1}{8.85} \approx 0.0113 + 0.113 = 0.1243 \]
Thus,
\[ C_{eq} = \frac{1}{0.1243} \approx 8.04 \, \text{pF} \]

### Final Answer
The equivalent capacitance of the capacitor with the dielectric slab is approximately:
\[ C_{eq} \approx 8 \, \text{pF} \]