A parallel plate capacitor has plate area `40 cm^2` and plates separation 2mm. The space between the plates is filled with a dielectric medium of a thickness 1 mm and dielectric constant 5. The capacitance of the system is :

Options

A. `10/3epsilon _0F`
B. `3/10epsilon _0F`
C. `24epsilon _0F`
D. `10epsilon _0F`

Answer

To solve the problem of finding the capacitance of a parallel plate capacitor with a dielectric medium, we can break it down into several steps.

### Step 1: Identify Given Values
- Plate area \( A = 40 \, \text{cm}^2 = 40 \times 10^{-4} \, \text{m}^2 \)
- Plate separation \( d = 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m} \)
- Thickness of dielectric \( d_1 = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \)
- Dielectric constant \( k = 5 \)

### Step 2: Calculate Capacitance with Dielectric
The capacitance \( C_1 \) of the section filled with the dielectric can be calculated using the formula:
\[ C_1 = \frac{k \epsilon_0 A}{d_1} \]
Where \( \epsilon_0 \) is the permittivity of free space, approximately \( 8.85 \times 10^{-12} \, \text{F/m} \).

Substituting the values:
\[ C_1 = \frac{5 \times (8.85 \times 10^{-12}) \times (40 \times 10^{-4})}{1 \times 10^{-3}} \]
Calculating \( C_1 \):
\[ C_1 = \frac{5 \times 8.85 \times 40 \times 10^{-16}}{1 \times 10^{-3}} = \frac{5 \times 8.85 \times 40}{1} \times 10^{-13} \]
\[ C_1 = 20 \times 8.85 \times 10^{-13} = 177 \times 10^{-13} \, \text{F} = 1.77 \times 10^{-12} \, \text{F} \]

### Step 3: Calculate Capacitance without Dielectric
The remaining gap \( d_2 \) is:
\[ d_2 = d - d_1 = 2 \times 10^{-3} - 1 \times 10^{-3} = 1 \times 10^{-3} \, \text{m} \]
The capacitance \( C_2 \) of the section without the dielectric is:
\[ C_2 = \frac{\epsilon_0 A}{d_2} \]
Substituting the values:
\[ C_2 = \frac{(8.85 \times 10^{-12}) \times (40 \times 10^{-4})}{1 \times 10^{-3}} \]
Calculating \( C_2 \):
\[ C_2 = \frac{8.85 \times 40 \times 10^{-16}}{1 \times 10^{-3}} = 354 \times 10^{-13} \, \text{F} = 3.54 \times 10^{-12} \, \text{F} \]

### Step 4: Combine Capacitances in Series
Since \( C_1 \) and \( C_2 \) are in series, the equivalent capacitance \( C_{eq} \) is given by:
\[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} \]
Substituting the values:
\[ \frac{1}{C_{eq}} = \frac{1}{1.77 \times 10^{-12}} + \frac{1}{3.54 \times 10^{-12}} \]
Calculating:
\[ \frac{1}{C_{eq}} = \frac{3.54 + 1.77}{1.77 \times 3.54} \times 10^{12} \]
Calculating the numerator:
\[ \frac{1}{C_{eq}} = \frac{5.31}{6.25 \times 10^{-24}} = 0.8496 \times 10^{12} \]
Thus:
\[ C_{eq} \approx \frac{1}{0.8496 \times 10^{12}} \approx 1.176 \times 10^{-12} \, \text{F} \]

### Final Step: Summary of Capacitance
The equivalent capacitance of the system is approximately:
\[ C_{eq} \approx \frac{10}{3} \epsilon_0 \, \text{F} \]