A dielectric (k=3.2) is placed in a capacitor of distance 1m and area `2 m^2`. The width of dielectric is 0.5m. If net capacitance is `x epsilon_0`

Answer

To solve the problem, we need to determine the net capacitance of a capacitor with a dielectric material inserted. Here are the steps to find the solution:

### Step-by-Step Solution:

1. **Identify the Parameters**:
- Dielectric constant \( k = 3.2 \)
- Distance between plates \( D = 1 \, \text{m} \)
- Area of the plates \( A = 2 \, \text{m}^2 \)
- Width of the dielectric \( d = 0.5 \, \text{m} \)

2. **Divide the Capacitor**:
The capacitor can be thought of as two capacitors in series:
- Capacitor \( C_1 \) (air gap): Distance = \( 0.5 \, \text{m} \)
- Capacitor \( C_2 \) (with dielectric): Distance = \( 0.5 \, \text{m} \)

3. **Calculate Capacitance of \( C_1 \)**:
The formula for capacitance is given by:
\[ C = \frac{A \epsilon_0 k}{D} \]
For \( C_1 \) (with air, \( k = 1 \)):
\[ C_1 = \frac{2 \epsilon_0 \cdot 1}{0.5} = 4 \epsilon_0 \]

4. **Calculate Capacitance of \( C_2 \)**:
For \( C_2 \) (with dielectric, \( k = 3.2 \)):
\[ C_2 = \frac{2 \epsilon_0 \cdot 3.2}{0.5} = 12.8 \epsilon_0 \]

5. **Combine Capacitors in Series**:
The total capacitance \( C_{\text{equiv}} \) for capacitors in series is given by:
\[ \frac{1}{C_{\text{equiv}}} = \frac{1}{C_1} + \frac{1}{C_2} \]
Substituting the values:
\[ \frac{1}{C_{\text{equiv}}} = \frac{1}{4 \epsilon_0} + \frac{1}{12.8 \epsilon_0} \]

6. **Calculate the Right Side**:
Finding a common denominator:
\[ \frac{1}{C_{\text{equiv}}} = \frac{3.2 + 1}{12.8 \epsilon_0} = \frac{4.2}{12.8 \epsilon_0} \]

7. **Find \( C_{\text{equiv}} \)**:
Inverting gives:
\[ C_{\text{equiv}} = \frac{12.8 \epsilon_0}{4.2} = 3.05 \epsilon_0 \]

8. **Determine Value of \( x \)**:
Since we have \( C_{\text{equiv}} = x \epsilon_0 \), we find:
\[ x = 3.05 \]

### Final Answer:
The value of \( x \) is approximately \( 3.05 \).