Calculate the capacitance of a parallel plate capacitor, with plate area A and distance between the plates d, when filled with a dielectric whose permittivity varies as
in(x)=in_(0)+kx(0 lt x lt (d)/(2)), in(x)=in_(0) +k(d-x)((d)/(2) lt x le d).

Answer

To calculate the capacitance of a parallel plate capacitor filled with a dielectric whose permittivity varies, we will follow these steps:

### Step 1: Understand the Configuration
We have a parallel plate capacitor with:
- Plate area \( A \)
- Distance between the plates \( d \)
- The dielectric permittivity varies as:
- For \( 0 < x < \frac{d}{2} \): \( \epsilon(x) = \epsilon_0 + kx \)
- For \( \frac{d}{2} < x \leq d \): \( \epsilon(x) = \epsilon_0 + k(d - x) \)

### Step 2: Set Up the Capacitance for Each Section
The total capacitance \( C \) of the capacitor can be considered as two capacitors in series:
- The first capacitor \( C_1 \) for the region \( 0 < x < \frac{d}{2} \)
- The second capacitor \( C_2 \) for the region \( \frac{d}{2} < x \leq d \)

### Step 3: Calculate the Differential Capacitance
For a small section \( dx \) of the capacitor, the differential capacitance \( dC \) can be expressed as:
\[ dC_1 = \frac{\epsilon(x) \cdot A}{dx} \]
For the first section:
\[ dC_1 = \frac{(\epsilon_0 + kx) \cdot A}{dx} \]
For the second section:
\[ dC_2 = \frac{(\epsilon_0 + k(d - x)) \cdot A}{dx} \]

### Step 4: Integrate to Find Total Capacitance
To find the total capacitance, we need to integrate \( dC_1 \) from \( 0 \) to \( \frac{d}{2} \) and \( dC_2 \) from \( \frac{d}{2} \) to \( d \).

#### For \( C_1 \):
\[ C_1 = \int_0^{\frac{d}{2}} \frac{A}{\epsilon_0 + kx} \, dx \]

#### For \( C_2 \):
\[ C_2 = \int_{\frac{d}{2}}^{d} \frac{A}{\epsilon_0 + k(d - x)} \, dx \]

### Step 5: Solve the Integrals
1. **For \( C_1 \)**:
\[ C_1 = A \int_0^{\frac{d}{2}} \frac{1}{\epsilon_0 + kx} \, dx \]
The integral evaluates to:
\[ C_1 = A \left[ \frac{1}{k} \ln(\epsilon_0 + kx) \right]_0^{\frac{d}{2}} = \frac{A}{k} \left( \ln(\epsilon_0 + k\frac{d}{2}) - \ln(\epsilon_0) \right) \]

2. **For \( C_2 \)**:
\[ C_2 = A \int_{\frac{d}{2}}^{d} \frac{1}{\epsilon_0 + k(d - x)} \, dx \]
The integral evaluates to:
\[ C_2 = A \left[ -\frac{1}{k} \ln(\epsilon_0 + k(d - x)) \right]_{\frac{d}{2}}^{d} = \frac{A}{k} \left( \ln(\epsilon_0 + k\frac{d}{2}) - \ln(\epsilon_0 + kd) \right) \]

### Step 6: Combine the Results
Since \( C_1 \) and \( C_2 \) are in series, the equivalent capacitance \( C \) is given by:
\[ \frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} \]

### Step 7: Final Expression
After simplifying, we arrive at the final expression for the equivalent capacitance \( C \):
\[ C = \frac{kA}{2 \ln\left(\frac{\epsilon_0 + k\frac{d}{2}}{\epsilon_0}\right)} \]