A parallel plate capacitor of capacitance has spacing between two plates having area The region between the plates is filled with dielectric layers, parallel to its plates, each with thickness `delta=d/N`. The dielectric constant of the `m^(th)` layer is `K_(m)=K(1+m/N)`. For a very large `N(gt10^(3))`, the capacitance `C is alpha((Kin_(0)A)/(dln2))`. Find the value of `alpha.[in_(0)` is the permittivity of free space]

Explanation

To solve the problem, we need to determine the value of \( \alpha \) in the expression for the capacitance of a parallel plate capacitor filled with multiple dielectric layers. Let's break down the steps systematically.

### Step-by-Step Solution

1. **Understanding the Configuration**:
- We have a parallel plate capacitor with a total separation \( D \) between the plates.
- The space between the plates is filled with \( N \) dielectric layers, each of thickness \( \delta = \frac{D}{N} \).
- The dielectric constant of the \( m^{th} \) layer is given by \( K_m = K \left(1 + \frac{m}{N}\right) \).

2. **Capacitance of Each Layer**:
- The capacitance \( C_m \) of the \( m^{th} \) dielectric layer can be expressed as:
\[ C_m = \frac{K_m \epsilon_0 A}{\delta} = \frac{K \left(1 + \frac{m}{N}\right) \epsilon_0 A}{\frac{D}{N}} = \frac{K \left(1 + \frac{m}{N}\right) N \epsilon_0 A}{D} \]

3. **Reciprocal of Capacitance**:
- Since the capacitors are in series, the reciprocal of the total capacitance \( C_{\text{eq}} \) is the sum of the reciprocals of the individual capacitances:
\[ \frac{1}{C_{\text{eq}}} = \sum_{m=1}^{N} \frac{1}{C_m} \]
- Substituting the expression for \( C_m \):
\[ \frac{1}{C_{\text{eq}}} = \sum_{m=1}^{N} \frac{D}{K \left(1 + \frac{m}{N}\right) N \epsilon_0 A} \]

4. **Simplifying the Summation**:
- The summation can be rewritten as:
\[ \frac{1}{C_{\text{eq}}} = \frac{D}{K N \epsilon_0 A} \sum_{m=1}^{N} \frac{1}{1 + \frac{m}{N}} \]
- Changing the variable \( x = \frac{m}{N} \) gives:
\[ \sum_{m=1}^{N} \frac{1}{1 + \frac{m}{N}} \approx N \int_{0}^{1} \frac{1}{1+x} \, dx \]
- Evaluating the integral:
\[ \int_{0}^{1} \frac{1}{1+x} \, dx = \ln(2) \]
- Thus, the summation approximates to \( N \ln(2) \).

5. **Final Expression for \( C_{\text{eq}} \)**:
- Substituting back, we get:
\[ \frac{1}{C_{\text{eq}}} = \frac{D \ln(2)}{K \epsilon_0 A} \]
- Therefore, the equivalent capacitance is:
\[ C_{\text{eq}} = \frac{K \epsilon_0 A}{D \ln(2)} \]

6. **Comparing with Given Expression**:
- The problem states that \( C_{\text{eq}} = \alpha \frac{K \epsilon_0 A}{D \ln(2)} \).
- By comparing both expressions, we find:
\[ \alpha = 1 \]

### Conclusion
The value of \( \alpha \) is \( 1 \).