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 find the value of α in the expression for the capacitance of a parallel plate capacitor filled with multiple dielectric layers. Let's break down the solution step by step.

### Step 1: Understand the Configuration
We have a parallel plate capacitor with:
- Area \( A \)
- Plate separation \( d \)
- The space between the plates is filled with \( N \) dielectric layers, each of thickness \( \delta = \frac{d}{N} \).

### Step 2: Dielectric Constant of Each Layer
The dielectric constant for the \( m^{th} \) layer is given by:
\[ K_m = K \left(1 + \frac{m}{N}\right) \]
where \( K \) is a constant.

### Step 3: Capacitance of Each Layer
The capacitance \( C_m \) of each dielectric layer can be expressed as:
\[ C_m = \frac{K_m \epsilon_0 A}{\delta} \]
Substituting \( \delta = \frac{d}{N} \):
\[ C_m = \frac{K_m \epsilon_0 A N}{d} \]

### Step 4: Substitute for \( K_m \)
Substituting \( K_m \) into the capacitance formula:
\[ C_m = \frac{K \left(1 + \frac{m}{N}\right) \epsilon_0 A N}{d} \]

### Step 5: Total Capacitance
Since the capacitors are in series, the total capacitance \( C \) can be found using:
\[ \frac{1}{C} = \sum_{m=1}^{N} \frac{1}{C_m} \]
Substituting for \( C_m \):
\[ \frac{1}{C} = \sum_{m=1}^{N} \frac{d}{K \left(1 + \frac{m}{N}\right) \epsilon_0 A N} \]

### Step 6: Simplifying the Summation
Factoring out constants:
\[ \frac{1}{C} = \frac{d}{K \epsilon_0 A N} \sum_{m=1}^{N} \frac{1}{1 + \frac{m}{N}} \]
Let \( x = \frac{m}{N} \), then the summation can be approximated by an integral as \( N \) becomes very large:
\[ \sum_{m=1}^{N} \frac{1}{1 + \frac{m}{N}} \approx N \int_{0}^{1} \frac{1}{1+x} \, dx \]
Calculating the integral:
\[ \int_{0}^{1} \frac{1}{1+x} \, dx = \ln(2) \]

### Step 7: Substitute Back into the Capacitance Formula
Thus, we have:
\[ \frac{1}{C} \approx \frac{d}{K \epsilon_0 A} \cdot \ln(2) \]
Taking the reciprocal gives:
\[ C \approx \frac{K \epsilon_0 A}{d \ln(2)} \]

### Step 8: Identify \( \alpha \)
From the problem statement, we have:
\[ C = \alpha \frac{K \epsilon_0 A}{d \ln(2)} \]
Comparing both expressions, we find:
\[ \alpha = 1 \]

### Final Answer
The value of \( \alpha \) is:
\[ \alpha = 1 \]