A dielectric slab is introduced inside an uncharged capacitor of plate area A and plate separation L. The surface area of the slab is A and its thickness is slightly less than L. Let the surfaces of the slab that are facing the plates of the capacitor be called its two “faces”. The dielectric constant of the slab varies with distance x from one of its faces as: `K=K_(0)(1+(X)/(L))`, where `K_(0)` is a constant. After the insertion of the slab, the capacitance of the capacitor is `n((K_(0)epsilon_(0)A)/(Llog_(e)2))`. The value of n is _________.

Explanation

To solve the problem, we need to find the value of \( n \) in the expression for the capacitance of a capacitor with a dielectric slab inserted. The dielectric constant \( K \) varies with distance \( x \) from one of the faces of the slab as given by:

\[ K = K_0 \left( 1 + \frac{x}{L} \right) \]

### Step-by-Step Solution:

1. **Understanding the Capacitor Configuration**:
- We have a parallel plate capacitor with plate area \( A \) and separation \( L \).
- A dielectric slab of thickness slightly less than \( L \) is inserted between the plates.

2. **Capacitance of a Differential Element**:
- Consider a small element of thickness \( dx \) at a distance \( x \) from one face of the slab.
- The dielectric constant at this position is:
\[ K(x) = K_0 \left( 1 + \frac{x}{L} \right) \]
- The capacitance \( dC \) of this small element can be expressed as:
\[ dC = \frac{A \cdot K(x) \cdot \epsilon_0}{dx} \]

3. **Substituting for \( K(x) \)**:
- Substitute \( K(x) \) into the expression for \( dC \):
\[ dC = \frac{A \cdot K_0 \left( 1 + \frac{x}{L} \right) \cdot \epsilon_0}{dx} \]

4. **Finding the Total Capacitance**:
- Since the dielectric slab is uniform and the capacitors formed by each differential element are in series, we need to find the equivalent capacitance \( C_{eq} \):
\[ \frac{1}{C_{eq}} = \int_0^L \frac{1}{dC} \]
- Substitute \( dC \) into the integral:
\[ \frac{1}{C_{eq}} = \int_0^L \frac{dx}{\frac{A \cdot K_0 \left( 1 + \frac{x}{L} \right) \cdot \epsilon_0}{dx}} = \int_0^L \frac{dx}{\frac{A \cdot K_0 \cdot \epsilon_0}{dx} + \frac{A \cdot K_0 \cdot x}{L \cdot \epsilon_0}} \]

5. **Evaluating the Integral**:
- The integral simplifies to:
\[ \frac{1}{C_{eq}} = \frac{1}{A \cdot K_0 \cdot \epsilon_0} \int_0^L \frac{dx}{1 + \frac{x}{L}} \]
- The integral can be evaluated:
\[ \int_0^L \frac{dx}{1 + \frac{x}{L}} = L \ln(2) \]

6. **Finding \( C_{eq} \)**:
- Thus, we have:
\[ C_{eq} = \frac{A \cdot K_0 \cdot \epsilon_0 \cdot L}{L \ln(2)} = \frac{A \cdot K_0 \cdot \epsilon_0}{\ln(2)} \]

7. **Comparing with Given Expression**:
- The problem states that the capacitance is given by:
\[ C = n \left( \frac{K_0 \epsilon_0 A}{L \ln(2)} \right) \]
- By comparing both expressions:
\[ n = 1 \]

### Final Answer:
The value of \( n \) is \( 1 \).