Space between the plates of a parallel plate capacitor is filled with a dielectric whose dielectric constant varies with distance as per the relation: `K(X)=K_(0)+lambdaX`,
(`lambda=` constant, `K_(0)=` constant, X is perpendicular distance from one plate to a point inside dielectric). The capacitance `C_(1)` of this capacitor, would be related to its vacuum capacitance `C_(0)` per the relation (d = plate separation):

Options

A. `C=(lambdad)/(ln(1+K_(0)lambdad))C_(0)`
B. `C=(lambda)/(d.ln(1+K_(0)lambdad))C_(0)`
C. `C=(lambdad)/(ln(1+lambdad//K_(0)))C_(0)`
D. `C=(lambda)/(d.ln(1+K_(0)//lambdad))C_(0)`

Explanation

To find the capacitance \( C_1 \) of a parallel plate capacitor filled with a dielectric whose dielectric constant varies with distance, we can follow these steps:

### Step 1: Understand the relationship of capacitance with dielectric
The capacitance \( C \) of a parallel plate capacitor filled with a dielectric is given by:
\[ C = \frac{K \epsilon_0 A}{d} \]
where \( K \) is the dielectric constant, \( \epsilon_0 \) is the permittivity of free space, \( A \) is the area of the plates, and \( d \) is the separation between the plates.

### Step 2: Define the varying dielectric constant
In this case, the dielectric constant \( K(x) \) varies with distance \( x \) as:
\[ K(x) = K_0 + \lambda x \]
where \( K_0 \) is a constant and \( \lambda \) is a constant that describes how the dielectric constant changes with distance.

### Step 3: Set up the integral for capacitance
To find the total capacitance \( C_1 \), we need to consider the varying dielectric constant across the distance \( d \). The capacitance can be expressed in terms of an integral:
\[ \frac{1}{C_1} = \int_0^d \frac{dx}{K(x) \epsilon_0 A} \]
Substituting \( K(x) \):
\[ \frac{1}{C_1} = \int_0^d \frac{dx}{(K_0 + \lambda x) \epsilon_0 A} \]

### Step 4: Solve the integral
The integral can be simplified:
\[ \frac{1}{C_1} = \frac{1}{\epsilon_0 A} \int_0^d \frac{dx}{K_0 + \lambda x} \]
This integral can be solved using the natural logarithm:
\[ \int \frac{dx}{K_0 + \lambda x} = \frac{1}{\lambda} \ln(K_0 + \lambda x) \]
Evaluating from 0 to \( d \):
\[ \frac{1}{C_1} = \frac{1}{\epsilon_0 A} \left[ \frac{1}{\lambda} \ln(K_0 + \lambda d) - \frac{1}{\lambda} \ln(K_0) \right] \]
\[ = \frac{1}{\epsilon_0 A \lambda} \left[ \ln(K_0 + \lambda d) - \ln(K_0) \right] \]
\[ = \frac{1}{\epsilon_0 A \lambda} \ln\left(\frac{K_0 + \lambda d}{K_0}\right) \]

### Step 5: Express \( C_1 \) in terms of \( C_0 \)
Now, we can express \( C_1 \):
\[ C_1 = \frac{\epsilon_0 A \lambda}{\ln\left(\frac{K_0 + \lambda d}{K_0}\right)} \]
The vacuum capacitance \( C_0 \) is given by:
\[ C_0 = \frac{\epsilon_0 A}{d} \]
Thus, we can relate \( C_1 \) to \( C_0 \):
\[ C_1 = C_0 \cdot \frac{\lambda d}{\ln\left(\frac{K_0 + \lambda d}{K_0}\right)} \]

### Final Result
The capacitance \( C_1 \) of the capacitor filled with a dielectric whose dielectric constant varies with distance is given by:
\[ C_1 = C_0 \cdot \frac{\lambda d}{\ln\left(\frac{K_0 + \lambda d}{K_0}\right)} \]