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

Explanation

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

### Step 1: Understand the Dielectric Constant
The dielectric constant \( K(x) \) is given by:
\[ K(x) = K_0 + \lambda x \]
where \( K_0 \) is a constant, \( \lambda \) is a constant, and \( x \) is the distance from one of the plates.

### Step 2: Write the Expression for Capacitance
The capacitance \( C \) of a parallel plate capacitor is given by:
\[ C = \frac{K \cdot A}{d} \]
where \( K \) is the dielectric constant, \( A \) is the area of the plates, and \( d \) is the separation between the plates.

### Step 3: Consider a Differential Element
To find the total capacitance when the dielectric constant varies, we consider a small thickness \( dx \) of the dielectric. The capacitance \( dC \) of this differential element can be expressed as:
\[ dC = \frac{K(x) \cdot A}{dx} \]
Substituting \( K(x) \):
\[ dC = \frac{(K_0 + \lambda x) \cdot A}{dx} \]

### Step 4: Relate Total Capacitance to Differential Capacitance
The total capacitance \( C_1 \) can be found by integrating the expression for \( dC \) from \( x = 0 \) to \( x = d \):
\[ \frac{1}{C_1} = \int_0^d \frac{dx}{dC} \]
This leads to:
\[ \frac{1}{C_1} = \int_0^d \frac{dx}{\frac{(K_0 + \lambda x) \cdot A}{dx}} = \frac{1}{A} \int_0^d \frac{dx}{K_0 + \lambda x} \]

### Step 5: Solve the Integral
The integral can be solved as follows:
\[ \int \frac{dx}{K_0 + \lambda x} = \frac{1}{\lambda} \ln |K_0 + \lambda x| \]
Evaluating from \( 0 \) to \( d \):
\[ \int_0^d \frac{dx}{K_0 + \lambda x} = \frac{1}{\lambda} \left( \ln(K_0 + \lambda d) - \ln(K_0) \right) = \frac{1}{\lambda} \ln \left( \frac{K_0 + \lambda d}{K_0} \right) \]

### Step 6: Substitute Back into the Capacitance Formula
Now substituting back into the expression for \( \frac{1}{C_1} \):
\[ \frac{1}{C_1} = \frac{1}{A} \cdot \frac{1}{\lambda} \ln \left( \frac{K_0 + \lambda d}{K_0} \right) \]
Taking the reciprocal gives:
\[ C_1 = \frac{A \lambda}{\ln \left( \frac{K_0 + \lambda d}{K_0} \right)} \]

### Step 7: Relate to Vacuum Capacitance
The vacuum capacitance \( C_0 \) is given by:
\[ C_0 = \frac{K_0 \cdot A}{d} \]
Thus, we can express \( C_1 \) in terms of \( C_0 \):
\[ C_1 = C_0 \cdot \frac{\lambda d}{\ln \left( \frac{K_0 + \lambda d}{K_0} \right)} \]

### Final Answer
The capacitance \( C_1 \) of the capacitor filled with the varying dielectric is related to its vacuum capacitance \( C_0 \) as:
\[ C_1 = C_0 \cdot \frac{\lambda d}{\ln \left( \frac{K_0 + \lambda d}{K_0} \right)} \]