A slab of material of dielectric constant K has the same area as the plates of a parallel capacitor, but has a thickness `((3)/(4) d)`,
where d is the separation of the plates. How is the capacitance changed when the slab is inserted between the plates

Options

A. `C=(epsi_(0)A)/(d)((K+3)/(4K))`
B. `C=(epsi_(0)A)/(d)((2K)/(K+3))`
C. `C=(epsi_(0)A)/(d)((K)/(K+3))`
D. `C=(epsi_(0)A)/(d)((4K)/(K+3))`

Explanation

To solve the problem of how the capacitance changes when a dielectric slab is inserted between the plates of a parallel plate capacitor, we can follow these steps:

### Step 1: Understand the Capacitance Formula
The capacitance \( C \) of a parallel plate capacitor without any dielectric is given by the formula:
\[ C_0 = \frac{A \epsilon_0}{d} \]
where:
- \( A \) is the area of the plates,
- \( \epsilon_0 \) is the permittivity of free space,
- \( d \) is the separation between the plates.

### Step 2: Insert the Dielectric Slab
When a dielectric slab of thickness \( t = \frac{3}{4}d \) and dielectric constant \( K \) is inserted between the plates, the effective capacitance can be calculated using the modified formula:
\[ C = \frac{A \epsilon_0}{d - t} + \frac{A \epsilon_0}{t/K} \]
This formula accounts for the two regions: one with the dielectric and one without.

### Step 3: Substitute the Values
Substituting \( t = \frac{3}{4}d \) into the formula:
\[ C = \frac{A \epsilon_0}{d - \frac{3}{4}d} + \frac{A \epsilon_0}{\frac{3}{4}d/K} \]
This simplifies to:
\[ C = \frac{A \epsilon_0}{\frac{1}{4}d} + \frac{A \epsilon_0 K}{\frac{3}{4}d} \]

### Step 4: Simplify Each Term
Calculating each term:
1. The first term becomes:
\[ C_1 = \frac{A \epsilon_0}{\frac{1}{4}d} = \frac{4A \epsilon_0}{d} \]
2. The second term becomes:
\[ C_2 = \frac{A \epsilon_0 K}{\frac{3}{4}d} = \frac{4A \epsilon_0 K}{3d} \]

### Step 5: Combine the Terms
Now, we can combine \( C_1 \) and \( C_2 \):
\[ C = \frac{4A \epsilon_0}{d} + \frac{4A \epsilon_0 K}{3d} \]
Factoring out \( \frac{4A \epsilon_0}{d} \):
\[ C = \frac{4A \epsilon_0}{d} \left(1 + \frac{K}{3}\right) \]

### Step 6: Final Expression
Thus, the final expression for the capacitance when the slab is inserted is:
\[ C = \frac{4A \epsilon_0}{d} \left(\frac{3 + K}{3}\right) \]

### Conclusion
The capacitance increases when the dielectric slab is inserted, and the final capacitance can be expressed as:
\[ C = \frac{4A \epsilon_0 (3 + K)}{3d} \]