Half of the space between parallel plate capacitor is filled with a medium of dielectric constant K parallel to the plates . if initially the capacity is C, then the new capacity will be :

Options

A. 2KC(1+K)
B. C(K+1)/2
C. CK/(1+K)
D. KC

Answer

To solve the problem of finding the new capacitance of a parallel plate capacitor when half of the space is filled with a dielectric medium, we can follow these steps:

### Step 1: Understand the Initial Capacitance
The initial capacitance \( C \) of a parallel plate capacitor is given by the formula:
\[ C = \frac{A \epsilon_0}{d} \]
where:
- \( A \) is the area of the plates,
- \( \epsilon_0 \) is the permittivity of free space,
- \( d \) is the distance between the plates.

### Step 2: Analyze the New Configuration
In the new configuration, half of the space between the plates is filled with a dielectric material of dielectric constant \( K \), while the other half remains air (or vacuum). This means that the area of the capacitor is now effectively divided into two parts:
- Area with dielectric: \( A_1 = \frac{A}{2} \)
- Area without dielectric: \( A_2 = \frac{A}{2} \)

### Step 3: Calculate the Capacitance of Each Section
1. **Capacitance of the section with dielectric \( C_1 \)**:
\[ C_1 = \frac{A_1 \cdot K \epsilon_0}{d} = \frac{\frac{A}{2} \cdot K \epsilon_0}{d} = \frac{K A \epsilon_0}{2d} \]

2. **Capacitance of the section without dielectric \( C_2 \)**:
\[ C_2 = \frac{A_2 \cdot \epsilon_0}{d} = \frac{\frac{A}{2} \cdot \epsilon_0}{d} = \frac{A \epsilon_0}{2d} \]

### Step 4: Combine the Capacitances
Since \( C_1 \) and \( C_2 \) are in parallel, the total capacitance \( C_{eq} \) can be calculated as:
\[ C_{eq} = C_1 + C_2 \]
Substituting the values we calculated:
\[ C_{eq} = \frac{K A \epsilon_0}{2d} + \frac{A \epsilon_0}{2d} \]
Factoring out the common terms:
\[ C_{eq} = \frac{A \epsilon_0}{2d} (K + 1) \]

### Step 5: Relate to the Initial Capacitance
Recall that the initial capacitance \( C \) is:
\[ C = \frac{A \epsilon_0}{d} \]
Thus, we can express \( C_{eq} \) in terms of \( C \):
\[ C_{eq} = \frac{1}{2} \frac{A \epsilon_0}{d} (K + 1) = \frac{C}{2} (K + 1) \]

### Final Result
The new capacitance \( C_{eq} \) is:
\[ C_{eq} = \frac{C}{2} (K + 1) \]