The capacity of a parallel plate condenser is `10 muF` without dielectric. Dielectric of constant `2` is used to fill half the distance between the plates, the new capacitance in `muF` is

Options

A. `10`
B. `20`
C. `15`
D. `13.33`

Explanation

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

### Step-by-Step Solution:

1. **Understand the Initial Conditions**:
- The initial capacitance \( C_0 \) of the parallel plate capacitor without any dielectric is given as \( 10 \, \mu F \).
- The formula for the capacitance of a parallel plate capacitor is:
\[ C = \frac{A \epsilon_0}{d} \]
where \( A \) is the area of the plates, \( \epsilon_0 \) is the permittivity of free space, and \( d \) is the separation between the plates.

2. **Identify the Configuration with Dielectric**:
- The dielectric constant \( k \) is given as \( 2 \).
- The distance between the plates is \( d \), and half of this distance \( \frac{d}{2} \) will be filled with the dielectric.

3. **Calculate the Capacitance with Dielectric**:
- The capacitor can be thought of as two capacitors in series:
- The first capacitor (without dielectric) has a capacitance \( C_1 \) for the distance \( \frac{d}{2} \):
\[ C_1 = \frac{A \epsilon_0}{\frac{d}{2}} = \frac{2A \epsilon_0}{d} \]
- The second capacitor (with dielectric) has a capacitance \( C_2 \) for the distance \( \frac{d}{2} \):
\[ C_2 = \frac{A k \epsilon_0}{\frac{d}{2}} = \frac{2A (2) \epsilon_0}{d} = \frac{4A \epsilon_0}{d} \]

4. **Combine the Capacitors in Series**:
- The total capacitance \( C_n \) of capacitors in series is given by:
\[ \frac{1}{C_n} = \frac{1}{C_1} + \frac{1}{C_2} \]
- Substituting the values:
\[ \frac{1}{C_n} = \frac{d}{2A \epsilon_0} + \frac{d}{4A \epsilon_0} \]
- Finding a common denominator:
\[ \frac{1}{C_n} = \frac{2d + d}{4A \epsilon_0} = \frac{3d}{4A \epsilon_0} \]
- Therefore, the total capacitance is:
\[ C_n = \frac{4A \epsilon_0}{3d} \]

5. **Relate New Capacitance to Initial Capacitance**:
- Since \( C_0 = \frac{A \epsilon_0}{d} = 10 \, \mu F \):
\[ C_n = \frac{4}{3} C_0 = \frac{4}{3} \times 10 \, \mu F = \frac{40}{3} \, \mu F \approx 13.33 \, \mu F \]

### Final Answer:
The new capacitance \( C_n \) is approximately \( 13.33 \, \mu F \).