A parallel plate air capacitor has a capacitance C. When it is half filled with a dielectric of dielectric constant 5, the percentage increase in the capacitance will be

Options

A. `400%`
B. `66.6%`
C. `33.3%`
D. `200%`

Explanation

To solve the problem of finding the percentage increase in the capacitance of a parallel plate air capacitor when it is half-filled with a dielectric, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Initial Capacitance:**
The initial capacitance \( C \) of a parallel plate capacitor filled with air (or vacuum) is given by:
\[ C = \frac{\varepsilon_0 A}{d} \]
where \( \varepsilon_0 \) is the permittivity of free space, \( A \) is the area of the plates, and \( d \) is the separation between the plates.

2. **Identify the Configuration After Filling with Dielectric:**
When the capacitor is half-filled with a dielectric of dielectric constant \( k = 5 \), we can consider the capacitor as two capacitors in series:
- The first capacitor \( C_1 \) (air part) has a distance of \( \frac{d}{2} \).
- The second capacitor \( C_2 \) (dielectric part) also has a distance of \( \frac{d}{2} \) but with a dielectric constant \( k = 5 \).

3. **Calculate the Capacitance of Each Section:**
- For the air section:
\[ C_1 = \frac{\varepsilon_0 A}{\frac{d}{2}} = \frac{2 \varepsilon_0 A}{d} \]
- For the dielectric section:
\[ C_2 = \frac{k \varepsilon_0 A}{\frac{d}{2}} = \frac{5 \varepsilon_0 A}{\frac{d}{2}} = \frac{10 \varepsilon_0 A}{d} \]

4. **Use the Series Capacitance Formula:**
The total capacitance \( C' \) for capacitors in series is given by:
\[ \frac{1}{C'} = \frac{1}{C_1} + \frac{1}{C_2} \]
Substituting the values of \( C_1 \) and \( C_2 \):
\[ \frac{1}{C'} = \frac{1}{\frac{2 \varepsilon_0 A}{d}} + \frac{1}{\frac{10 \varepsilon_0 A}{d}} \]
This simplifies to:
\[ \frac{1}{C'} = \frac{d}{2 \varepsilon_0 A} + \frac{d}{10 \varepsilon_0 A} = \frac{5d + d}{10 \varepsilon_0 A} = \frac{6d}{10 \varepsilon_0 A} \]
Therefore:
\[ C' = \frac{10 \varepsilon_0 A}{6d} = \frac{5 \varepsilon_0 A}{3d} \]

5. **Calculate the Percentage Increase in Capacitance:**
The percentage increase in capacitance can be calculated using the formula:
\[ \text{Percentage Increase} = \left( \frac{C' - C}{C} \right) \times 100 \]
Substituting \( C' \) and \( C \):
\[ \text{Percentage Increase} = \left( \frac{\frac{5 \varepsilon_0 A}{3d} - \frac{\varepsilon_0 A}{d}}{\frac{\varepsilon_0 A}{d}} \right) \times 100 \]
Simplifying this gives:
\[ = \left( \frac{\frac{5}{3} - 1}{1} \right) \times 100 = \left( \frac{2}{3} \right) \times 100 = \frac{200}{3} \approx 66.67\% \]

### Final Answer:
The percentage increase in capacitance is approximately **66.67%**.