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 capacitance when a parallel plate air capacitor is half-filled with a dielectric of dielectric constant 5, we can follow these steps:

### Step 1: Understand the Initial Capacitance
The initial capacitance \( C \) of a parallel plate capacitor filled with air (where the dielectric constant \( k = 1 \)) 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 Configuration After Filling with Dielectric
When the capacitor is half-filled with a dielectric of dielectric constant \( k = 5 \), we can treat the capacitor as two capacitors in series:
1. **Capacitor 1 (C1)**: The portion filled with the dielectric (height = \( \frac{d}{2} \)).
2. **Capacitor 2 (C2)**: The portion filled with air (height = \( \frac{d}{2} \)).

### Step 3: Calculate the Capacitance of Each Section
For **Capacitor 1 (C1)**:
\[ C_1 = \frac{A \epsilon_0 k_1}{d_1} = \frac{A \epsilon_0 \cdot 5}{\frac{d}{2}} = \frac{10A \epsilon_0}{d} \]

For **Capacitor 2 (C2)**:
\[ C_2 = \frac{A \epsilon_0 k_2}{d_2} = \frac{A \epsilon_0 \cdot 1}{\frac{d}{2}} = \frac{2A \epsilon_0}{d} \]

### Step 4: Find the Equivalent Capacitance
Since \( C_1 \) and \( C_2 \) are in series, the equivalent capacitance \( C_e \) is given by:
\[ \frac{1}{C_e} = \frac{1}{C_1} + \frac{1}{C_2} \]
Substituting the values:
\[ \frac{1}{C_e} = \frac{1}{\frac{10A \epsilon_0}{d}} + \frac{1}{\frac{2A \epsilon_0}{d}} \]
\[ = \frac{d}{10A \epsilon_0} + \frac{d}{2A \epsilon_0} \]
Finding a common denominator (which is 10):
\[ = \frac{d}{10A \epsilon_0} + \frac{5d}{10A \epsilon_0} = \frac{6d}{10A \epsilon_0} \]
Thus,
\[ C_e = \frac{10A \epsilon_0}{6d} = \frac{5A \epsilon_0}{3d} \]

### Step 5: Calculate the Percentage Increase in Capacitance
Now, we can find the percentage increase in capacitance:
\[ \text{Percentage Increase} = \left( \frac{C_e - C}{C} \right) \times 100 \]
Substituting \( C = \frac{A \epsilon_0}{d} \) and \( C_e = \frac{5A \epsilon_0}{3d} \):
\[ \text{Percentage Increase} = \left( \frac{\frac{5A \epsilon_0}{3d} - \frac{A \epsilon_0}{d}}{\frac{A \epsilon_0}{d}} \right) \times 100 \]
\[ = \left( \frac{\frac{5}{3} - 1}{1} \right) \times 100 = \left( \frac{5 - 3}{3} \right) \times 100 = \left( \frac{2}{3} \right) \times 100 \approx 66.67\% \]

### Conclusion
The percentage increase in capacitance when the capacitor is half-filled with a dielectric of dielectric constant 5 is approximately **66.67%**.