A parallel plate condenser has initially air medium between the plates. If a slab of dielectric constant `5` having thickness half the distance of separation between the plates is introduced, the percentage increase in its capacity is.

Options

A. `33.3%`
B. `66.7 %`
C. 0.5
D. 0.75

Explanation

To solve the problem of finding the percentage increase in the capacitance of a parallel plate capacitor when a dielectric slab is introduced, we can follow these steps:

### Step 1: Determine the initial capacitance (C₀)
The capacitance of a parallel plate capacitor with air (or vacuum) as the dielectric medium is given by the formula:

\[ C_0 = \frac{A \epsilon_0}{D} \]

Where:
- \(C_0\) = initial capacitance
- \(A\) = area of the plates
- \(\epsilon_0\) = permittivity of free space
- \(D\) = distance between the plates

### Step 2: Introduce the dielectric slab
When a dielectric slab of thickness \(t = \frac{D}{2}\) and dielectric constant \(K = 5\) is introduced between the plates, the capacitor can be considered as two capacitors in series:
1. The first capacitor has the dielectric slab with thickness \(t = \frac{D}{2}\).
2. The second capacitor has air with thickness \(D - t = \frac{D}{2}\).

### Step 3: Calculate the capacitance with the dielectric slab (C')
The capacitance of the first part (with the dielectric) is:

\[ C_1 = \frac{A K \epsilon_0}{t} = \frac{A \cdot 5 \epsilon_0}{\frac{D}{2}} = \frac{10A \epsilon_0}{D} \]

The capacitance of the second part (with air) is:

\[ C_2 = \frac{A \epsilon_0}{D - t} = \frac{A \epsilon_0}{\frac{D}{2}} = \frac{2A \epsilon_0}{D} \]

### Step 4: Find the total capacitance (C')
Since the two capacitances are in series, the total capacitance \(C'\) can be calculated using the formula for capacitors in series:

\[ \frac{1}{C'} = \frac{1}{C_1} + \frac{1}{C_2} \]

Substituting the values:

\[ \frac{1}{C'} = \frac{1}{\frac{10A \epsilon_0}{D}} + \frac{1}{\frac{2A \epsilon_0}{D}} \]

This simplifies to:

\[ \frac{1}{C'} = \frac{D}{10A \epsilon_0} + \frac{D}{2A \epsilon_0} \]

Finding a common denominator (10Aε₀):

\[ \frac{1}{C'} = \frac{D}{10A \epsilon_0} + \frac{5D}{10A \epsilon_0} = \frac{6D}{10A \epsilon_0} = \frac{3D}{5A \epsilon_0} \]

Thus,

\[ C' = \frac{5A \epsilon_0}{3D} \]

### Step 5: Calculate the percentage increase in capacitance
Now, we can find the percentage increase in capacitance:

\[ \text{Percentage Increase} = \left( \frac{C' - C_0}{C_0} \right) \times 100 \]

Substituting \(C'\) and \(C_0\):

\[ \text{Percentage Increase} = \left( \frac{\frac{5A \epsilon_0}{3D} - \frac{A \epsilon_0}{D}}{\frac{A \epsilon_0}{D}} \right) \times 100 \]

This simplifies to:

\[ = \left( \frac{\frac{5}{3} - 1}{1} \right) \times 100 = \left( \frac{2}{3} \right) \times 100 = 66.67\% \]

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

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