An air capacitor, a capacitor with a dielectric and a capacitor with a conducting slab (thickness one half the separation introducing between the plates of parallel plate air capacitor in both case) has capacity `C_1, C_2 and C_3` respectively then :

Options

A. `C_(1) gt C_(2) gt C_(3)`
B. `C_(2) gt C_(3) gt C_(1)`
C. `C_(3) gt C_(2) gt C_(1)`
D. `C_(3) gt C_(1) gt C_(2)`

Answer

To solve the problem of comparing the capacitance of three types of capacitors: an air capacitor (C1), a capacitor with a dielectric slab (C2), and a capacitor with a conducting slab (C3), we will derive the expressions for each capacitance and then compare them.

### Step 1: Capacitance of the Air Capacitor (C1)

The formula for the capacitance of a parallel plate capacitor filled with air (or vacuum) is given by:

\[ C_1 = \frac{\varepsilon_0 A}{d} \]

Where:
- \( \varepsilon_0 \) is the permittivity of free space,
- \( A \) is the area of the plates,
- \( d \) is the separation between the plates.

### Step 2: Capacitance of the Capacitor with a Dielectric Slab (C2)

When a dielectric slab of thickness \( \frac{d}{2} \) is inserted between the plates, the capacitor can be treated as two capacitors in series: one with the dielectric and one with air.

1. The capacitance of the portion with the dielectric slab is:

\[ C_{dielectric} = \frac{k \varepsilon_0 A}{\frac{d}{2}} = \frac{2k \varepsilon_0 A}{d} \]

2. The capacitance of the portion with air (the remaining \( \frac{d}{2} \)) is:

\[ C_{air} = \frac{\varepsilon_0 A}{\frac{d}{2}} = \frac{2 \varepsilon_0 A}{d} \]

3. The total capacitance \( C_2 \) of the system is given by the formula for capacitors in series:

\[ \frac{1}{C_2} = \frac{1}{C_{dielectric}} + \frac{1}{C_{air}} \]

Substituting the values we calculated:

\[ \frac{1}{C_2} = \frac{d}{2k \varepsilon_0 A} + \frac{d}{2 \varepsilon_0 A} \]

Finding a common denominator:

\[ \frac{1}{C_2} = \frac{d}{2 \varepsilon_0 A} \left(\frac{1}{k} + 1\right) \]

Thus,

\[ C_2 = \frac{2 \varepsilon_0 A}{d \left(\frac{1}{k} + 1\right)} = \frac{2k \varepsilon_0 A}{d(k + 1)} \]

### Step 3: Capacitance of the Capacitor with a Conducting Slab (C3)

When a conducting slab of thickness \( \frac{d}{2} \) is placed between the plates, the electric field inside the conductor is zero. The effective distance between the plates is reduced to \( d - \frac{d}{2} = \frac{d}{2} \).

Thus, the capacitance \( C_3 \) is given by:

\[ C_3 = \frac{\varepsilon_0 A}{\frac{d}{2}} = \frac{2 \varepsilon_0 A}{d} \]

### Step 4: Comparing the Capacitances

Now we have:
- \( C_1 = \frac{\varepsilon_0 A}{d} \)
- \( C_2 = \frac{2k \varepsilon_0 A}{d(k + 1)} \)
- \( C_3 = \frac{2 \varepsilon_0 A}{d} \)

Since \( k > 1 \), we can deduce that:

1. \( C_2 > C_3 \) because \( \frac{2k}{k + 1} > 2 \).
2. \( C_3 > C_1 \) because \( \frac{2 \varepsilon_0 A}{d} > \frac{\varepsilon_0 A}{d} \).

Thus, the order of capacitance is:

\[ C_2 > C_3 > C_1 \]

### Final Answer:
The capacitance values can be compared as follows:
\[ C_2 > C_3 > C_1 \]