The two metallic plates of radius `r` are placed at a distance `d` apart and its capacity is `C`. If a plate of radius `r//2` and thickness `d`of dielectric constant `6` is placed between the plates of the condenser, then its capacity will be

Options

A. 7C/2
B. 3C/7
C. 7C/3
D. 9C/4

Answer

To solve the problem step by step, we need to analyze the situation where a dielectric slab is inserted between the plates of a capacitor. We will find the new capacitance after inserting the dielectric.

### Step 1: Understand the initial capacitance
The initial capacitance \( C \) of a parallel plate capacitor is given by the formula:
\[ C = \frac{\varepsilon_0 A}{d} \]
where:
- \( \varepsilon_0 \) is the permittivity of free space,
- \( A \) is the area of the plates,
- \( d \) is the distance between the plates.

For plates of radius \( r \), the area \( A \) is:
\[ A = \pi r^2 \]
Thus, the initial capacitance can be expressed as:
\[ C = \frac{\varepsilon_0 \pi r^2}{d} \]

### Step 2: Calculate the capacitance of the smaller capacitor with dielectric
When a dielectric slab of radius \( \frac{r}{2} \) and thickness \( d \) with dielectric constant \( K = 6 \) is inserted, we need to calculate the capacitance of this smaller capacitor \( C_1 \).

The area of the smaller plate is:
\[ A_1 = \pi \left(\frac{r}{2}\right)^2 = \frac{\pi r^2}{4} \]
The capacitance \( C_1 \) of this smaller capacitor with the dielectric is:
\[ C_1 = \frac{K \varepsilon_0 A_1}{d} = \frac{6 \varepsilon_0 \left(\frac{\pi r^2}{4}\right)}{d} = \frac{3 \pi r^2 \varepsilon_0}{2d} \]
Since \( C = \frac{\varepsilon_0 \pi r^2}{d} \), we can express \( C_1 \) in terms of \( C \):
\[ C_1 = \frac{3}{2} C \]

### Step 3: Calculate the capacitance of the remaining area
The remaining area \( A_2 \) of the capacitor (the area not covered by the dielectric) is:
\[ A_2 = \pi r^2 - A_1 = \pi r^2 - \frac{\pi r^2}{4} = \frac{3\pi r^2}{4} \]
The capacitance \( C_2 \) of this remaining area without the dielectric is:
\[ C_2 = \frac{\varepsilon_0 A_2}{d} = \frac{\varepsilon_0 \left(\frac{3\pi r^2}{4}\right)}{d} = \frac{3\pi r^2 \varepsilon_0}{4d} \]
Again, expressing \( C_2 \) in terms of \( C \):
\[ C_2 = \frac{3}{4} C \]

### Step 4: Combine the capacitances
Since \( C_1 \) and \( C_2 \) are in parallel, the total capacitance \( C' \) is given by:
\[ C' = C_1 + C_2 = \frac{3}{2} C + \frac{3}{4} C \]
To combine these fractions, we find a common denominator:
\[ C' = \frac{6}{4} C + \frac{3}{4} C = \frac{9}{4} C \]

### Final Answer
Thus, the new capacitance after inserting the dielectric slab is:
\[ C' = \frac{9}{4} C \]