The separation between the plates of a capacitor of capacitance `2muF` is 4mm. Initially there is air between the paltes. If two dielectric slabs of area of cross-section same as the capacitor, thickness 2mm each, and dielectric constant 3 and 5 respectively are introduced, the capacitance becomes (in `muF`)

Options

A. 4
B. 7.5
C. 8
D. 15

Explanation

To solve the problem, we need to find the new capacitance of the capacitor after introducing two dielectric slabs. Here’s a step-by-step solution:

### Step 1: Understand the initial conditions
The initial capacitance \( C_0 \) of the capacitor is given as \( 2 \, \mu F \) with a separation \( d = 4 \, mm \) between the plates. The area \( A \) of the plates is not provided, but we will keep it as a variable since it will cancel out later.

### Step 2: Determine the configuration with dielectrics
Two dielectric slabs are introduced, each with a thickness of \( 2 \, mm \), filling the entire space between the plates. The first dielectric has a dielectric constant \( K_1 = 3 \) and the second has \( K_2 = 5 \).

### Step 3: Calculate the potential difference across each dielectric
The potential difference across the capacitor can be expressed as the sum of the potential differences across each dielectric slab.

The electric field \( E \) in a dielectric is given by:
\[ E = \frac{Q}{A \epsilon_0 K} \]
where \( K \) is the dielectric constant.

For the first dielectric slab (thickness \( 2 \, mm \)):
\[ E_1 = \frac{Q}{A \epsilon_0 K_1} = \frac{Q}{A \epsilon_0 \cdot 3} \]

For the second dielectric slab (thickness \( 2 \, mm \)):
\[ E_2 = \frac{Q}{A \epsilon_0 K_2} = \frac{Q}{A \epsilon_0 \cdot 5} \]

### Step 4: Calculate the potential difference across the entire capacitor
The total potential difference \( V \) across the capacitor is given by:
\[ V = E_1 \cdot d_1 + E_2 \cdot d_2 \]
where \( d_1 = d_2 = 2 \, mm \).

Substituting the expressions for \( E_1 \) and \( E_2 \):
\[ V = \left( \frac{Q}{A \epsilon_0 \cdot 3} \cdot 2 \times 10^{-3} \right) + \left( \frac{Q}{A \epsilon_0 \cdot 5} \cdot 2 \times 10^{-3} \right) \]

### Step 5: Simplify the expression for \( V \)
Combining the terms:
\[ V = \frac{Q \cdot 2 \times 10^{-3}}{A \epsilon_0} \left( \frac{1}{3} + \frac{1}{5} \right) \]

Calculating \( \frac{1}{3} + \frac{1}{5} \):
\[ \frac{1}{3} + \frac{1}{5} = \frac{5 + 3}{15} = \frac{8}{15} \]

So,
\[ V = \frac{Q \cdot 2 \times 10^{-3}}{A \epsilon_0} \cdot \frac{8}{15} \]

### Step 6: Find the new capacitance
The capacitance \( C \) is defined as:
\[ C = \frac{Q}{V} \]

Substituting the expression for \( V \):
\[ C = \frac{Q}{\frac{Q \cdot 2 \times 10^{-3}}{A \epsilon_0} \cdot \frac{8}{15}} = \frac{A \epsilon_0 \cdot 15}{2 \times 10^{-3} \cdot 8} \]

### Step 7: Relate to the original capacitance
We know the original capacitance \( C_0 \):
\[ C_0 = \frac{A \epsilon_0}{d} = \frac{A \epsilon_0}{4 \times 10^{-3}} \]

Now, we can express \( C \) in terms of \( C_0 \):
\[ C = C_0 \cdot \frac{15}{16} \]

Given \( C_0 = 2 \, \mu F \):
\[ C = 2 \, \mu F \cdot \frac{15}{16} = \frac{30}{16} \, \mu F = 1.875 \, \mu F \]

### Final Answer
Thus, the new capacitance becomes:
\[ \boxed{1.875 \, \mu F} \]