A slab of a material of dielectric constaant 2 is placed between two identical parallel plates. One of the plates carries total charge density +2mC/`m^(2)` and the other plate carries total charge density -4mC/`m^(2)` the magnitude of electric field inside the dielectric in (in `10^(11)N//C`) Take `epsi_(0)=(80)/(9)xx10^(-12)Nm^(2)//C^(2)`)

Options

A. 2.25
B. 3.38
C. 4.5
D. 6.75

Explanation

To solve the problem, we need to find the magnitude of the electric field inside a dielectric slab placed between two parallel plates with different charge densities. Let's break down the solution step by step.

### Step 1: Identify the Charge Densities
We have two plates:
- Plate 1 has a charge density \( \sigma_1 = +2 \, \text{mC/m}^2 = 2 \times 10^{-3} \, \text{C/m}^2 \)
- Plate 2 has a charge density \( \sigma_2 = -4 \, \text{mC/m}^2 = -4 \times 10^{-3} \, \text{C/m}^2 \)

### Step 2: Calculate the Net Charge Density
The net charge density \( \sigma \) between the plates can be calculated as:
\[ \sigma = \sigma_1 + |\sigma_2| = 2 \times 10^{-3} + 4 \times 10^{-3} = 6 \times 10^{-3} \, \text{C/m}^2 \]

### Step 3: Use the Formula for Electric Field in a Dielectric
The electric field \( E \) inside a dielectric material can be calculated using the formula:
\[ E = \frac{\sigma}{\varepsilon} \]
where \( \varepsilon \) is the permittivity of the dielectric material, given by:
\[ \varepsilon = K \varepsilon_0 \]
Here, \( K \) is the dielectric constant (which is 2 in this case) and \( \varepsilon_0 \) is the permittivity of free space, given as:
\[ \varepsilon_0 = \frac{80}{9} \times 10^{-12} \, \text{N m}^2/\text{C}^2 \]

### Step 4: Calculate the Permittivity
Now, we can calculate \( \varepsilon \):
\[ \varepsilon = 2 \times \left(\frac{80}{9} \times 10^{-12}\right) = \frac{160}{9} \times 10^{-12} \, \text{N m}^2/\text{C}^2 \]

### Step 5: Substitute Values into the Electric Field Formula
Now, substitute \( \sigma \) and \( \varepsilon \) into the electric field formula:
\[ E = \frac{6 \times 10^{-3}}{\frac{160}{9} \times 10^{-12}} = 6 \times 10^{-3} \times \frac{9}{160} \times 10^{12} \]

### Step 6: Simplify the Expression
Calculating the above expression:
\[ E = \frac{54 \times 10^{9}}{160} = 0.3375 \times 10^{11} \, \text{N/C} \]

### Step 7: Final Result
To express this in the required format:
\[ E = 3.375 \times 10^{10} \, \text{N/C} \]

### Step 8: Convert to the Required Units
Since the problem asks for the answer in \( 10^{11} \, \text{N/C} \):
\[ E = 0.3375 \times 10^{11} \, \text{N/C} \]

### Final Answer
Thus, the magnitude of the electric field inside the dielectric is:
\[ E \approx 0.3375 \times 10^{11} \, \text{N/C} \]

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