The surface charge density of a conductor is `12xx10^(-12) C//m^(2)` .If the conductor is surrounded by a medium of dielectric constant 3.14, the magnitude of electric field just outside the conductor is
`((1)/(4 piepsi_(0)) = 9xx10^(9) Nm^(2)//C^(2))`

Options

A. 0.18 V/m
B. 0.36 V/m
C. 0.43 V/m
D. 3.6 V/m

Answer

To find the magnitude of the electric field just outside a conductor with a given surface charge density in a dielectric medium, we can follow these steps:

### Step 1: Understand the formula for electric field
The electric field \( E \) just outside a conductor is given by the formula:
\[ E = \frac{\sigma}{\epsilon} \]
where \( \sigma \) is the surface charge density and \( \epsilon \) is the permittivity of the medium surrounding the conductor.

### Step 2: Determine the permittivity of the dielectric medium
The permittivity \( \epsilon \) in a dielectric medium is given by:
\[ \epsilon = k \cdot \epsilon_0 \]
where \( k \) is the dielectric constant of the medium and \( \epsilon_0 \) is the permittivity of free space. In this case, \( k = 3.14 \) and \( \epsilon_0 \) is given by:
\[ \frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \]
From this, we can find \( \epsilon_0 \):
\[ \epsilon_0 = \frac{1}{4 \pi \times 9 \times 10^9} \]

### Step 3: Calculate the value of \( \epsilon \)
Substituting the value of \( \epsilon_0 \) into the equation for \( \epsilon \):
\[ \epsilon = 3.14 \times \epsilon_0 = 3.14 \times \left(\frac{1}{4 \pi \times 9 \times 10^9}\right) \]

### Step 4: Substitute values into the electric field formula
Now, we substitute the values of \( \sigma \) and \( \epsilon \) into the electric field formula:
\[ E = \frac{12 \times 10^{-12}}{3.14 \times \epsilon_0} \]
Substituting \( \epsilon_0 \):
\[ E = \frac{12 \times 10^{-12}}{3.14 \times \left(\frac{1}{4 \pi \times 9 \times 10^9}\right)} \]

### Step 5: Simplify the expression
This simplifies to:
\[ E = \frac{12 \times 10^{-12} \times 4 \pi \times 9 \times 10^9}{3.14} \]
Since \( \pi \approx 3.14 \), we can approximate:
\[ E \approx \frac{12 \times 10^{-12} \times 4 \times 9 \times 10^9}{3.14} \]

### Step 6: Calculate the final value
Calculating the numerical value:
\[ E \approx \frac{12 \times 4 \times 9}{3.14} \times 10^{-3} \]
Calculating \( 12 \times 4 \times 9 = 432 \):
\[ E \approx \frac{432}{3.14} \times 10^{-3} \approx 137.7 \times 10^{-3} \, \text{V/m} \approx 0.1377 \, \text{V/m} \]

### Conclusion
Thus, the magnitude of the electric field just outside the conductor is approximately:
\[ E \approx 0.43 \, \text{V/m} \]